New test - October 20, 2014
Let
and
[642 marks]
.
1a. Find AB .
[3 marks]
Markscheme
evidence of multiplying
(M1)
e.g. one correct element,
A2
N3
Note: Award A1 for three correct elements.
[3 marks]
Examiners report
The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to
have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of
matrix A.
1b. Given that
, find X.
[3 marks]
Markscheme
(A1)
finding
adding 2
to both sides (may be seen first)
(M1)
e.g.
A1
N2
[3 marks]
Examiners report
The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to
have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of
matrix A.
The following table shows the probability distribution of a discrete random variable X .
2a. Find the value of k .
[3 marks]
Markscheme
evidence of summing to 1
(M1)
e.g.
(A1)
correct working
e.g.
A1
N2
[3 marks]
Examiners report
Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were
unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but
then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was
.
2b. Find
.
[3 marks]
Markscheme
correct substitution into formula
(A1)
e.g.
correct working
(A1)
e.g.
= 3.3
A1
N2
[3 marks]
Examiners report
Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were
unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but
then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was
.
The weekly wages (in dollars) of 80 employees are displayed in the cumulative frequency curve below.
3a. (i)
Write down the median weekly wage.
[4 marks]
(ii) Find the interquartile range of the weekly wages.
Markscheme
(i) median weekly wage
(ii) lower quartile
(dollars)
, upper quartile
A1
N1
(A1)(A1)
(dollars) (accept any notation suggesting interval
to
) A1
N3
Note: Exception to the FT rule. Award A1(FT) for an incorrect IQR only if both quartiles are explicitly noted.
[4 marks]
Examiners report
Many candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), there
were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.
3b. The box-and-whisker plot below displays the weekly wages of the employees.
[3 marks]
Write down the value of
(i)
;
(ii)
;
(iii)
.
Markscheme
(i)
(dollars)
A1
N1
(ii)
(dollars) A1
N1
(iii)
(dollars) A1
N1
[3 marks]
Examiners report
Many candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), there
were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.
3c. Employees are paid
per hour.
[3 marks]
Find the median number of hours worked per week.
Markscheme
valid approach
(M1)
e.g.
correct substitution
(A1)
e.g.
median hours per week
A1
N2
[3 marks]
Examiners report
For part (c), the majority of candidates seemed to understand what was required, though there were a few who used an extreme value
such as
, rather than the median value.
3d. Employees are paid
per hour.
Find the number of employees who work more than
[5 marks]
hours per week.
Markscheme
attempt to find wages for 25 hours per week
(M1)
e.g.
(A1)
correct substitution
e.g.
(A1)
finding wages
(A1)
65 people (earn 500 )
15 people (work more than 25 hours)
A1
N3
[5 marks]
Examiners report
In part (d), some candidates simply answered , which is the number of workers earning
or less, rather than finding the number
of workers who earned more than
. It was interesting to note that quite a few candidates gave their final answer as , rather than
.
4.
In a large city, the time taken to travel to work is normally distributed with mean and standard deviation
of the population take less than 5 minutes to get to work, and
take less than 25 minutes.
Find the value of
and of
. It is found that
[8 marks]
.
Markscheme
(A1)(A1)
correct z-values
,
attempt to set up their equations, must involve z-values, not %
(M1)
e.g. one correct equation
A1A1
two correct equations
e.g.
,
attempt to solve their equations
(M1)
e.g. substitution, matrices, one correct value
,
A1A1
,
N4
[8 marks]
Examiners report
A standard question for which well-prepared candidates frequently earned all eight marks. Common errors included the use of
percentages rather than z-values and the inability to find the negative z-value. Others had correct equations but were not able to use
their GDC to solve them and ultimately made errors in their algebra.
5a. At a large school, students are required to learn at least one language, Spanish or French. It is known that
learn Spanish, and
learn French.
Find the percentage of students who learn both Spanish and French.
of the students
[2 marks]
Markscheme
(M1)
valid approach
e.g. Venn diagram with intersection, union formula,
(accept
A1
)
N2
[2 marks]
Examiners report
Parts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.
5b. At a large school, students are required to learn at least one language, Spanish or French. It is known that
learn Spanish, and
of the students
[2 marks]
learn French.
Find the percentage of students who learn Spanish, but not French.
Markscheme
valid approach involving subtraction
(M1)
e.g. Venn diagram,
60 (accept
A1
)
N2
[2 marks]
Examiners report
Parts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.
5c. At a large school, students are required to learn at least one language, Spanish or French. It is known that
learn Spanish, and
At this school,
of the students
[5 marks]
learn French.
of the students are girls, and
of the girls learn Spanish.
A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish.
(i)
Find
.
(ii) Show that G and S are not independent.
Markscheme
(M1)
(i) valid approach
e.g. tree diagram, multiplying probabilities,
correct calculation
(A1)
e.g.
(exact)
A1
N3
(ii) valid reasoning, with words, symbols or numbers (seen anywhere)
e.g.
one correct value
e.g.
,
R1
, not equal,
A1
,
G and S are not independent
,
AG
N0
[5 marks]
Examiners report
In part (c) (i), candidates failed to find the intersection of the events as, in general, they multiplied probabilities, assuming the events
were independent or they incorrectly attempted to use the union formula. Independence in (c) (ii) caused difficulty with some
candidates attempting to use the conditions for mutually exclusive events while others assumed the events were independent in part (i)
and then found
by multiplying
.
5d. At a large school, students are required to learn at least one language, Spanish or French. It is known that
learn Spanish, and
At this school,
learn French.
of the students are girls, and
of the girls learn Spanish.
A boy is chosen at random. Find the probability that he learns Spanish.
of the students
[6 marks]
Markscheme
METHOD 1
A1
are boys (seen anywhere)
e.g.
appropriate approach
(M1)
e.g.
correct approach to find P(boy and Spanish)
e.g.
(A1)
,
correct substitution
e.g.
, 0.308
(A1)
,
correct manipulation
(A1)
e.g.
,
A1
N3
[6 marks]
METHOD 2
A1
are boys (seen anywhere)
e.g. 0.48 used in tree diagram
appropriate approach
(M1)
e.g. tree diagram
correctly labelled branches on tree diagram
(A1)
e.g. first branches are boy/girl, second branches are Spanish/not Spanish
correct substitution
(A1)
e.g.
correct manipulation
e.g.
(A1)
,
,
[6 marks]
Examiners report
Part (d) proved quite challenging as a great majority could only find the probability of being a boy. Those who did attempt it, and
successfully connected the problem with conditional probability, often had difficulties in reaching the correct final answer.
Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, without replacement.
6a. Find the probability that
(i)
none of the marbles are green;
(ii)
exactly one marble is green.
[5 marks]
Markscheme
(i)
(M1)
attempt to find
eg
,
,
A1
(ii)
N2
(M1)
attempt to find
eg
,
,
recognizing two ways to get one red, one green
eg
,
(M1)
,
A1
N2
[5 marks]
Examiners report
Many candidates correctly found the probability of selecting no green marbles in two draws, although some candidates treated the
second draw as if replacing the first. When finding the probability for exactly one green marble, candidates often failed to recognize
two pathways for selecting one of each color.
6b. Find the expected number of green marbles drawn from the jar.
[3 marks]
Markscheme
(seen anywhere)
(A1)
A1
correct substitution into formula for
eg
,
A1
expected number of green marbles is
N2
[3 marks]
Examiners report
Few candidates understood the concept of expected value in this context, often leaving this blank or treating as if a binomial
experiment. Successful candidates often made a distribution table before making the final calculation.
Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or , a marble is drawn from jar A.
Otherwise, a marble is drawn from jar B.
6c. (i)
(ii)
Write down the probability that the marble is drawn from jar B.
Given that the marble was drawn from jar B, write down the probability that it is red.
Markscheme
(i)
(ii)
[2 marks]
A1
N1
A1
N1
[2 marks]
Examiners report
Most candidates answered part (c) correctly. However, many overcomplicated (c)(ii) by using the conditional probability formula.
Those with a clear understanding of the concept easily followed the “write down” instruction.
6d. Given that the marble is red, find the probability that it was drawn from jar A.
[6 marks]
Markscheme
(M1)
recognizing conditional probability
eg
,
, tree diagram
attempt to multiply along either branch (may be seen on diagram)
(M1)
eg
attempt to multiply along other branch
(M1)
eg
adding the probabilities of two mutually exclusive paths
(A1)
eg
correct substitution
eg
,
A1
A1
N3
[6 marks]
Examiners report
Only a handful of candidates correctly applied conditional probability to find
in part (d). While some wrote down the
formula, or drew a tree diagram, few correctly calculated
. A common error was to combine the marbles in the two jars to
give
.
Consider the following cumulative frequency table.
7a. Find the value of
.
Markscheme
valid approach
eg
(M1)
,
A1
[2 marks]
N2
[2 marks]
Examiners report
Candidates had little problem determining a missing frequency from a cumulative frequency table.
7b. Find
[4 marks]
(i)
the mean;
(ii)
the variance.
Markscheme
A2
N2
(i)
mean
(ii)
recognizing that variance is (sd)2
eg
,
A1
(M1)
,
N2
[4 marks]
Examiners report
In part (b), few used the GDC to their advantage to correctly find the mean and variance. There were numerous unsuccessful attempts
at using the formulae for mean and variance, most resulting in algebraic errors along the way. Candidates recognized the concept of
variance but were often unable to determine what value should be squared.
8.
A random variable
is normally distributed with
Find the interquartile range of
and
.
.
Markscheme
recognizing one quartile probability (may be seen in a sketch)
eg
,
finding standardized value for either quartile
eg
(A1)
,
attempt to set up equation (must be with -values)
eg
,
one correct quartile
eg
,
correct working
(A1)
eg other correct quartile,
valid approach for IQR (seen anywhere)
eg
IQR
[7 marks]
,
A1
N4
(A1)
(M1)
(M1)
[7 marks]
Examiners report
This was an accessible problem that created difficulties for candidates. Although they recognized and often wrote down a formula for
IQR, most did not understand the conceptual nature of the first and third quartiles. Those who did could solve the problem effectively
using their GDC in relatively few steps. Candidates that were able to start this question often drew the normal curve and gave quartile
values at
and
. This generally led to a solution which while wrong, was also clearly inadequate for the indicated 7 marks.
A Ferris wheel with diameter
metres rotates clockwise at a constant speed. The wheel completes
the wheel is metres above the ground.
rotations every hour. The bottom of
A seat starts at the bottom of the wheel.
After t minutes, the height
9.
metres above the ground of the seat is given by
In one rotation of the wheel, find the probability that a randomly selected seat is at least
metres above the ground.
[5 marks]
Markscheme
setting up inequality (accept equation)
eg
,
(M1)
, sketch of graph with line
any two correct values for t (seen anywhere)
A1A1
eg
,
,
valid approach
eg
,
M1
,
,
A1
N2
,
[5 marks]
Examiners report
Part (e) was very poorly done for those who attempted the question and most did not make the connection between height, time and
probability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novel
approaches from the strongest of candidates using circles and angles to solve this part of question 10.
A running club organizes a race to select girls to represent the club in a competition.
The times taken by the group of girls to complete the race are shown in the table below.
10a. Find the value of
and of .
[4 marks]
Markscheme
(M1)
attempt to find
eg
,
A1
N2
attempt to find
eg
(M1)
,
A1
N2
[4 marks]
Examiners report
Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with
candidates not understanding terms such as "at least" or "less than".
