COMSATS Institute of Information Technology Islamabad
MTH 161 Introduction to Statistics
Assignment 4 − Solution
Program: BBA
Due Date: Dec 12, 2012
Total Points: 50
Instructor: Shahzad Fazal Bhatti
Note: Solve all the question, however only five randomly selected questions will be graded.
Question # 01: (5 points)
A person is said to be overweight if his or her body mass index (BMI) is between 25 and 29, inclusive; a
person is said to be obese if his or her BMI is 30 or greater. From the document Utah Behavioral Risk Factor
Surveillance System (BRFSS) Local Health District Report, issued by the Utah Department of Health, we
obtained the following table. The first two columns of the table provide an age distribution for adults living in
Utah. The third column gives the percentage of adults in each age group who are either obese or overweight.
Age (yr)
% adults
% overweight
42.5
28.5
16.4
12.6
41.1
57.9
68.2
55.3
18 − 34
35 − 49
50 − 64
65 & over
a. What percentage of Utah adults are overweight or obese?
Let A1 = event that an adult is aged 18 − 34, A2 = event that an adult is aged 35 − 49
A3 = event that an adult is aged 50 − 64, A4 = event that an adult is aged 65 & over
O = event that an adult is obese
P (O) = P (O|A1 )P (A1 ) + P (O|A2 )P (A2 ) + P (O|A3 )P (A3 ) + P (O|A4 )P (A4 )
= (0.411)(0.425) + (0.579)(0.285) + (0.682)(0.164) + (0.553)(0.126) = 0.521
So the percentage obese Utah adults is 52.1%.
b. Of those Utah adults who are between 35 and 49 years old, inclusive, what percentage are overweight or
obese?
The percentage of obese adults aged 35 − 49 is 57.9%.
c. Of those Utah adults who are overweight or obese, what percentage are between 35 and 49 years old, inclusive?
P (A2 |O) =
(0.579)(0.285)
P (O|A2 )P (A2 )
=
= 0.317
P (O)
(0.521)
d. Interpret your answers to parts (a)(c) in terms of percentages.
The percentage of obese adults in Utah is 52.1%.
Of those Utah adults who are between 35 and 49 years old, inclusive, the percentage of adults who are
1
obese is 57.9%.
Of those Utah adults who are overweight or obese, the percentage of obese adults between 35 and 49 years
old, inclusive is 31.7%.
Question # 02: (5 points)
Textbook publishers must estimate the sales of new (first-edition) books. The records of one major publishing
company indicate that 10% of all new books sell more than projected, 30% sell close to the number projected,
and 60% sell less than projected. Of those that sell more than projected, 70% are revised for a second edition,
as are 50% of those that sell close to the number projected and 20% of those that sell less than projected.
a. What percentage of books published by this publishing company go to a second edition?
Let S = event that a book goes to a second edition, M P = event that a book sells more than the projected
CP = event that a book sells close to the projected, LP = event that a book sells less than the projected
P (S) = P (S|M P )P (M P ) + P (S|CP )P (CP ) + P (S|LP )P (LP )
= (0.70)(0.10) + (0.50)(0.30) + (0.20)(0.60) = 0.34
The percentage of books that go to a second edition is 34%.
b. What percentage of books published by this publishing company that go to a second edition sold less than
projected in their first edition?
P (LP |S) =
P (S|LP )P (LP )
(0.20)(0.60)
=
= 0.353
P (S)
P (0.34)
Of those book that go to a second edition, 35.3% are sold more than the projected.
Question # 03: (5 points)
In the United States, telephone numbers consist of a three-digit area code followed by a seven digit local
number. Suppose neither the first digit of an area code nor the first digit of a local number can be a zero but that
all other choices are acceptable.
a. How many different area codes are possible?
Using BCR
m1 × m2 × m3 = 9 × 10 × 10 = 900
b. For a given area code, how many local telephone numbers are possible?
m1 × m2 × m3 × m4 × m5 × m6 × m7 = 9 × 10 × 10 × 10 × 10 × 10 × 10
= 9, 000, 000
c. How many telephone numbers are possible?
m1 × m2 = 900 × 9, 000, 000 = 8, 100, 000, 000
Question # 04: (5 points)
At a lottery, 100 tickets were sold and three prizes are to be given. How many possible outcomes are there if
a. the three prizes are equivalent?
Since it does not matter in what order the prizes are obtained so
100 C3
=
100!
= 161, 700
3!(100 − 3)!
2
b. there is a first, second, and third prize?
Now, we are given specific order which the prizes should have so
100 C3
=
100!
= 970, 200
(100 − 3)!
