IB Math – High Level Math – Probability Continuous RVs
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Continuous Random Variable Practice
1.
A factory has a machine designed to produce 1 kg bags of sugar. It is found that the average weight
of sugar in the bags is 1.02 kg. Assuming that the weights of the bags are normally distributed, find
the standard deviation if 1.7% of the bags weigh below 1 kg.
Give your answer correct to the nearest 0.1 gram.
Working:
Answer:
…………………………………………..
(Total 4 marks)
2.
The random variable X is distributed normally with mean 30 and standard deviation 2.
Find p(27  X  34).
Working:
Answer:
....……………………………………..........
3.
A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of 110
g and standard deviation of 1.142 g. If the weight of a bag of salt is less than 108 g, the bag is
rejected. With these settings, 4% of the bags are rejected.
The settings of the machine are altered and it is found that 7% of the bags are rejected.
(a) (i)
If the mean has not changed, find the new standard deviation, correct to three decimal
places.
(Total 4 marks)
(4)
The machine is adjusted to operate with this new value of the standard deviation.
(ii) Find the value, correct to two decimal places, at which the mean should be set so that
only 4% of the bags are rejected.
(4)
(b)
With the new settings from part (a), it is found that 80% of the bags of salt have a weight
which lies between A g and B g, where A and B are symmetric about the mean. Find the values
of A and B, giving your answers correct to two decimal places.
(4)
(Total 12 marks)
4.
The lifetime of a particular component of a solar cell is Y years, where Y is a continuous random
variable with probability density function
0 when y  0

f ( y)  
- y/2
when y  0.
0.5e
(a) Find the probability, correct to four significant figures, that a given component fails within six
months.
(3)
Each solar cell has three components which work independently and the cell will continue to run if at
least two of the components continue to work.
(b) Find the probability that a solar cell fails within six months.
(4)
(Total 7 marks)
5.
The diameters of discs produced by a machine are normally distributed with a mean of 10 cm and
standard deviation of 0.1 cm. Find the probability of the machine producing a disc with a diameter
smaller than 9.8 cm.
Working:
Answer:
…………………………………………..
(Total 3 marks)
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IB Math – High Level Math – Probability Continuous RVs
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6.
Z is the standardized normal random variable with mean 0 and variance 1. Find the value of a such
that P(  Z   a) = 0.75.
Working:
Answer:
..........................................................................
7.
A continuous random variable X has probability density function
4

, for 0  x  1,

f ( x)    (1  x 2 )

0,
elsewhere

(Total 3 marks)
Find E(X).
Working:
Answer:
..........................................................................
(Total 3 marks)
8.
The weights of a certain species of bird are normally distributed with mean 0.8 kg and standard
deviation 0.12 kg. Find the probability that the weight of a randomly chosen bird of the species lies
between 0.74 kg and 0.95 kg.
Working:
Answer:
..........................................................................
9.
The probability density function f (x), of a continuous random variable X is defined by
1
 x(4 – x 2 ), 0  x  2
f (x) =  4

0, otherwise.