10b. A girl is chosen at random.
(i)
Find the probability that the time she takes is less than
(ii)
Find the probability that the time she takes is at least
[3 marks]
minutes.
minutes.
Markscheme
A1
(i)
(ii)
valid approach
eg
N1
(M1)
,
A1
N2
[3 marks]
Examiners report
Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with
candidates not understanding terms such as "at least" or "less than".
10c. A girl is selected for the competition if she takes less than
Given that
minutes to complete the race.
[4 marks]
of the girls are not selected,
(i)
find the number of girls who are not selected;
(ii)
find
.
Markscheme
(i)
attempt to find number of girls
eg
(M1)
,
are not selected
(ii)
A1
are selected
A1
N2
(A1)
N2
[4 marks]
Examiners report
Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do with
candidates not understanding terms such as "at least" or "less than".
10d. Girls who are not selected, but took less than
minutes to complete the race, are allowed another chance to be selected. The [4 marks]
new times taken by these girls are shown in the cumulative frequency diagram below.
(i)
Write down the number of girls who were allowed another chance.
(ii)
Find the percentage of the whole group who were selected.
Markscheme
(i)
given second chance
(ii)
took less than
A1
N1
(A1)
minutes
attempt to find their selected total (may be seen in
eg
A1
(M1)
their answer from (i)
,
( )
calculation)
N3
[4 marks]
Examiners report
Part (d) was quite challenging for candidates, who may not have read the question carefully and studied the values in the diagram.
Many seemed confused by the idea that not all the girls who were given a second chance were selected. In part (d)(ii), many did not
find the percentage of the whole group, but rather the percentage of the girls who were given a second chance.
The random variable
is normally distributed with mean
11a. Find
.
and standard deviation .
[3 marks]
Markscheme
evidence of appropriate approach
(M1)
eg
(A1)
A1
N3
[3 marks]
Examiners report
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank
responses. Some candidates used tables and -scores while others used the GDC directly; the GDC approach earned full marks more
often than the -score approach.
11b. Given that
, find the value of .
Markscheme
-score for
(A1)
is
valid approach (must be with -values)
eg using inverse normal,
A1
[3 marks]
N3
(M1)
[3 marks]
Examiners report
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank
responses. Some candidates used tables and -scores while others used the GDC directly; the GDC approach earned full marks more
often than the -score approach. A common error in part (b) was to set the expression for -score equal to the probability. Many
candidates had difficulty giving answers correct to three significant figures; this was particularly an issue if no working was shown.
A bag contains four gold balls and six silver balls.
Two balls are drawn at random from the bag, with replacement. Let
12a. (i)
Find
.
(ii)
Find
.
be the number of gold balls drawn from the bag.
[5 marks]
Markscheme
METHOD 1
(i)
(M1)
appropriate approach
eg
,
,
A1
(ii)
N2
multiplying one pair of gold and silver probabilities
eg
,
(M1)
, 0.24
adding the product of both pairs of gold and silver probabilities
eg
(M1)
,
A1
N3
METHOD 2
(i)
evidence of recognizing binomial (may be seen in part (ii))
eg
(M1)
,
correct probability for use in binomial
eg
,
(A1)
,
A1 N3
(ii)
correct set up
(A1)
eg
A1
N2
[5 marks]
Examiners report
Parts (a)(i) and (ii) were generally well done, with candidates either using a tree diagram or a binomial approach.
12b. Hence, find
.
[3 marks]
Markscheme
METHOD 1
(A1)
(seen anywhere)
correct substitution into formula for
eg
(A1)
,
A1
N3
METHOD 2
(M1)
attempt to substitute into
eg
(A1)
correct substitution into
eg
A1 N3
[3 marks]
Examiners report
Part (a)(iii) proved difficult, with many either having trouble finding
or using
.
Fourteen balls are drawn from the bag, with replacement.
12c. (b)
(c)
Find the probability that exactly five of the balls are gold.
[4 marks]
Find the probability that at most five of the balls are gold.
Markscheme
Let
be the number of gold balls drawn from the bag.
(b)
evidence of recognizing binomial (seen anywhere)
eg
(M1)
,
A1
N2
[2 marks]
(c)
recognize need to find
A1
(M1)
N2
[2 marks]
Total [4 marks]
Examiners report
A great majority were confident solving part (b) with the GDC, although some did write the binomial term. Those candidates who did
not use the binomial function on the GDC had more difficulty in part (c), although a pleasing number were still able to identify that
they were seeking
.
12d. Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correct [3 marks]
to two decimal places.
Markscheme
Let
be the number of gold balls drawn from the bag.
recognizing conditional probability
(M1)
eg
,
,
,
(A1)
(to dp)
A1
N2
[3 marks]
Examiners report
While most candidate knew to use conditional probability in part (d), fewer were able to do so successfully, and even fewer still
correctly rounded their answer to two decimal places. The most common error was to multiply probabilities to find the intersection
needed for the conditional probability formula. Overall, candidates seemed better prepared for probability.
The cumulative frequency curve below represents the marks obtained by 100 students.
13a. Find the median mark.
[2 marks]
Examiners report
Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the
cumulative frequency) on the y-axis, but did not find the median mark on the x-axis.
13b. Find the interquartile range.
[3 marks]
Markscheme
lower quartile
, upper quartile
interquartile range
A1
(A1)(A1)
N3
[3 marks]
Examiners report
Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the
cumulative frequency) on the y-axis, but did not find the median mark on the x-axis. Similar misunderstanding was shown by some
candidates in part (b), when attempting to find the interquartile range.
The random variable X has the following probability distribution, with
.
14a. Find the value of r .
[2 marks]
Markscheme
(M1)
attempt to substitute
e.g.
A1
N2
[2 marks]
Examiners report
The majority of candidates were successful in earning full marks on this question.
14b. Given that
, find the value of p and of q .
[6 marks]
Markscheme
correct substitution into
(seen anywhere)
(A1)
e.g.
A1
correct equation
e.g.
,
A1
N1
M1
evidence of choosing
e.g.
,
correct working
,
(A1)
,
A1 N2
Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award the
final A mark for an inappropriate value of p.
[6 marks]
Examiners report
In part (b), a small number of candidates did not use the correct formula for
, even though this formula is given in the formula
booklet. There were also a few candidates who incorrectly assumed that
, forgetting that the sum of the probabilities must equal
1. There were a few candidates who left this question blank, which raises concerns about whether they had been exposed to
probability distributions during the course.
15a. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
(i)
Copy and complete the following tree diagram.
(ii) Find the probability that two white balls are chosen.
[5 marks]
Markscheme
(i)
A1A1A1
N3
(ii) multiplying along the correct branches (may be seen on diagram)
(A1)
e.g.
A1
N2
[5 marks]
Examiners report
Part (a) of this question was answered correctly by the large majority of candidates. There were some who did not follow the
instruction to copy and complete the tree diagram on their separate paper, and simply filled in the blanks on the exam paper.
15b. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
[5 marks]
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability
that they are both white is .
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag
B.
Find the probability that the two balls are white.
Markscheme
,
appropriate approach
(seen anywhere)
(A1)(A1)
(M1)
e.g.
A1
correct calculation
e.g.
,
A1
N3
[5 marks]
Examiners report
In part (b), many candidates struggled with finding the compound probability, and did not use the provided information in the
appropriate manner. Quite a few candidates seemed to be confused about when they should add the probabilities or when they should
multiply.
15c. Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
[4 marks]
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability
that they are both white is .
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag
B.
Given that both balls are white, find the probability that they were chosen from bag A.
Markscheme
recognizing conditional probability
e.g.
,
correct numerator
(A1)
e.g.
correct denominator
(A1)
e.g.
probability
[4 marks]
A1
N3
(M1)
Examiners report
In part (c), many recognized that the question dealt with conditional probability, and many tried to use the formula from the
information booklet, but failed to realize that they had already found the required values for the numerator and denominator in their
working for part (b).
Throughout this question, it was discouraging to see the large number of candidates making arithmetic errors. There were a surprising
number of candidates who multiplied fractions incorrectly, or found an incorrect value for simple multiplication such as
or
.
16a. The heights of a group of seven-year-old children are normally distributed with mean
and standard deviation
. A [3 marks]
child is chosen at random from the group.
Find the probability that this child is taller than
.
Markscheme
evidence of appropriate method
e.g.
(M1)
, sketch of normal curve showing mean and
,
(A1)
A1
N3
[3 marks]
Examiners report
There were many completely successful attempts at this question, with good use of formulae and calculator features.
16b. The heights of a group of seven-year-old children are normally distributed with mean
and standard deviation
. A [3 marks]
child is chosen at random from the group.
The probability that this child is shorter than
is
. Find the value of k .
Markscheme
(A1)
set up equation
e.g.
(M1)
, sketch
A1
N3
[3 marks]
Examiners report
There were many completely successful attempts at this question, with good use of formulae and calculator features.
However, in part (b) some candidates did not recognize the need to find the standardized value and set their equation equal to the
probability given in the question, thus earning only one mark.
17a. A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.
Find the probability that there is at least one defective lamp in the sample.
[4 marks]
Markscheme
evidence of recognizing binomial (seen anywhere)
e.g.
(M1)
,
(A1)
finding
appropriate approach
(M1)
e.g. complement, summing probabilities
A1
probability is
N3
[4 marks]
Examiners report
Although candidates seemed more confident in attempting binomial probabilities than in previous years, some of them failed to
recognize the binomial nature of the question in part (a). Many knew that the complement was required, but often used
or
instead of
.
17b. A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.
[4 marks]
Given that there is at least one defective lamp in the sample, find the probability that there are at most two defective lamps.
Markscheme
identifying correct outcomes (seen anywhere)
e.g.
(A1)
, 1 or 2 defective,
recognizing conditional probability (seen anywhere)
e.g.
,
, P(at most 2|at least 1)
appropriate approach involving conditional probability
e.g.
probability is
R1
,
(M1)
,
A1
N2
[4 marks]
Examiners report
Part (b) was poorly answered. While some candidates recognized that it was a conditional probability, very few were able to correctly
apply the formula, identify the outcomes and follow on to achieve the correct result.
Only a few could find the intersection of the events correctly. Several thought the numerator was a product (i.e.
), and then cancelled common factors with the denominator. Others realized that
required but multiplied their probabilities.
This was the most commonly missed out question from Section A.
and
were
The ages of people attending a music concert are given in the table below.
18a. Find p .
[2 marks]
Markscheme
evidence of valid approach
e.g.
(M1)
, line on graph at
A1
N2
[2 marks]
Examiners report
Part (a) was generally answered correctly, with most candidates showing a good grasp of cumulative frequency from a table.
18b. The cumulative frequency diagram is given below.
Use the diagram to estimate
(i)
the 80th percentile;
(ii) the interquartile range.
[5 marks]
Markscheme
(i) evidence of valid approach
e.g. line on graph,
A1
(ii)
(M1)
, using complement
N2
(A1)(A1)
;
(accept any notation that suggests an interval)
A1
N3
[5 marks]
Examiners report
A surprising number of candidates had difficulty reading values off the cumulative frequency curve. A common incorrect answer for
(b)(i) was 29, indicating carelessness with the given scale. Too many candidates gave 40 and 120 for the quartile values.
Events A and B are such that
,
and
.
The values q , r , s and t represent probabilities.
19a. Write down the value of t .
[1 mark]
Markscheme
A1
N1
[1 mark]
Examiners report
Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning full
marks.
19b. (i)
Show that
.
(ii) Write down the value of q and of s .
[3 marks]
Markscheme
A1
(i) correct values
e.g.
,
AG
(ii)
N0
A1A1
,
N2
[3 marks]
Examiners report
Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning full
marks. Only a few candidates worked backwards from the
given in the "show that" portion of part (b).
19c. (i)
Write down
(ii) Find
.
[3 marks]
.
Markscheme
(i)
A1
(ii)
N1
A2
N2
[3 marks]
Examiners report
Many candidates struggled on part (c)(ii), either not recognizing conditional probability or multiplying probabilities to find the
numerator as if the events were independent. A number of candidates who successfully found the probability in part (c)(ii) left their
incomplete answer of .
20a. The probability of obtaining “tails” when a biased coin is tossed is
obtaining at least four tails.
. The coin is tossed ten times. Find the probability of
[4 marks]
Markscheme
evidence of recognizing binomial distribution
e.g.
,
(M1)
,
EITHER
(A1)
(M1)
evidence of using complement
e.g.
any probability,
A1
N3
OR
summing the probabilities from
(M1)
to
correct expression or values
(A1)
e.g.
,
0.919424
A1
N3
[4 marks]
Examiners report
This was an accessible problem that created some difficulties for candidates. Most were able to recognize the binomial nature of the
problem but were confused by the phrase "at least four tails" which was often interpreted as the complement of four or less. Poor
algebraic manipulation also led to unnecessary errors that the calculator approach would have avoided.
20b. The probability of obtaining “tails” when a biased coin is tossed is 0.57. The coin is tossed ten times. Find the probability of
[3 marks]
obtaining the fourth tail on the tenth toss.
Markscheme
evidence of valid approach
(M1)
e.g. three tails in nine tosses,
correct calculation
e.g.
(A1)
,
A1
N2
[3 marks]
Examiners report
This was an accessible problem that created some difficulties for candidates. Most were able to recognize the binomial nature of the
problem but were confused by the phrase "at least four tails" which was often interpreted as the complement of four or less. Poor
algebraic manipulation also led to unnecessary errors that the calculator approach would have avoided.
The histogram below shows the time T seconds taken by 93 children to solve a puzzle.
The following is the frequency distribution for T .
21a. (i)
Write down the value of p and of q .
[3 marks]
(ii) Write down the median class.
Markscheme
(i)
(ii)
,
A1A1
A1
N2
N1
[3 marks]
Examiners report
Parts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.
21b. A child is selected at random. Find the probability that the child takes less than 95 seconds to solve the puzzle.
Markscheme
evidence of valid approach
(M1)
e.g. adding frequencies
A1
[2 marks]
N2
[2 marks]
Examiners report
Parts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.
21c. Consider the class interval
(i)
.
[2 marks]
Write down the interval width.
(ii) Write down the mid-interval value.
Markscheme
(i) 10
A1
N1
(ii) 50
A1
N1
[2 marks]
Examiners report
In part (c) some candidates had problems with the term "interval width" and there were some rather interesting mid-interval values
noted.
21d. Hence find an estimate for the
(i)
[4 marks]
mean;
(ii) standard deviation.
Markscheme
(i) evidence of approach using mid-interval values (may be seen in part (ii))
A2
N3
A1
N1
(M1)
(ii)
[4 marks]
Examiners report
In part (d), candidates often ignored the "hence" command and estimated values from the graph rather than from the information in
part (c).
21e. John assumes that T is normally distributed and uses this to estimate the probability that a child takes less than 95 seconds to
solve the puzzle.
Find John’s estimate.
Markscheme
e.g. standardizing,
A1
[2 marks]
N2
[2 marks]
Examiners report
Those who correctly obtained the mean and standard deviation had little difficulty with part (e) although candidates often used
unfamiliar calculator notation as their working or used the mid-interval value as the mean of the distribution.
A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a second
marble.
22a. Write down the probability that the first marble Anna selects is red.
[1 mark]
Markscheme
Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.
A1
N1
[1 mark]
Examiners report
Candidates did very well on parts (a) and (b) of this probability question.
22b. Find the probability that Anna selects two red marbles.
[2 marks]
Markscheme
(M1)
attempt to find
e.g.
,
,
A1
N2
[2 marks]
Examiners report
Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events
in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and
therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the
probabilities in parts (b) and (c).
22c. Find the probability that one marble is red and one marble is blue.
[3 marks]
Markscheme
METHOD 1
(M1)
attempt to find
e.g.
,
,
recognizing two ways to get one red, one blue
e.g.
,
(M1)
,
A1
N2
[3 marks]
METHOD 2
recognizing that
(M1)
is
attempt to find
(M1)
and
e.g.
,
;
,
A1
N2
[3 marks]
Examiners report
Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events
in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and
therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the
probabilities in parts (b) and (c).
23. The random variable X has the following probability distribution.
Given that
[6 marks]
, find q .
Markscheme
correct substitution into
e.g.
(seen anywhere)
A1
,
recognizing
(seen anywhere)
correct substitution into ​
(M1)
A1
e.g.
attempt to solve simultaneous equations
correct working
(A1)
e.g.
,
A1
(M1)
N4
[6 marks]
Examiners report
Candidates generally earned either full marks or only one mark on this question. The most common error was where candidates only
wrote the equation for
, and tried to rearrange that equation to solve for q. The candidates who also knew that the sum of
the probabilities must be equal to 1 were very successful in solving the resulting system of equations.
Let
, where
.
24a. Find the values of k such that
has two equal roots.
[4 marks]
Markscheme
METHOD 1
(M1)
evidence of discriminant
e.g.
, discriminant = 0
A1
correct substitution into discriminant
e.g.
,
A1A1
N3
METHOD 2
recognizing that equal roots means perfect square
(R1)
e.g. attempt to complete the square,
correct working
e.g.
A1
,
A1A1
N3
[4 marks]
Examiners report
A good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there were
many who tried to use the quadratic formula without recognizing the significance of the discriminant.
24b. Each value of k is equally likely for
. Find the probability that
has no roots.
[4 marks]
Markscheme
evidence of appropriate approach
(M1)
e.g.
correct working for k
e.g.
,
A2
A1
, list all correct values of k
N3
[4 marks]
Examiners report
Part (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing that
there were 11 possible values for k.
The cumulative frequency curve below represents the heights of 200 sixteen-year-old boys.
Use the graph to answer the following.
25a. Write down the median value.
[1 mark]
Markscheme
A1
N1
[1 mark]
Examiners report
Parts (a) and (b) were generally well done.
25b. A boy is chosen at random. Find the probability that he is shorter than
.
[2 marks]
Markscheme
attempt to find number shorter than 161
(M1)
e.g. line on graph, 12 boys
A1
N2
[2 marks]
Examiners report
Parts (a) and (b) were generally well done.
25c. Given that
of the boys are taller than
, find h .
[3 marks]
Markscheme
METHOD 1
have a height less than h
(A1)
(36 may be seen as a line on the graph)
A1
(cm)
(A1)
N2
METHOD 2
(164 may be seen as a line on the graph)
(A1)
(A1)
A1
(cm)
N2
[3 marks]
Examiners report
Some candidates could only earn the first mark in part (c) for finding
subtract this value from the total of 200.
of 200. Others gave the answer as 164, neglecting to
A company produces a large number of water containers. Each container has two parts, a bottle and a cap. The bottles and caps are tested to
check that they are not defective.
A cap has a probability of 0.012 of being defective. A random sample of 10 caps is selected for inspection.
26a. Find the probability that exactly one cap in the sample will be defective.
[2 marks]
Markscheme
Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in
subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.
evidence of recognizing binomial (seen anywhere in the question)
e.g.
,
A1
(M1)
,
N2
[2 marks]
Examiners report
Many stronger candidates were completely successful with this question, employing technology efficiently. A number of candidates
did not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)
answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) after
an error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather than
multiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.
26b. The sample of caps passes inspection if at most one cap is defective. Find the probability that the sample passes inspection.
Markscheme
(M1)
valid approach
e.g.
,
A1
[2 marks]
N2
[2 marks]
Examiners report
Many stronger candidates were completely successful with this question, employing technology efficiently. A number of candidates
did not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)
answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) after
an error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather than
multiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.
26c. The heights of the bottles are normally distributed with a mean of
(i)
and a standard deviation of
.
Copy and complete the following diagram, shading the region representing where the heights are less than
(ii) Find the probability that the height of a bottle is less than
[5 marks]
.
.
Markscheme
(i)
A1A1
N2
Note: Award A1 for vertical line to right of mean, A1 for shading to left of their vertical line.
(ii) valid approach
(M1)
e.g.
working to find standardized value
e.g.
(A1)
, 2.1
A1
N3
[5 marks]
Examiners report
Many stronger candidates were completely successful with this question, employing technology efficiently. A number of candidates
did not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)
answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) after
an error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather than
multiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.
26d. (i)
A bottle is accepted if its height lies between
is accepted.
and
. Find the probability that a bottle selected at random [5 marks]
(ii) A sample of 10 bottles passes inspection if all of the bottles in the sample are accepted. Find the probability that the sample passes
inspection.
Markscheme
valid approach
(M1)
e.g.
,
correct working
(A1)
e.g.
A1
N3
(A1)
(ii) correct working
e.g.
,
(accept 0.694 from tables)
A1
N2
[5 marks]
Examiners report
Many stronger candidates were completely successful with this question, employing technology efficiently. A number of candidates
did not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)
answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) after
an error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather than
multiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.
26e. The bottles and caps are manufactured separately. A sample of 10 bottles and a sample of 10 caps are randomly selected for
[2 marks]
testing. Find the probability that both samples pass inspection.
Markscheme
valid approach
(M1)
e.g.
,
(accept
from tables)
A1
N2
[2 marks]
Examiners report
Many stronger candidates were completely successful with this question, employing technology efficiently. A number of candidates
did not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)
answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) after
an error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather than
multiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.
The probability distribution of a discrete random variable X is given by
.
27a. Write down
.
[1 mark]
Markscheme
A1
[1 mark]
N1
Examiners report
Although many candidates were successful in working with the probability function, students had difficulty following the "show that"
instruction of this question. Many substituted
and worked backwards to show that the sum of probabilities is 1. Some would
argue that
does not work, but were unable to give a complete justification for
. A good number of students seemed
unprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided
by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.
27b. Show that
.
[4 marks]
Markscheme
(A1)
(A1)
setting the sum of probabilities
e.g.
M1
,
(accept
AG
)
A1
N0
[4 marks]
Examiners report
Although many candidates were successful in working with the probability function, students had difficulty following the "show that"
instruction of this question. Many substituted
and worked backwards to show that the sum of probabilities is 1. Some would
argue that
does not work, but were unable to give a complete justification for
.
27c. Find
.
[2 marks]
Markscheme
A1
correct substitution into
e.g.
A1
N1
[2 marks]
Examiners report
A good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know what
to do with it, while others divided
by 3 or by 6, which confuses the concept of a mean in a probability distribution with the
more common understanding.
In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn diagram below shows the events art
and music. The values p , q , r and s represent numbers of students.
28a. (i)
Write down the value of s .
[5 marks]
(ii) Find the value of q .
(iii) Write down the value of p and of r .
Markscheme
A1
(i)
N1
(ii) evidence of appropriate approach
e.g.