Calculate the median value of X.
Working:
Answer:
..........................................................................
10.
(a)
At a building site the probability, P(A), that all materials arrive on time is 0.85. The
probability, P(B), that the building will be completed on time is 0.60. The probability that the
materials arrive on time and that the building is completed on time is 0.55.
(i)
Show that events A and B are not independent.
(ii) All the materials arrive on time. Find the probability that the building will not be
completed on time.
(b)
There was a team of ten people working on the building, including three electricians and two
plumbers. The architect called a meeting with five of the team, and randomly selected people
to attend. Calculate the probability that exactly two electricians and one plumber were called
to the meeting.
(c)
The number of hours a week the people in the team work is normally distributed with a mean
of 42 hours. 10% of the team work 48 hours or more a week. Find the probability that both
plumbers work more than 40 hours in a given week.
(Total 6 marks)
(Total 6 marks)
(5)
(2)
(8)
(Total 15 marks)
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IB Math – High Level Math – Probability Continuous RVs
11.
12.
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The random variable X is normally distributed and
P(X  10) = 0.670
P(X  12) = 0.937.
Find E(X).
Working:
Answer:
.........................................................................
A random variable X is normally distributed with mean  and standard deviation σ, such that
P(X > 50.32) = 0.119, and P(X < 43.56) = 0.305.
(a) Find  and .
(b)
(Total 6 marks)
(5)
Hence find P(|X – | < 5).
(2)
(Total 7 marks)
13.
The following diagram shows the probability density function for the random variable X, which is
normally distributed with mean 250 and standard deviation 50.
f(x)
180
250
280
x
Find the probability represented by the shaded region.
Working:
Answer:
.........................................................................
(Total 6 marks)
14.
The discrete random variable X has the following probability distribution.
k
 , x  1, 2, 3, 4
P(X = x) =  x
0, otherwise
Calculate
(a) the value of the constant k;
(b) E(X).
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
15.
Let f (x) be the probability density function for a random variable X, where
kx 2 , for 0  x  2
f (x) 
0, otherwise
(Total 6 marks)
3
.
8
(a)
Show that k =
(b)
Calculate
(i)
E(X);
(ii) the median of X.
(2)
(6)
(Total 8 marks)
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IB Math – High Level Math – Probability Continuous RVs
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16.
A continuous random variable X has probability density function given by
f (x) = k (2x – x2), for 0  x  2
f (x) = 0,
elsewhere.
(a) Find the value of k.
(b) Find P(0.25  x  0.5).
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
17.
Ian and Karl have been chosen to represent their countries in the Olympic discus throw. Assume that
the distance thrown by each athlete is normally distributed. The mean distance thrown by Ian in the
past year was 60.33 m with a standard deviation of 1.95 m.
(a) In the past year, 80% of Ian’s throws have been longer than x metres.
Find x, correct to two decimal places.
(Total 6 marks)
(3)
(b)
In the past year, 80% of Karl’s throws have been longer than 56.52 m. If the mean distance of
his throws was 59.39 m, find the standard deviation of his throws, correct to two decimal
places.
(3)
(c)
This year, Karl’s throws have a mean of 59.50 m and a standard deviation of 3.00 m. Ian’s
throws still have a mean of 60.33 m and standard deviation 1.95 m. In a competition an athlete
must have at least one throw of 65 m or more in the first round to qualify for the final round.
Each athlete is allowed three throws in the first round.
(i)
Determine which of these two athletes is more likely to qualify for the final on their first
throw.
(ii) Find the probability that both athletes qualify for the final.
(11)
(Total 17 marks)
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IB Math – High Level Math – Probability Continuous RVs Markscheme
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Continuous Random Variable Practice - MarkScheme
1.
Let (z) = 0.017
then (–z) = 1 – 0.017 = 0.983
z = –2.12
x   1  1.02

But z =
where x = 1 kg

Therefore
1  1.02

(M1)
(A1)

= –2.12
(M1)
  = 0.00943 kg = 9.4 g (to the nearest 0.1 g)
(A1) (C4)
[4]
2.
34  30 
 27  30
p(27  X  34) = p 
Z 

2 
 2
= p(–1.5  Z  2)
= 0.433… + 0.477...
= 0.910 (3 sf) (accept 0.911)
(M1)
(A1)
(M1)
(A1) (C4)
[4]
3.
Note: In all 3 parts, award (A2) for correct answers with no working.
Award (M2)(A2) for correct answers with written evidence of the
correct use of a GDC (see GDC examples).
(a) (i)
Let X be the random variable “the weight of a bag of salt”.
Then X ~ N(110, σ2), where σ is the new standard deviation.
X  110
Given P(X < 108) = 0.07, let Z =

108  110 

Then P  Z 
 = 0.07



2
Therefore,
= –1.476

(M1)
(M1)(A1)
Therefore, σ = 1.355.
(A1) 4
GDC Example: Graphing of normal cdf with σ as the variable, and
finding the intersection with p = 0.07.
(ii)
(b)
Let the new mean be µ, then X ~ N(µ, 1.3552).
108   

Then P  Z 
(M1)
 = 0.04
1.355 

108  
Therefore,
= 1.751
(M1)(A1)
1.355
Therefore, µ = 110.37
(A1) 4
GDC Example: Graphing of normal cdf with µ as the variable and
finding intersection with p = 0.04.
If the mean is 110.37 g then X ~ N(110.37, 1.3552), P(A < X < B) = 0.8
Then P(X < A) = 0.1, and P(X < B) = 0.9.
A  110.37
Therefore,
= –1.282
1.355
A = 108.63
B  110.37
Therefore,
= 1.282
1.355
B = 112.11
GDC Example: Graphing of normal cdf with X as the variable and
finding intersection with p = 0.1 and 0.9.
(M1)
(A1)
(M1)
(A1) 4
[12]
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IB Math – High Level Math – Probability Continuous RVs Markscheme
4.
(a)
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Required probability
= P(Y  12 )
=