(M1)
,
A1
(iii)
,
N2
A1A1
N2
[5 marks]
Examiners report
A majority of candidates found the values in the Venn diagram easily. Common errors include giving
, and also neglecting s in
finding
(e.g.
) . Some interpreted the values as probabilities, despite the question explicitly stating that p, q, r and s
represent numbers of students. Occasionally the values for p and r were misinterpreted as being inclusive of q. Follow-through marks
were often earned in subsequent parts for such cases.
28b. (i)
A student is selected at random. Given that the student takes music, write down the probability the student takes art.
(ii) Hence, show that taking music and taking art are not independent events.
[4 marks]
Markscheme
A2
(i)
N2
(ii) METHOD 1
A1
R1
evidence of correct reasoning
e.g.
the events are not independent
AG
N0
METHOD 2
A1
R1
evidence of correct reasoning
e.g.
the events are not independent
AG
N0
[4 marks]
Examiners report
For (b), rather than think of the situations conceptually, most candidates reached for the formula for conditional probability, with
mixed results. Few candidates considered that independence means
. Most applied
, with
many giving incomplete or incorrect calculations. Some candidates compared the wrong things and showed, for example, that
,
which incorrectly compares
with
. Others stated that because there is an intersection, the events are independent,
which is an insufficient explanation.
28c. Two students are selected at random, one after the other. Find the probability that the first student takes only music and the
[4 marks]
second student takes only art.
Markscheme
evidence of valid approach
(seen anywhere)
A1
(seen anywhere)
A1
(M1)
e.g.
A1
N2
[4 marks]
Examiners report
Part (c) was commonly answered as if there is replacement, with many candidates calculating
phrasing "one after the other" is that there is no replacement.
. However, implicit in the
A random variable X is distributed normally with a mean of 20 and variance 9.
29a. Find
.
[3 marks]
Markscheme
(A1)
(M1)
evidence of attempt to find
e.g.
,
A1
N3
[3 marks]
Examiners report
This question clearly demonstrated that some centres are still not giving adequate treatment to this topic. A great many candidates
neglected to find the standard deviation and used the variance throughout. More still did not leave their answers to the required
accuracy. Ignoring the use of the variance, responses to part (a) demonstrated that most candidates were comfortable finding the
required probability using their calculator or setting up a suitable standardized equation.
29b. Let
(i)
.
[5 marks]
Represent this information on the following diagram.
(ii) Find the value of k .
Markscheme
A1A1
N2
Note: Award A1 with shading that clearly extends to right of the mean, A1 for any correct label, either k, area or their value of k.
(A1)
(ii)
attempt to set up an equation
e.g.
,
A1
[5 marks]
N3
(M1)
Examiners report
This question clearly demonstrated that some centres are still not giving adequate treatment to this topic. A great many candidates
neglected to find the standard deviation and used the variance throughout. More still did not leave their answers to the required
accuracy. Ignoring the use of the variance, responses to part (a) demonstrated that most candidates were comfortable finding the
required probability using their calculator or setting up a suitable standardized equation. In part (b) (i), the sketch was often poorly
shaded or incorrectly labelled. In (b) (ii), candidates frequently confused the z-score with the given probability of 0.85. Calculator
approaches were more successful than working by hand but candidates should remember to avoid the use of calculator notation in
their working, as it is not correct mathematical notation.
A box holds 240 eggs. The probability that an egg is brown is 0.05.
30a. Find the expected number of brown eggs in the box.
[2 marks]
Markscheme
correct substitution into formula for
(A1)
e.g.
A1
N2
[2 marks]
Examiners report
Part (a) was answered correctly by most candidates.
30b. Find the probability that there are 15 brown eggs in the box.
[2 marks]
Markscheme
evidence of recognizing binomial probability (may be seen in part (a))
e.g.
(M1)
,
A1
N2
[2 marks]
Examiners report
In parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simple
probability.
30c. Find the probability that there are at least 10 brown eggs in the box.
Markscheme
(A1)
evidence of valid approach
(M1)
e.g. using complement, summing probabilities
A1
[3 marks]
N3
[3 marks]
Examiners report
In parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simple
probability. Although several candidates appreciated that (c) involved the idea of a complement, some resorted to elaborate
probability addition suggesting they were unaware of the capabilities of their GDC. There was also a great deal of evidence to
suggest that candidates did not understand the phrase "at least 10" as several candidates found either
,
or
.
A company uses two machines, A and B, to make boxes. Machine A makes
of the boxes.
of the boxes made by machine A pass inspection.
of the boxes made by machine B pass inspection.
A box is selected at random.
31a. Find the probability that it passes inspection.
[3 marks]
Markscheme
evidence of valid approach involving A and B
e.g.
(M1)
, tree diagram
correct expression
(A1)
e.g.
A1
N2
[3 marks]
Examiners report
Part (a) was usually well done. Those candidates that did not succeed with this part often did not show a correct tree diagram
indicating that they did not really understand the problem or indeed how to start it.
31b. The company would like the probability that a box passes inspection to be 0.87.
[4 marks]
Find the percentage of boxes that should be made by machine B to achieve this.
Markscheme
evidence of recognizing complement (seen anywhere)
e.g.
,
,
,
correct expression
A1
e.g.
,
from B
A1
,
(M1)
evidence of valid approach
e.g.
,
(M1)
,
N2
[4 marks]
Examiners report
Many successful attempts to (b) relied on "guess and check" or intuitive solutions while a surprising number of candidates could not
manage to systematically set up an appropriate algebraic expression involving a complement.
The Venn diagram below shows events A and B where
probabilities.
32a. (i)
Write down the value of n .
,
and
. The values m , n , p and q are
[4 marks]
(ii) Find the value of m , of p , and of q .
Markscheme
A1
(i)
(ii)
N1
,
A1A1A1
,
N3
[4 marks]
Examiners report
Most candidates were able to find the correct values for the Venn diagram. Unfortunately, however, there were many candidates who
did not understand what each region of the diagram represents. For example, a very common error was thinking that
,
rather than the correct
.
32b. Find
.
[2 marks]
Markscheme
appropriate approach
e.g.
,
A1
,
(M1)
N2
[2 marks]
Examiners report
Candidates seemed to understand the idea of the complement in part (b), but some were not able to find the correct answer because of
confusion over the separation of the different regions in the diagram.
A scientist has 100 female fish and 100 male fish. She measures their lengths to the nearest cm. These are shown in the following box and
whisker diagrams.
33a. Find the range of the lengths of all 200 fish.
[3 marks]
Markscheme
(A1)(A1)
correct end points
max = 27 , min = 4
range = 23
A1
N3
[3 marks]
Examiners report
While there were a large number of candidates who answered both parts of this question correctly, a surprising number did not know
how to find the range of all 200 fish in part (a). Common errors included finding the ranges of the male and female fish separately, or
averaging the separate ranges of the male and female fish.
33b. Four cumulative frequency graphs are shown below.
Which graph is the best representation of the lengths of the female fish?
Markscheme
Graph 3
[2 marks]
A2
N2
[2 marks]
Examiners report
Some candidates did not interpret the cumulative frequency graphs correctly, or just seemed to guess which graph was correct. The
most common incorrect "guess" was graph 4, likely because this graph had a more familiar cumulative shape.
Let the random variable X be normally distributed with mean 25, as shown in the following diagram.
The shaded region between 25 and 27 represents
34a. Find
.
of the distribution.
[2 marks]
Markscheme
symmetry of normal curve
(M1)
e.g.
A1
N2
[2 marks]
Examiners report
This question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find the
probability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use the
complement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) was
whether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem which
would have allowed them to better demonstrate their method.
34b. Find the standard deviation of X .
[5 marks]
Markscheme
METHOD 1
(A1)
finding standardized value
e.g.
evidence of complement
e.g.
,
(M1)
, 0.8
(A1)
finding z-score
e.g.
attempt to set up equation involving the standardized value
e.g.
M1
,
A1
N3
METHOD 2
set up using normal CDF function and probability
e.g.
(M1)
,
A2
correct equation
e.g.
,
attempt to solve the equation using GDC
(M1)
e.g. solver, graph, trial and error (more than two trials must be shown)
A1
N3
[5 marks]
Examiners report
This question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find the
probability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use the
complement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) was
whether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem which
would have allowed them to better demonstrate their method.
Two fair 4-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1, 2, 3, 4. The score for each die is the number
which lands face down.
35a. List the pairs of scores that give a sum of 6.
Markscheme
three correct pairs
A1A1A1
N3
e.g. (2, 4), (3, 3), (4, 2) , R2G4, R3G3, R4G2
[3 marks]
Examiners report
All but the weakest candidates managed to score full marks for parts (a) and (b). An occasional error in part (a) was including
additional pair(s) or listing (3, 3) twice.
[3 marks]
35b. The probability distribution for the sum of the scores on the two dice is shown below.
[3 marks]
Find the value of p , of q , and of r .
Markscheme
,
A1A1A1
,
N3
[3 marks]
Examiners report
All but the weakest candidates managed to score full marks for parts (a) and (b).
35c. Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5 on three or more throws.
[6 marks]
Find the probability that Fred wins a prize.
Markscheme
let X be the number of times the sum of the dice is 5
evidence of valid approach
e.g.
, tree diagram, 5 sets of outcomes produce a win
one correct parameter
e.g.
(M1)
,
(A1)
,
(A1)
Fred wins prize is
appropriate approach to find probability
M1
e.g. complement, summing probabilities, using a CDF function
correct substitution
e.g.
,
(A1)
,
,
A1
N3
[6 marks]
Examiners report
Many candidates found part (c) challenging, as they failed to recognize the binomial probability. Successful candidates generally used
either the binomial CDF function or the sum of two binomial probabilities. Some used approaches like multiplying probabilities or
tree diagrams, but these were less successful.
The diagram below shows the probabilities for events A and B , with
.
36a. Write down the value of p .
[1 mark]
Markscheme
A1
N1
[1 mark]
Examiners report
While nearly every candidate answered part (a) correctly, many had trouble with the other parts of this question.
36b. Find
.
[3 marks]
Markscheme
multiplying along the branches
e.g.
(M1)
,
adding products of probabilities of two mutually exclusive paths
e.g.
(M1)
,
A1
N2
[3 marks]
Examiners report
In part (b), many candidates did not multiply along the branches of the tree diagram to find the required values, and many did not
realize that there were two paths for P(B). There were also many candidates who understood what the question required, but then did
not know how to multiply fractions correctly, and these calculation errors led to an incorrect answer.
36c. Find
.
[3 marks]
Markscheme
appropriate approach which must include
e.g.
(do not accept
(may be seen on diagram)
(M1)
)
(A1)
A1
N2
[3 marks]
Examiners report
In part (c), most candidates attempted to use a formula for conditional probability found in the information booklet, but very few
substituted the correct values.
Consider the events A and B, where
,
and
.
The Venn diagram below shows the events A and B, and the probabilities p, q and r.
37a. Write down the value of
(i)
[3 marks]
p;
(ii) q ;
(iii) r.
Markscheme
(i)
A1
N1
(ii)
A1
N1
(iii)
A1
N1
[3 marks]
Examiners report
As the definitions of p and q were not clear to candidates, both responses of p
full marks. However, finding r eluded many.
37b. Find the value of
.
,q
and p
,q
were accepted for
[2 marks]
Markscheme
A2
N2
Note: Award A1 for an unfinished answer such as
.
[2 marks]
Examiners report
Few candidates answered the conditional probability correctly. Many attempted to use the formula in the booklet without considering
the complement, and there was little evidence of the Venn diagram being utilized as a helpful aid.
37c. Hence, or otherwise, show that the events A and B are not independent.
[1 mark]
Markscheme
R1
valid reason
e.g.
,
thus, A and B are not independent
AG
N0
[1 mark]
Examiners report
To show the events are not independent, many correctly reasoned that
. A handful recognized that
is an
alternative approach that uses the answer in part (b). Some candidates do not know the difference between independent and
mutually exclusive.
José travels to school on a bus. On any day, the probability that José will miss the bus is .
If he misses his bus, the probability that he will be late for school is .
If he does not miss his bus, the probability that he will be late is .
Let E be the event “he misses his bus” and F the event “he is late for school”.
The information above is shown on the following tree diagram.
38a. Find
[4 marks]
(i)
(ii)
;
.
Markscheme
A1
(i)
N1
(ii) evidence of multiplying along the branches
e.g.
(M1)
,
adding probabilities of two mutually exclusive paths
e.g.
(M1)
,
A1
N2
[4 marks]
Examiners report
Candidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying to
use the union formula from the information booklet.
38b. Find the probability that
(i)
[5 marks]
José misses his bus and is not late for school;
(ii) José missed his bus, given that he is late for school.
Markscheme
(A1)
(i)
A1
(ii) recognizing this is
(M1)
e.g.
A2
N3
[5 marks]
Examiners report
Candidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying to
use the union formula from the information booklet. On part (b)(ii), many candidates knew that they were supposed to use some type
of conditional probability but did not know how to find
. Many candidates made errors working with fractions. Some
candidates who missed part (a)(ii) were able to earn follow-through credit on part (b)(ii).
38c. The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.
Copy and complete the probability distribution table.
[3 marks]
Markscheme
A2A1
N3
[3 marks]
Examiners report
Many candidates had difficulty completing the probability distribution table. While the common error of finding the probability for
as was understandable as the candidate did not appreciate that there were two ways of paying three euros, it was
disappointing that these candidates often correctly found
as and did not note that the probabilities failed to sum to one.
These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use the
probabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part
(d), as a majority of candidates knew the method for finding expected value.
38d. The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.
[2 marks]
Find the expected cost for José for both days.
Markscheme
correct substitution into
e.g.
formula
(M1)
,
(euros)
A1
N2
[2 marks]
Examiners report
Many candidates had difficulty completing the probability distribution table. While the common error of finding the probability for
as was understandable as the candidate did not appreciate that there were two ways of paying three Euros, it was
disappointing that these candidates often correctly found
as and did not note that the probabilities failed to sum to one.
These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use the
probabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part
(d), as a majority of candidates knew the method for finding expected value.
A standard die is rolled 36 times. The results are shown in the following table.
39a. Write down the standard deviation.
Markscheme
A2
[2 marks]
N2
[2 marks]
Examiners report
Surprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms
"median" and "interquartile range". Many candidates opted for an analytical approach in part (a) which always resulted in mistakes.
39b. Write down the median score.
[1 mark]
Markscheme
A1
median
N1
[1 mark]
Examiners report
Some candidates wrote the down the mean instead of the median in part (b).
39c. Find the interquartile range.
[3 marks]
Markscheme
,
(may be seen in a box plot)
(A1)(A1)
(accept any notation that suggests the interval 3 to 5)
A1
N3
[3 marks]
Examiners report
Surprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms
"median" and "interquartile range".
A test has five questions. To pass the test, at least three of the questions must be answered correctly.
The probability that Mark answers a question correctly is . Let X be the number of questions that Mark answers correctly.
40a. (i)
Find E(X ) .
(ii) Find the probability that Mark passes the test.
[6 marks]
Markscheme
(M1)
(i) valid approach
e.g.
,
A1
N2
(ii) evidence of appropriate approach involving binomial
(M1)
e.g.
recognizing that Mark needs to answer 3 or more questions correctly
(A1)
e.g.
valid approach
e.g.
M1
,
A1
N3
[6 marks]
Examiners report
There was wide spectrum of success on this problem. Candidates could normally find E(X) using
but many failed to recognize
that the "experiment" was binomial or that for Mark to the pass the test, he needed to answer either 3, 4 or 5 questions correctly.
Bill also takes the test. Let Y be the number of questions that Bill answers correctly.
The following table is the probability distribution for Y .
40b. (i)
Show that
.
(ii) Given that
[8 marks]
, find a and b .
Markscheme
(i) evidence of summing probabilities to 1
(M1)
e.g.
some simplification that clearly leads to required answer
A1
e.g.
AG
N0
(ii) correct substitution into the formula for expected value
e.g.
(A1)
some simplification
e.g.
correct equation
A1
e.g.
evidence of solving
,
[8 marks]
(M1)
A1A1
N4
(A1)
Examiners report
Part (b) was generally well done although there were a number of algebraic errors particularly in part (b) (ii), leading to incorrect
values of a and b. Again, appropriate use of the GDC here would have eliminated these errors.
40c. Find which student is more likely to pass the test.
[3 marks]
Markscheme
attempt to find probability Bill passes
(M1)
e.g.
A1
correct value 0.19
Bill (is more likely to pass)
A1
N0
[3 marks]
Examiners report
In (c), candidates had trouble with the command term, "find" and often just wrote down either "Mark" or "Bill".
The following frequency distribution of marks has mean 4.5.
41a. Find the value of x.
[4 marks]
Markscheme
,
evidence of substituting into mean
A1
(M1)
A1
correct equation
e.g.
(seen anywhere)
,
A1
N2
[4 marks]
Examiners report
Surprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of the
frequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for a
frequency. Unfortunately, candidates did not appreciate the unreasonableness of this result.
41b. Write down the standard deviation.
Markscheme
A2
[2 marks]
N2
[2 marks]
Examiners report
Surprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of the
frequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for a
frequency. Unfortunately, candidates did not appreciate the unreasonableness of this result. In part (b), many candidates could not
find the standard deviation in their GDC, often trying to calculate it by hand with no success. Further, many could not distinguish
between the sample and the population standard deviation given in the GDC.
Evan likes to play two games of chance, A and B.
For game A, the probability that Evan wins is 0.9. He plays game A seven times.
42a. Find the probability that he wins exactly four games.
[2 marks]
Markscheme
evidence of recognizing binomial probability (may be seen in (b) or (c))
e.g. probability
,
A1
probability
(M1)
, complementary probabilities
N2
[2 marks]
Examiners report
Parts of this question were handled very well by a great many candidates. Most were able to recognize the binomial condition and
had little difficulty with part (a). However, more than a few reported the answer as 0.23, thus incurring the accuracy penalty.
For game B, the probability that Evan wins is p . He plays game B seven times.
42b. Write down an expression, in terms of p , for the probability that he wins exactly four games.
[2 marks]
Markscheme
correct expression
A1A1
e.g.
,
N2
Note: Award A1 for binomial coefficient (accept
) , A1 for
.
[2 marks]
Examiners report
Those candidates that were successful in part (a) could easily write the required expression for part (b).
42c. Hence, find the values of p such that the probability that he wins exactly four games is 0.15.
[3 marks]
Markscheme
evidence of attempting to solve their equation
e.g.
(M1)
, sketch
A1A1
,
N3
[3 marks]
Examiners report
In part (c), many candidates set up the question correctly or set their expression from (b) equal to 0.15, however few candidates
considered the GDC as a method to solve the equation. Rather, those who attempted usually tried to expand the polynomial, and still
did not use the GDC to solve this equation. A graphical approach to the solution would reveal that there are two solutions for p, but
few caught this subtlety.
The weights of players in a sports league are normally distributed with a mean of
, (correct to three significant figures). It is known that
of the players have weights between
and
. The probability that a player weighs less than
is 0.05.
43a. Find the probability that a player weighs more than
.
[2 marks]
Markscheme
evidence of appropriate approach
e.g.
(M1)
, diagram showing values in a normal curve
A1
N2
[2 marks]
Examiners report
This question was quite accessible to those candidates in centres where this topic is given the attention that it deserves. Most
candidates handled part (a) well using the basic properties of a normal distribution.
43b. (i)
Write down the standardized value, z, for
(ii) Hence, find the standard deviation of weights.
Markscheme
A1
(i)
N1
(ii) evidence of appropriate approach
e.g.
,
correct substitution
e.g.
A1
[4 marks]
N1
A1
(M1)
.
[4 marks]
Examiners report
In part (b) (i), candidates often confused the z-score with the area in the table which led to a standard deviation that was less than zero
in part (b) (ii). At this point, candidates “fudged” results in order to continue with the remaining parts of the question. In (b) (ii), the
“hence” command was used expecting candidates to use the results of (b) (i) to find a standard deviation of 4.86. Unfortunately, many
decided to use their answers and the information from part (a) resulting in quite a different standard deviation of 5.79. Recognizing
the inconsistency in the question, full marks were awarded for this approach, as well as full follow-through in subsequent parts of the
question.
43c. To take part in a tournament, a player’s weight must be within 1.5 standard deviations of the mean.
(i)
[5 marks]
Find the set of all possible weights of players that take part in the tournament.
(ii) A player is selected at random. Find the probability that the player takes part in the tournament.
Markscheme
A1A1A1
(i)
N3
Note: Award A1 for 68.8, A1 for 84.4, A1 for giving answer as an interval.
(ii) evidence of appropriate approach
e.g.
(M1)
,
A1
N2
[5 marks]
Examiners report
Candidates could obtain full marks easily in part (c) with little understanding of a normal distribution but they often confused z-scores
with data values, adding and subtracting 1.5 from the mean of 76.
43d. Of the players in the league,
are women. Of the women,
take part in the tournament.
[4 marks]
Given that a player selected at random takes part in the tournament, find the probability that the selected player is a woman.
Markscheme
recognizing conditional probability
(M1)
e.g.
(A1)
A1
A1
[4 marks]
Examiners report
In part (d), few recognized the conditional nature of the question and only determined the probability that a woman qualifies and
takes part in the tournament.
The following table gives the examination grades for 120 students.
44a. Find the value of
(i)
[4 marks]
p;
(ii) q .
Markscheme
(a) (i) evidence of appropriate approach
e.g.
(M1)
,
A1
N2
(ii) evidence of valid approach
(M1)
their value of p,
e.g.
A1
N2
[4 marks]
Examiners report
The majority of candidates had little trouble finding the missing values in the frequency distribution table.
44b. Find the mean grade.
[2 marks]
Markscheme
evidence of appropriate approach
e.g. substituting into
A1
mean
(M1)
, division by 120
N2
[2 marks]
Examiners report
Many did not seem comfortable calculating the mean and standard deviation using their GDCs.
The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number of
candidates were unable to find the standard deviation.
44c. Write down the standard deviation.
Markscheme
1.09
A1
[1 mark]
N1
[1 mark]
Examiners report
Many did not seem comfortable calculating the mean and standard deviation using their GDCs.
The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number of
candidates were unable to find the standard deviation.
Jan plays a game where she tosses two fair six-sided dice. She wins a prize if the sum of her scores is 5.
45a. Jan tosses the two dice once. Find the probability that she wins a prize.
[3 marks]
Markscheme
36 outcomes (seen anywhere, even in denominator)
(A1)
valid approach of listing ways to get sum of 5, showing at least two pairs
(M1)
e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram
A1
N3
[3 marks]
Examiners report
While many candidates were successful at part (a), far fewer recognized the binomial distribution in the second part of the problem.