1/ 2
0
0.5e – y / 2 dy
(M2)
= 0.2212.
OR
(G1)
Required probability =

1/ 2
0

0.5e – y / 2 dy
–y/2
=– e

1/ 2
0
(M1)
(M1)
–1/4
=1–e
= 0.2212 (4 sf)
(b)
Required probability
= P(2 or 3 of the components fail in six months)
 3
=   (0.2212)2(0.7788) + (0.2212)3
 2
= 0.125.
(A1) 3
(M1)
(M2)
(G1) 4
[7]
5.
Let X be the random variable “the diameter of the disc,” then X ~ N(10, 0.12).
X 
Let Z =
, the standardised normal variable (using formulae and

statistical tables).
Then P(X < 9.8) = P(Z < –2)
= 1 – (2)
= 1 – 0.9773
= 0.0227
(using formulae and statistical tables)
OR
Using a graphic display calculator,
P(X < 9.8) = 0.0228
Note: A different answer is obtained if a graphic display calculator
is used
(A1)
(A1)
(A1)
(A3)
[3]
6.
0.75
1–(a)
–a
a
From the diagram 1 – 2(1 – (a)) 0.75
2(a) = 1.75
a = 1.15
(M1)
(A1)
(A1) (C3)
[3]
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IB Math – High Level Math – Probability Continuous RVs Markscheme
7.
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METHOD 1
E(X) =

1
0
4x
dx
π(1  x 2 )
(M1)
= 0.441.
METHOD 2
E(X) =

(G2) (C3)
4x
dx
0 π (1  x 2 )
1
(M1)
2
[ln (1  x 2 )]10
π
2
 ln 4 
= (ln 2) or
.
π
π 

=
(M1)
(A1) (C3)
[3]
8.
 0.95 – 0.8 
 0.74 – 0.8 
 – 

 0.12 
 0.12 
P(0.74 < X < 0.95) =  
=  (1.25) –  (–0.5)
= 0.8944 – (l – 0.6915)
= 0.586
(M1)(A1)
(A1)
(A1)(A1)
(A1)
(C6)
[6]
9.
Let m be the median.
m

Then
0
1
x (4 – x2)dx = 0.5.
4
 4 x – x dx
(M1)
m
=>
3
=2
(A1)
0
1 4 m
x ]0 = 2
4
1
=> 2m2 – m4 = 2
4
=> [2x2 –
(M1)
=> m4 – 8m2 + 8 = 0
m = 1.08
OR
(A1)
(G2)
8  64 – 32 8  32
=
= 4  8 (4  2 2 )
2
2
=> m = 4 – 8  4 – 2 2 


m2 =
(M1)
(A1)
(C6)
Note: Award (C5) if other solutions to the equation
m4 – 8m2 + 8 = 0 appear in the answer box.
10.
(a)
(i)
To be independent P(A  B) = P(A) × P(B)
P(A) × P(B) = (0.85)(0.60)
= 0.51
but P(A  B) = 0.55
P(A  B)  P(A) × P(B)
Hence A and B are not independent.
[6]
(R1)
(A1)
(AG)
(ii)
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IB Math – High Level Math – Probability Continuous RVs Markscheme
A
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B
0.30
0.55
0.05
0.10
P(BA) =
P ( B' A)
P(A)
(M1)
0.30
0.85
6
=
(= 0.353)
17
=
(b)
(M1)
(A1) 5
 3  2  5 
   
2 1 2
Probability of 2 electricians and 1 plumber =    
10 
 
5
=
60  5

 0.238 

252  21

(M1)
(A1)
OR
5!  3  2  2  5  4 
     
2!2!  10  9  8  7  6 
5
=
(= 0.238)
21
Probability of 2 electricians and 1 plumber =
(c)
X = number of hours worked.
X ~ N (42, σ2)
P(X  48) = 0.10
P(X < 48) = 0.90
(z) = 0.90
z = 1.28
(z = 1.28155)
(Answers given to more than 3 significant figures will be accepted.)
z=
X –

=> 1.28 =
48 – 42
(M1)
(A1) 2
(AG)
(M1)
(A1)
(M1)

=>  = 4.69 (Accept  = 4.68)
40 – 42 

P(X > 40) = P  Z 

4.69 

= 0.665
OR
P(X > 40) = 0.665
Therefore, the probability that one plumber works more than 40 hours
per week is 0.665.
The probability that both plumbers work more than 40 hours per week
= (0.665)2
= 0.443 (Accept 0.442 or 0.444)
(A1)
(M1)
(A1)
(G2)
(M1)
(A1) 8
[15]
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IB Math – High Level Math – Probability Continuous RVs Markscheme
11.
Let E(X) = .
From tables, z1 = 0.44, z2 = 1.53
10 =  + 0.44
12 =  + 1.53
=
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(A1)(A1)
(A1)
(A1)
1.53  10 – 0.44  12
1.53 – 0.44
(M1)
 E(X) = 9.19
(A1)
(C6)
[6]
12.
(a)
30.5%
11.9%
43.56
50.32
(z) = 0.305  z = –0.51
and (z) = 0.881  z = 1.18
50.32  
43.56  
= 1.18 and
= –0.51