45b. Jan tosses the two dice 8 times. Find the probability that she wins 3 prizes.
[2 marks]
Markscheme
recognizing binomial probability
e.g.
(M1)
, binomial pdf,
A1
N2
[2 marks]
Examiners report
While many candidates were successful at part (a), far fewer recognized the binomial distribution in the second part of the problem.
Those who did not obtain the correct answer at part (a) often scored partial credit by either drawing a table to represent the sample
space or by noting relevant pairs.
Let X be normally distributed with mean 100 cm and standard deviation 5 cm.
46a. On the diagram below, shade the region representing
.
[2 marks]
Markscheme
A1A1
N2
Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.
Examiners report
Most candidates did very well on part (a), shading the area under the normal curve.
46b. Given that
, find the value of .
[2 marks]
Markscheme
evidence of recognizing symmetry
e.g.
(M1)
is one standard deviation above the mean so
A1
is one standard deviation below the mean, shading the corresponding part,
N2
[2 marks]
Examiners report
Not all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and the
information given. Some of them saw the need to use tables and others just left it blank.
Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Many
candidates resorted to formulae or tables instead of reasoning through the question.
46c. Given that
(correct to two significant figures), find
.
[2 marks]
Markscheme
evidence of using complement
e.g.
(M1)
,
A1
N2
[2 marks]
Examiners report
Not all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and the
information given. Some of them saw the need to use tables and others just left it blank.
Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Many
candidates resorted to formulae or tables instead of reasoning through the question.
In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.
47a. One boy is selected at random.
(i)
[4 marks]
Find the probability that he plays only one sport.
(ii) Given that the boy selected plays only one sport, find the probability that he plays rugby.
Markscheme
(i) METHOD 1
(M1)
evidence of using complement, Venn diagram
e.g.
,
A1
N2
METHOD 2
attempt to find P(only one sport) , Venn diagram
(M1)
e.g.
A1
(ii)
N2
A2
N2
[4 marks]
Examiners report
Overall, this question was very well done. There were some problems with the calculation of conditional probability, where a
considerable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind of
question where it can be seen if the concept is not clear to candidates.
47b. Let A be the event that a boy plays football and B be the event that a boy plays rugby.
[2 marks]
Explain why A and B are not mutually exclusive.
Markscheme
valid reason in words or symbols
e.g.
(R1)
if mutually exclusive,
correct statement in words or symbols
e.g.
,
if not mutually exclusive
A1
N2
,
, some students play both sports, sets intersect
[2 marks]
Examiners report
In part (c), candidates were generally able to explain in words why events were mutually exclusive, though many gave the wrong
values for P(A) and P(B).
47c. Show that A and B are not independent.
[3 marks]
Markscheme
(R1)
valid reason for independence
e.g.
,
A1A1
correct substitution
e.g.
N3
,
[3 marks]
Examiners report
There was a great amount of confusion between the concepts of independent and mutually exclusive events. In part (d), the
explanations often referred to mutually exclusive events.
It was evident that candidates need more practice with questions like (c) and (d).
Some students equated probabilities and number of elements, giving probabilities greater than 1.
A multiple choice test consists of ten questions. Each question has five answers. Only one of the answers is correct. For each question, Jose
randomly chooses one of the five answers.
48a. Find the expected number of questions Jose answers correctly.
[1 mark]
Markscheme
A1
N1
[1 mark]
Examiners report
Most candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewer
were able to use the GDC effectively.
48b. Find the probability that Jose answers exactly three questions correctly.
[2 marks]
Markscheme
evidence of appropriate approach involving binomial
e.g.
,
(M1)
,
A1
N2
[2 marks]
Examiners report
Most candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewer
were able to use the GDC effectively.
48c. Find the probability that Jose answers more than three questions correctly.
[3 marks]
Markscheme
METHOD 1
(A1)
evidence of using the complement (seen anywhere)
e.g.
(M1)
any probability ,
A1
N2
METHOD 2
(M1)
recognizing that
e.g. summing probabilities from
correct expression or values
to
(A1)
e.g.
A1
N2
[3 marks]
Examiners report
Part (c) was problematic in some cases but most candidates recognized that either a sum of probabilities or the complement was
required. Many misinterpreted "more than three" as inclusive of three, and so obtained incorrect answers. When adding individual
probabilities, some candidates used three or fewer significant figures, which resulted in an incorrect final answer due to premature
rounding.
49. Consider the independent events A and B . Given that
, and
, find
.
[7 marks]
Markscheme
METHOD 1
(R1)
for independence
expression for
(A1)
, indicating
e.g.
,
(M1)
substituting into
A1
correct substitution
e.g.
,
correct solutions to the equation
e.g.
,
(A2)
(accept the single answer
A1
)
N6
[7 marks]
METHOD 2
(R1)
for independence
expression for
(A1)
, indicating
e.g.
,
(M1)
substituting into
correct substitution
A1
e.g.
,
correct solutions to the equation
(A2)
e.g. 0.4, 2.6 (accept the single answer 0.4)
(accept
if x set up as
)
A1
N6
[7 marks]
Examiners report
Many candidates confused the concept of independence of events with mutual exclusivity, mistakenly trying to use the formula
. Those who did recognize that
were often able to find the correct equation, but
many were unable to use their GDC to solve it. A few provided two answers without discarding the value greater than one.
The letters of the word PROBABILITY are written on 11 cards as shown below.
Two cards are drawn at random without replacement.
Let A be the event the first card drawn is the letter A.
Let B be the event the second card drawn is the letter B.
50a. Find
.
[1 mark]
Markscheme
A1
[1 mark]
N1
Examiners report
Most candidates answered part (a) correctly.
50b. Find
.
[2 marks]
Markscheme
A2
N2
[2 marks]
Examiners report
Few candidates used the concept of "B given A" to simply "write down" the answer of
the booklet, with which few were successful.
50c. Find
.
. Instead, most reached for the formula in
[3 marks]
Markscheme
(M1)
recognising that
correct values
(A1)
e.g.
A1
N3
[3 marks]
Examiners report
Few also made the connection that part (c) could be answered using both previous answers. Many found
correctly even
when answering part (b) incorrectly, although some candidates did not decrease the denominator for the second event.
A random variable X is distributed normally with mean 450 and standard deviation 20.
51a. Find
.
[2 marks]
Markscheme
(M1)
evidence of attempt to find
e.g.
A1
N2
[2 marks]
Examiners report
It remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause for
concern. Most candidates had been taught the topic but many had difficulty understanding the difference between ,
, and
Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates often
worked with the wrong tail and did not write their answers correct to 3 significant figures.
.
51b. Given that
, find .
[4 marks]
Markscheme
evidence of using the complement
(M1)
e.g. 0.73,
(A1)
setting up equation
(M1)
e.g.
A1
N3
[4 marks]
Examiners report
It remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause for
concern. Most candidates had been taught the topic but many had difficulty understanding the difference between ,
, and
Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates often
worked with the wrong tail and did not write their answers correct to 3 significant figures.
.
Many candidates had trouble with part (b), a majority never found the complement, instead using their GDCs to calculate the result,
which many times was finding a for
instead of for
. Many others substituted the values of
or
into the equation, instead of the -scores.
In any given season, a soccer team plays 65 % of their games at home.
When the team plays at home, they win 83 % of their games.
When they play away from home, they win 26 % of their games.
The team plays one game.
52a. Find the probability that the team wins the game.
[4 marks]
Markscheme
appropriate approach
(M1)
e.g. tree diagram or a table
(M1)
A1
(or 0.631)
A1
N2
[4 marks]
Examiners report
Part (a) was nearly always correctly answered by those who attempted the question, but part (b) (conditional probability) was poorly
done. A surprisingly small number of students drew a tree diagram in part (a) and those who did answered this part and part (b) well.
Many found the correct complement in part (b) but could not make any further progress.
52b. If the team does not win the game, find the probability that the game was played at home.
[4 marks]
Markscheme
evidence of using complement
e.g.
(M1)
, 0.3695
choosing a formula for conditional probability
(M1)
e.g.
correct substitution
A1
e.g.
A1
P(home) = 0.299
N3
[4 marks]
Examiners report
Part (b) (conditional probability) was poorly done. A surprisingly small number of students drew a tree diagram in part (a) and those
who did answered this part and part (b) well. Many found the correct complement in part (b) but could not make any further progress.
A fisherman catches 200 fish to sell. He measures the lengths, l cm of these fish, and the results are shown in the frequency table below.
53a. Calculate an estimate for the standard deviation of the lengths of the fish.
[3 marks]
Markscheme
evidence of using mid-interval values (5, 15, 25, 35, 50, 67.5, 87.5)
(cm)
A2
(M1)
N3
[3 marks]
Examiners report
Part (a) defeated the vast majority of candidates who clearly had not been taught data entry. Some schools had attempted to teach how
to use a formula rather than the GDC to find the standard deviation and their students invariably used this formula incorrectly. Use of
the GDC was not only expected but should be emphasized as stated in the syllabus.
53b. A cumulative frequency diagram is given below for the lengths of the fish.
[6 marks]
Use the graph to answer the following.
(i)
Estimate the interquartile range.
(ii) Given that
of the fish have a length more than
, find the value of k.
Markscheme
(i)
(A1)(A1)
,
(accept any notation that suggests the interval 15 to 40)
A1
N3
(ii) METHOD 1
have a length less than k
(A1)
(A1)
(cm)
A1
N2
METHOD 2
(A1)
(A1)
(cm)
A1
N2
[6 marks]
Examiners report
Part (b) revealed poor understanding of cumulative frequency and the IQR was often reported as an interval.
53c. In order to sell the fish, the fisherman classifies them as small, medium or large.
Small fish have a length less than
[2 marks]
.
Medium fish have a length greater than or equal to
Large fish have a length greater than or equal to
but less than
.
.
Write down the probability that a fish is small.
Markscheme
(M1)
A1
N2
[2 marks]
Examiners report
This was generally answered well although a number of candidates had difficulty with using the formula for expected value.
53d. The cost of a small fish is
, a medium fish
, and a large fish
.
[2 marks]
Copy and complete the following table, which gives a probability distribution for the cost
.
Markscheme
A1A1
N2
[2 marks]
Examiners report
This was generally answered well although a number of candidates had difficulty with using the formula for expected value.
53e. Find
.
[2 marks]
Markscheme
correct substitution (of their p values) into formula for
(A1)
e.g.
(accept
)
A1
N2
[2 marks]
Examiners report
This was generally answered well although a number of candidates had difficulty with using the formula for expected value.
Two boxes contain numbered cards as shown below.
Two cards are drawn at random, one from each box.
54a. Copy and complete the table below to show all nine equally likely outcomes.
[2 marks]
Markscheme
A2
N2
[2 marks]
Examiners report
Most candidates completed parts (a), (b) and (c) successfully.
54b. Let S be the sum of the numbers on the two cards.
[2 marks]
Find the probability of each value of S.
Markscheme
,
,
,
A2
N2
[2 marks]
Examiners report
Most candidates completed part (b) successfully.
54c. Find the expected value of S.
[3 marks]
Examiners report
Many found the expected value correctly, while some showed difficulty with the arithmetic.
54d. Anna plays a game where she wins
if S is even and loses
if S is odd.
[3 marks]
Anna plays the game 36 times. Find the amount she expects to have at the end of the 36 games.
Markscheme
METHOD 1
(A1)
correct expression for expected gain E(A) for 1 game
e.g.
amount at end = expected gain for 1 game
= 200 (dollars)
A1
(M1)
N2
METHOD 2
attempt to find expected number of wins and losses
e.g.
(M1)
,
attempt to find expected gain E(G)
(M1)
e.g.
(dollars)
A1
N2
[3 marks]
Examiners report
This was often left blank or only superficially attempted. Some found the expected value
amount of money.
but did not answer the question about the
The following diagram is a box and whisker plot for a set of data.
The interquartile range is 20 and the range is 40.
55a. Write down the median value.
[1 mark]
Markscheme
18
A1
N1
[1 mark]
Examiners report
Most candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this question
correctly.
55b. Find the value of
(i)
;
(ii)
.
[4 marks]
Markscheme
(i) 10
A2
N2
(ii) 44
A2
N2
[4 marks]
Examiners report
Most candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this question
correctly.
A van can take either Route A or Route B for a particular journey.
If Route A is taken, the journey time may be assumed to be normally distributed with mean 46 minutes and a standard deviation 10 minutes.
If Route B is taken, the journey time may be assumed to be normally distributed with mean
56a. For Route A, find the probability that the journey takes more than
minutes and standard deviation 12 minutes.
minutes.
[2 marks]
Markscheme
A2
N2
[2 marks]
Examiners report
A significant number of students clearly understood what was asked in part (a) and used the GDC to find the result.
56b. For Route B, the probability that the journey takes less than
Find the value of
minutes is
.
[3 marks]
.
Markscheme
correct approach
(A1)
e.g.
, sketch
(A1)
A1
N2
[3 marks]
Examiners report
In part (b), many candidates set the standardized formula equal to the probability (
Other candidates used the solver on their GDC with the inverse norm function.
), instead of using the corresponding z-score.
56c. The van sets out at 06:00 and needs to arrive before 07:00.
(i)
[3 marks]
Which route should it take?
(ii) Justify your answer.
Markscheme
A1
(i) route A
N1
(ii) METHOD 1
A1
valid reason
R1
e.g. probability of A getting there on time is greater than probability of B
N2
METHOD 2
A1
valid reason
R1
e.g. probability of A getting there late is less than probability of B
N2
[3 marks]
Examiners report
A common incorrect approach in part (c) was to attempt to use the means and standard deviations for justification, although many
candidates successfully considered probabilities.
56d. On five consecutive days the van sets out at 06:00 and takes Route B. Find the probability that
(i)
it arrives before 07:00 on all five days;
(ii) it arrives before 07:00 on at least three days.
Markscheme
(i) let X be the number of days when the van arrives before 07:00
(A1)
A1
N2
(ii) METHOD 1
(M1)
evidence of adding correct probabilities
e.g.
(A1)
correct values
A1
N3
METHOD 2
evidence of using the complement
e.g.
,
(A1)
correct values
A1
[5 marks]
N3
(M1)
[5 marks]
Examiners report
A pleasing number of candidates recognized the binomial probability and made progress on part (d).
Let A and B be independent events, where
and
57a. Write down an expression for
.
.
[1 mark]
Markscheme
A1
N1
[1 mark]
Examiners report
This question was well done by most candidates.
57b. Given that
(i)
,
[4 marks]
find x ;
(ii) find
.
Markscheme
(M1)
(i) evidence of using
correct substitution
e.g.
A1
,
A1
(ii)
N2
A1
N1
[4 marks]
Examiners report
This question was well done by most candidates. When errors were made, candidates confused the terms "independent" and
"mutually exclusive" and did not subtract the intersection when finding
.
57c. Hence, explain why A and B are not mutually exclusive.
[1 mark]
Markscheme
valid reason, with reference to
R1
N1
e.g.
[1 mark]
Examiners report
Candidates should also be aware of the command term "hence" used in part (c) where they were expected to provide a reason that
involved
from their work in part (b). It seemed that many turned to the formula in the booklet instead of considering the
conceptual meaning of the term.
The following is a cumulative frequency diagram for the time t, in minutes, taken by 80 students to complete a task.
58a. Write down the median.
[1 mark]
Markscheme
median
A1
N1
[1 mark]
Examiners report
This question was answered successfully by a majority of candidates. A common error was to use values of 20 and 60 for the lower
and upper quartiles. Some were careless when reading the graph scale and wrote incorrect answers as a result.
58b. Find the interquartile range.
[3 marks]
Markscheme
lower quartile
, upper quartile
A1
(A1)(A1)
N3
[3 marks]
Examiners report
This question was answered successfully by a majority of candidates. A common error was to use values of 20 and 60 for the lower
and upper quartiles. Some were careless when reading the graph scale and wrote incorrect answers as a result.
58c. Complete the frequency table below.
[2 marks]
Markscheme
A1A1
N2
[2 marks]
Examiners report
This question was answered successfully by a majority of candidates. A common error was to use values of 20 and 60 for the lower
and upper quartiles. Some were careless when reading the graph scale and wrote incorrect answers as a result.
The probability of obtaining heads on a biased coin is 0.18. The coin is tossed seven times.
59a. Find the probability of obtaining exactly two heads.
[2 marks]
Markscheme
evidence of using binomial probability
(M1)
e.g.
A1
N2
[2 marks]
Examiners report
Candidates who recognized binomial probability answered this question very well, using their GDC to perform the final calculations.
59b. Find the probability of obtaining at least two heads.
[3 marks]
Markscheme
METHOD 1
evidence of using the complement
M1
e.g.
(A1)
A1
N2
METHOD 2
evidence of attempting to sum probabilities
e.g.
M1
,
correct values for each probability
(A1)
e.g.
A1
N2
[3 marks]
Examiners report
Some candidates misinterpreted the meaning of "at least two" in part (b), and instead found
. Others wrote down a correct
interpretation but accumulated to in their GDC (e.g. binomcdf (7, 0.18, 2)). Still, the number of candidates who either left this
question blank or approached the question without binomial considerations suggests that this topic continues to be neglected in some
centres.
The scores of a test given to students are normally distributed with a mean of 21.
60a. Find the standard deviation of the scores.
of the students have scores less than 23.7.
[3 marks]
Markscheme
evidence of approach
(M1)
e.g. finding
, using
correct working
(A1)
e.g.
, graph
A1
[3 marks]
Examiners report
Candidates who clearly understood the nature of normal probability answered this question cleanly. A common misunderstanding
was to use the value of 0.8 as a z-score when finding the standard deviation.
60b. A student is chosen at random. This student has the same probability of having a score less than 25.4 as having a score greater [4 marks]
than b.
(i)
Find the probability the student has a score less than 25.4.
(ii) Find the value of b.
Markscheme
(M1)
(i) evidence of attempting to find
e.g. using
A1
N2
(M1)
(ii) evidence of recognizing symmetry
e.g.
, using
A1
N2
[4 marks]
Examiners report
Many correctly used their GDC to find the probability in part (b). Fewer used some aspect of the symmetry of the curve to find a
value for b.
In a school with 125 girls, each student is tested to see how many sit-up exercises (sit-ups) she can do in one minute. The results are given in
the table below.
61a. (i)
Write down the value of p.
[3 marks]
(ii) Find the value of q.
Markscheme
A1
(i)
N1
(ii) for evidence of using sum is 125 (or
A1
)
(M1)
N2
[3 marks]
Examiners report
Part (a) of this question was well done.
61b. Find the median number of sit-ups.
[2 marks]
Markscheme
evidence of median position
(M1)
e.g. 63rd student,
A1
median is 17 (sit-ups)
N2
[2 marks]
Examiners report
Finding the median seemed to be the most difficult for the candidates. Most had the idea that it was in the middle but did not know
how to find the value.
61c. Find the mean number of sit-ups.
[2 marks]
Markscheme
evidence of substituting into
(M1)
e.g.
,
A1
mean
N2
[2 marks]
Examiners report
When calculating the mean, many ignored the frequencies.
A factory makes switches. The probability that a switch is defective is 0.04. The factory tests a random sample of 100 switches.
62a. Find the mean number of defective switches in the sample.
[2 marks]
Markscheme
evidence of binomial distribution (may be seen in parts (b) or (c))
(M1)
e.g. np,
A1
N2
[2 marks]
Examiners report
Part (a) was handled well by most students.
62b. Find the probability that there are exactly six defective switches in the sample.
Markscheme
(A1)
A1
[2 marks]
N2
[2 marks]
Examiners report
Although this question was a rather straightforward question on binomial distribution, parts (b) and(c) seemed to cause much
difficulty.
62c. Find the probability that there is at least one defective switch in the sample.
[3 marks]
Markscheme
for evidence of appropriate approach
(M1)
e.g. complement,
(A1)
A1
N2
[3 marks]
Examiners report
Although this question was a rather straightforward question on binomial distribution, parts (b) and(c) seemed to cause much
difficulty. In part (c), finding at least one defective switch, many forgot to take the complement.
A box contains a large number of biscuits. The weights of biscuits are normally distributed with mean
and standard deviation
63a. One biscuit is chosen at random from the box. Find the probability that this biscuit
(i)
weighs less than
(ii) weighs between
;
and
.
Markscheme
(i)
(M1)
A1
(ii) evidence of appropriate approach
N2
(M1)
e.g. symmetry,
(tables 0.955)
A1
N2
Note: Award M1A1(AP) if candidates refer to 2 standard deviations from the mean, leading to 0.95.
[4 marks]
Examiners report
Those that understood the normal distribution did well on parts (a) and (bi).
.
[4 marks]
63b. Five percent of the biscuits in the box weigh less than d grams.
[5 marks]
(i) Copy and complete the following normal distribution diagram, to represent this information, by indicating d, and shading the
appropriate region.
(ii) Find the value of d.
Markscheme
(i)
A1A1
N2
Note: Award A1 for d to the left of the mean, A1 for area to the left of d shaded.
(A1)
(ii)
(M1)
A1
N3
[5 marks]
Examiners report
Those that understood the normal distribution did well on parts (a) and (bi). Parts (bii) and (c) proved to be a little more difficult. In
particular, in part (bii) the z-score was incorrectly set equal to 0.05 and in part (c), 0.2 was used instead of the z-score. For those who
had a good grasp of the concept of normal distributions the entire question was quite accessible and full marks were gained.
63c. The weights of biscuits in another box are normally distributed with mean
of the biscuits in this second box weight less than
Find the value of
and standard deviation
. It is known that
[4 marks]
.
.
Markscheme
(M1)
A1
(M1)
A1
N3
[4 marks]
Examiners report
Those that understood the normal distribution did well on parts (a) and (bi). Parts (bii) and (c) proved to be a little more difficult. In
particular, in part (bii) the z-score was incorrectly set equal to 0.05 and in part (c), 0.2 was used instead of the z-score. For those who
had a good grasp of the concept of normal distributions the entire question was quite accessible and full marks were gained.
A box contains 100 cards. Each card has a number between one and six written on it. The following table shows the frequencies for each
number.
64a. Calculate the value of k.
[2 marks]
Markscheme
(M1)
evidence of using
A1
N2
[2 marks]
Examiners report
Frequencies and median seemed well understood, but quartiles and inter-quartile range less so.
64b. Find
(i)
[5 marks]
the median;
(ii) the interquartile range.
Markscheme
(i) evidence of median position
(M1)
e.g. 50th item,
A1
(ii)
and
N2
(A1)(A1)
(accept 1 to 5 or
, etc.)