(b)
13.
Solving simultaneously
 50.32 =  + 1.18σ and 43.56 = µ – 0.51σ
 1.69σ = 6.76
 σ = 4  µ = 45.6
P(X – < 5) = P(40.6 < x < 50.6)
= 0.789
(A1)
(A1)
(M1)
(A1)(A1)
5
(M1)
(G1) 2
[7]
z1 = 0.6  Φ(z1) = 0.7257
z2 = –1.4  Φ(z2) = 0.0808
Probability = 0.7257 – 0.0808 = 0.6449 = 0.645 (3 sf)
(A1)(A1)
(A1)(A1)
(M1)(A1)
(C6)
[6]
14.
(a)
(b)
1 1 1

k 1     = 1
2 3 4

12
k=
(= 0.48)
25
12
6
4
3
 2
 3
 4
E(X) = 1 ×
25
25
25
25
48
=
( = 1.92)
25
(M1)(A1)
(A1)
(C3)
(M1)(A1)
(A1) (C3)
[6]
15.
(a)
k

2
0
x 2 dx = 1
(M1)
2
 x3 
k8
k   
=1
 3 0 3
3
k=
8
(A1)
(AG)
2
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IB Math – High Level Math – Probability Continuous RVs Markscheme
(b)
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2
3  x4 
xx dx (   )
8  4 0
(i)
3
E(X) =
8
(ii)
3
2
The median m must be a number such that
3 m 2
1 3 2 2 
x dx   or
x dx 
8 0
2 8 m


2
0
2
(M1)
=


(A1)
(M1)(A1)
m
 1
3  x3 
3  m3
 0  
   
8  3 0 8  3
 2
(A1)
m3 1
  m3 = 4.
8
2
 m = 3 4 (= 1.59 to 3 sf) .
(A1)
6
[8]
16.
(a)
Since X is a continuous r.v. 
2
 k (2 x – x
0
2
)dx  1
(M1)
2

x3 
 k  x 2 –  1
3 0

(A1)
 8 

 4 –  – 0  1
 3 

3
k 
4
(A1)
(C3)
0.5
(b)
P(0.25  x  0.5) =
 f ( x) dx
(M1)
0.25

29
 0.113
256
(A2)
(C3)
[6]
17.
(a)
(b)
X = length of Ian’s throw.
X  N (60.33, 1.952)
P(X > x) = 0.80  z = –0.8416
x – 60.33
– 0.8416
1.95
x = 58.69 m
Y = length of Karl’s throw.
Y  N (59.39, 2)
P(Y > 56.52) = 0.80  z = –0.8416
56.52 – 59.39
– 0.8416
(A1)
(M1)
(A1) (N3)

 = 3.41 (accept 3.42)
(c)
(i)
3
(A1)
(M1)
(A1) 3
Y  N (59.50, 3.00 )
X  N (60.33, 1.952)
EITHER
P(Y  65) = 0.0334 P(X  65) = 0.00831 (no (AP) here)
OR
65 – 59.50 

P(Y  65)  P  Z 

3.00 

= P(Z  1.833)
2
(A2)(A2)
(M1)
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IB Math – High Level Math – Probability Continuous RVs Markscheme
(ii)
= 0.0334 (accept 0.0336)
(A1)
65 – 60.33 

P(X  65)  P  Z 

1.95 

= P(Z  2.395)
= 0.0083 (accept 0.0084)
(A1)
THEN
Karl is more likely to qualify since P(Y  65)  P(X  65)
(R1)
Note: Award full marks if probabilities are not calculated but the
correct conclusion is reached with the reason 1.833 < 2.395.
If p represents the probability that an athlete throws 65 metres
or more then with 3 throws the probability of qualifying
for the final is
1 – (1 – p)3, or p + (1 – p)p + (1 – p)2p,
or p3 + 3(1 – p)p2 + 3(1 – p)2p
(M1)
3
Therefore P(Ian qualifies) = 1 – (1 – 0.00831)
= 0.0247
P (Karl qualifies) = 1 – (1 – 0.0334)3
= 0.0969
(A1)
Assuming independence
P(both qualify) = (0.0247)(0.0969)
= 0.00239
Note: Depending on use of tables or gdc answers may vary from
0.00239 to 0.00244.
Alei - Desert Academy
(M1)
(A1)
(R1)
(M1)
(A1) 11
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C:\Users\Bob\Documents\Dropbox\IB Curriculum\HL\5 StatProb\TestsQuizzesPractice\HLProb.ContRV.Practice.docx on 03/04/2017 at 11:17 AMPage 7 of 7