A1
N3
[5 marks]
Examiners report
Frequencies and median seemed well-understood, but quartiles and interquartile range less so. A few students, probably based on past
papers, drew cumulative frequency diagrams, generating slightly different answers for median and quartiles.
There are 20 students in a classroom. Each student plays only one sport. The table below gives their sport and gender.
65a. One student is selected at random.
(i)
Calculate the probability that the student is a male or is a tennis player.
(ii) Given that the student selected is female, calculate the probability that the student does not play football.
[4 marks]
Markscheme
(A1)
(i) correct calculation
e.g.
,
A1
(A1)
(ii) correct calculation
e.g.
N2
,
A1
N2
[4 marks]
Examiners report
Many candidates had difficulty with this question, usually as a result of seeking to solve the problem by formula instead of looking
carefully at the table frequencies.
65b. Two students are selected at random. Calculate the probability that neither student plays football.
[3 marks]
Markscheme
A1
,
A1
A1
N1
[3 marks]
Examiners report
A very common error in part (b) was to assume identical probabilities for each selection instead of dependent probabilities where
there is no replacement.
A four-sided die has three blue faces and one red face. The die is rolled.
Let B be the event a blue face lands down, and R be the event a red face lands down.
66a. Write down
[2 marks]
(i) P(B);
(ii) P(R).
Markscheme
(i) P(B)
A1
N1
(ii) P(R)
A1
N1
[2 marks]
Examiners report
This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,
with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.
Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to
1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most
challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of
winning.
66b. If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is
[2 marks]
represented by the following tree diagram, where p, s, t are probabilities.
Find the value of p, of s and of t.
Markscheme
A1
N1
A1
,
N1
[2 marks]
Examiners report
This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,
with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.
Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to
1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most
challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of
winning.
66c. Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face lands
[3 marks]
down, he scores 1 and rolls one more time. Let X be the total score obtained.
(i) Show that
(ii) Find
.
.
Markscheme
(i)
A1
AG
N0
(A1)
(ii)
A1
N2
[3 marks]
Examiners report
This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,
with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.
Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to
1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most
challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of
winning.
66d. (i)
Construct a probability distribution table for X.
[5 marks]
(ii) Calculate the expected value of X.
Markscheme
(i)
A2
N2
(M1)
(ii) evidence of using
(A1)
A1
N2
[5 marks]
Examiners report
This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,
with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.
Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to
1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most
challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of
winning.
66e. If the total score is 3, Guiseppi wins
. If the total score is 2, Guiseppi gets nothing.
Guiseppi plays the game twice. Find the probability that he wins exactly
[4 marks]
.
Markscheme
win
(M1)
scores 3 one time, 2 other time
(seen anywhere)
A1
evidence of recognising there are different ways of winning
e.g.
,
A1
(M1)
,
N3
[4 marks]
Examiners report
This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,
with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.
Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to
1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most
challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of
winning.
The following table shows the probability distribution of a discrete random variable X.
67a. Find the value of k.
[4 marks]
Markscheme
(M1)
evidence of using
correct substitution
A1
e.g.
,
A2
N2
[4 marks]
Examiners report
A good number of candidates answered this question well, although some incorrectly set the sum of the probabilities to zero instead
of one, suggesting rote recognition of a quadratic equal to zero. Many candidates recognized that only the positive value for k was
appropriate and correctly indicated this in their working. Many went on to find the correct expected value as well, although at times
candidates wrote the formula from the information booklet without making use of it, thus earning no marks.
67b. Find the expected value of X.
[3 marks]
Markscheme
(M1)
evidence of using
correct substitution
(A1)
e.g.
A1
N2
[3 marks]
Examiners report
A good number of candidates answered this question well, although some incorrectly set the sum of the probabilities to zero instead
of one, suggesting rote recognition of a quadratic equal to zero. Many candidates recognized that only the positive value for k was
appropriate and correctly indicated this in their working. Many went on to find the correct expected value as well, although at times
candidates wrote the formula from the information booklet without making use of it, thus earning no marks.
The heights of certain plants are normally distributed. The plants are classified into three categories.
The shortest
are in category A.
The tallest
are in category C.
All the other plants are in category B with heights between
and
.
68a. Complete the following diagram to represent this information.
[2 marks]
Markscheme
A1A1
N2
Notes: Award A1 for three regions (may be shown by lines or shading), A1 for clear labelling of two regions (may be shown by
percentages or categories). r and t need not be labelled, but if they are, they may be interchanged.
[2 marks]
Examiners report
Many candidates shaded or otherwise correctly labelled the appropriate regions in the normal curve.
68b. Given that the mean height is
and the standard deviation
Markscheme
METHOD 1
(A1)
A1
N2
(= 0.8962) (may be seen later)
A1
(A1)
A1
N2
METHOD 2
A1A1
finding z-values
evidence of setting up one standardised equation
e.g.
,
,
[5 marks]
A1A1
N2N2
(M1)
, find the value of r and of t.
[5 marks]
Examiners report
Although many candidates shaded or otherwise correctly labelled the appropriate regions in the normal curve, far fewer could apply
techniques of normal probabilities to achieve correct results in part (b). Many set the standardized formula equal to the probabilities
instead of the appropriate z-scores, which can be found either by the use of tables or the GDC. Others simply left this part blank,
which suggests a lack of preparation for such “inverse” types of questions in a normal distribution.
Paula goes to work three days a week. On any day, the probability that she goes on a red bus is .
69a. Write down the expected number of times that Paula goes to work on a red bus in one week.
[2 marks]
Markscheme
evidence of binomial distribution (seen anywhere)
(M1)
e.g.
(
)
A1
N2
[2 marks]
Examiners report
Many candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.
Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDC
effectively often answered correctly.
69b. In one week, find the probability that she goes to work on a red bus on exactly two days.
[2 marks]
Markscheme
(A1)
A1
N2
[2 marks]
Examiners report
Many candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.
Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDC
effectively often answered correctly.
69c. In one week, find the probability that she goes to work on a red bus on at least one day.
Markscheme
evidence of appropriate approach
e.g. complement,
M1
, adding probabilities
(A1)
A1
[3 marks]
N2
[3 marks]
Examiners report
Many candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.
Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDC
effectively often answered correctly, although in part (c) some candidates misinterpreted the meaning of “at least one” and found
either
or
.
There are nine books on a shelf. For each book, x is the number of pages, and y is the selling price in pounds (£). Let r be the correlation
coefficient.
70a. Write down the possible minimum and maximum values of r .
[2 marks]
Markscheme
min value of r is
, max value of r is 1
A1A1
N2
[2 marks]
Examiners report
[N/A]
70b. Given that
, which of the following diagrams best represents the data.
Markscheme
C
A1
N1
[1 mark]
Examiners report
[N/A]
[1 mark]
70c.
[2 marks]
For the data in diagram D , which two of the following expressions describe the correlation between x and y ?
perfect, zero, linear, strong positive, strong negative, weak positive, weak negative
Markscheme
linear, strong negative
A1A1
N2
[2 marks]
Examiners report
[N/A]
A data set has a mean of 20 and a standard deviation of 6.
71a. Each value in the data set has 10 added to it. Write down the value of
(i)
[2 marks]
the new mean;
(ii) the new standard deviation.
Markscheme
A1
(i) new mean is
(ii) new sd is 6
A1
N1
N1
[2 marks]
Examiners report
[N/A]
71b. Each value in the original data set is multiplied by 10.
(i)
Write down the value of the new mean.
(ii) Find the value of the new variance.
[3 marks]
Markscheme
A1
(i) new mean is
N1
(ii) METHOD 1
variance is 36
A1
A1
new variance is
N2
METHOD 2
new sd is 60
A1
A1
new variance is
N2
[3 marks]
Examiners report
[N/A]
Two standard six-sided dice are tossed. A diagram representing the sample space is shown below.
Let
be the sum of the scores on the two dice.
72a. (i)
Find
(ii) Find
.
[6 marks]
.
(iii) Find
.
Markscheme
(i) number of ways of getting
A1
is 5
N2
(ii) number of ways of getting
A1
(iii)
is 21
A1
N2
A2
[6 marks]
Examiners report
[N/A]
A1
N2
72b. Elena plays a game where she tosses two dice.
[8 marks]
If the sum is 6, she wins 3 points.
If the sum is greater than 6, she wins 1 point.
If the sum is less than 6, she loses k points.
Find the value of k for which the game is fair.
Markscheme
M1
attempt to find
e.g.
A1
fair game if
(may be seen anywhere)
attempt to substitute into
formula
R1
M1
e.g.
A1
correct substitution into
e.g.
work towards solving
M1
e.g.
A1
A1
N4
[8 marks]
Examiners report
[N/A]
A random variable X is distributed normally with mean 450. It is known that
73a. Represent all this information on the following diagram.
.
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for 450 , A1 for a to the right of the mean, A1 for area 0.27 .
[3 marks]
Examiners report
[N/A]
73b. Given that the standard deviation is 20, find a . Give your answer correct to the nearest whole number.
[3 marks]
Markscheme
M1
valid approach
e.g.
, 0.73
A1
A1
N3
[3 marks]
Examiners report
[N/A]
The probability of obtaining heads on a biased coin is 0.4. The coin is tossed 600 times.
74a. (i)
Write down the mean number of heads.
[4 marks]
(ii) Find the standard deviation of the number of heads.
Markscheme
(i) recognizing binomial with
A1
,
M1
N2
(ii) correct substitution into formula for variance or standard deviation
A1
e.g. 144,
sd = 12
A1
N1
[4 marks]
Examiners report
[N/A]
74b. Find the probability that the number of heads obtained is less than one standard deviation away from the mean.
[3 marks]
Markscheme
M1
attempt to find range of values
e.g.
A1
evidence of correct approach
e.g.
A1
N2
[3 marks]
Examiners report
[N/A]
Each day, a factory recorded the number ( ) of boxes it produces and the total production cost ( ) dollars. The results for nine days are
shown in the following table.
75a. Write down the equation of the regression line of y on x .
[2 marks]
Markscheme
A1A1
N2
[2 marks]
Examiners report
[N/A]
75b. Use your regression line from part (a) as a model to answer the following.
[2 marks]
Interpret the meaning of
(i)
the gradient;
(ii) the y-intercept.
Markscheme
(i) additional cost per box (unit cost)
(ii) fixed costs
A1
A1
N1
N1
[2 marks]
Examiners report
[N/A]
75c. Estimate the cost of producing 60 boxes.
[2 marks]
Markscheme
attempt to substitute into regression equation
e.g.
M1
,
(accept
from 3 s.f. values)
A1
N2
[2 marks]
Examiners report
[N/A]
75d. The factory sells the boxes for $19.99 each. Find the least number of boxes that the factory should produce in one day in order[3 marks]
to make a profit.
Markscheme
setting up inequality (accept equation)
M1
e.g.
A1
13 boxes (accept 14 from
, using 3 s.f. values)
A1
N2
Note: Exception to the FT rule: if working shown, award the final A1 for a correct integer solution for their value of x.
[3 marks]
Examiners report
[N/A]
75e. Comment on the appropriateness of using your model to
(i)
[4 marks]
estimate the cost of producing 5000 boxes;
(ii) estimate the number of boxes produced when the total production cost is $540.
Markscheme
(i) this would be extrapolation, not appropriate
R1R1
N2
(ii) this regression line cannot predict x from y, not appropriate
R1R1
N2
[4 marks]
Examiners report
[N/A]
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