Question # 05: (5 points)
A student takes a multiple choice exam with 10 questions, each with four possible selections for the answer. A
passing grade is 60% or better. Suppose that the student was unable to find time to study for the exam and just
guesses at each question. Find the probability that the student
a. gets at least one question correct.
Let X = number of question that the student guesses correct out of 10 question.
Each question has one correct choice out of four possible choices, so p = 1/4 = 0.25
Total questions = n = 10
P (X ≥ 1) = 1 − P (X < 1) = 1 − P (X = 0)
10
=1−
(0.25)0 (0.75)10 = 1 − 0.056 = 0.944
0
b. passes the exam.
P (X ≥ 6) = P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)
10
10
10
7
3
6
4
(0.25)8 (0.75)2 +
(0.25) (0.75) +
(0.25) (0.75) +
=
8
7
6
10
10
9
1
(0.25)10 (0.75)0
(0.25) (0.75) +
10
9
= 0.01622 + 0.00309 + 0.00039 + 0.00003 + 0 = 0.01973
c. receives an A on the exam (90% or better).
P (X ≥ 9) = P (X = 9) + P (X = 10)
10
10
9
1
=
(0.25) (0.75) +
(0.25)10 (0.75)0
10
9
= 0.00003 + 0 = 0.00003
d. How many questions would you expect the student to get correct?
Mean = µ = np = (10)(0.25) = 2.5
e. Obtain the standard deviation of the number of questions that the student gets correct.
p
√
Standard Deviation = σ = np(1 − p) = 1.85 = 1.369
Question # 06: (5 points)
An article titled “You’re Eating That?”, published in the New York Times, discussed consumer perception of
food safety. The article cited research by the Food Marketing Institute, which indicates that 66% of consumers
in the United States are confident that the food they buy is safe. Suppose that six consumers in the United
States are randomly sampled and asked whether they are confident that the food they buy is safe. Determine
the probability that the number answering in the affirmative is
3
a. exactly two.
Let X = number of people who are confident that the food they buy is safe out of 6 people.
p = 66/100 = 0.66, n = 6
6
P (X = 2) =
(0.66)2 (0.34)4 = 0.0873
2
b. exactly four.
6
4
P (X = 4) =
(0.66)4 (0.34)2 = 0.329
c. at least two.
P (X ≥ 2) = 1 − P (X < 2) = 1 − [P (X = 0) + P (X = 1)]
6
6
0
6
1
5
=1−
(0.66) (0.34) +
(0.66) (0.34)
0
1
= 1 − [0.0015 + 0.018] = 0.981
Question # 07: (5 points)
From past records, the owner of a fast-food restaurant knows that, on average, 2.4 cars use the drive-through
window between 3:00 P.M. and 3:15 P.M. Furthermore, the number, X, of such cars has a Poisson distribution.
Determine the probability that, between 3:00 P.M. and 3:15 P.M.,
a. exactly two cars use the drive-through window.
Let X = number of cars using the drive-through between 3:00 P.M. and 3:15 P.M
λ = 2.4
(2.4)2
= 0.261
P (X = 2) = e−2.4
2!
b. at least three cars use the drive-through window.
P (X ≥ 3) = 1 − P (X < 3) = 1 − [P (X = 0) + P (X = 1) + P (X = 2)]
0
1
2
−2.4 (2.4)
−2.4 (2.4)
−2.4 (2.4)
=1− e
+e
+e
0!
1!
2!
0
1
2
(2.4)
(2.4)
(2.4)
= 1 − e−2.4
+
+
0!
1!
2!
= 1 − e−2.4 [1 + 2.4 + 2.88]
= 1 − e−2.4 (6.28) = 1 − 0.57 = 0.43
c. Construct a table of probabilities for the random variable X. Compute the probabilities until they are zero
to two decimal places.
x=
P (X = x) =
0
1
2
3
4
5
6
7
0.091
0.218
0.261
0.209
0.125
0.060
0.024
0.008
d. Draw a histogram of the probabilities in part (c).
4
Question # 08: (5 points)
In the paper “The Distribution of Wars in Time” (Journal of the Royal Statistical Society, Vol. 107, No. 3/4,
pp. 242 − 250), L. F. Richardson analyzed the distribution of wars in time. From the data, we determined that
the number of wars that begin during a given calendar year has roughly a Poisson distribution with parameter
λ = 0.7. If a calendar year is selected at random, find the probability that the number, X, of wars that begin
during that calendar year will be
a. zero.
Let X = number of wars that begin during a calendar year.
λ = 0.7
(0.7)0
P (X = 0) = e−0.7
= 0.497
0!
b. at most two.
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
(0.7)0
(0.7)1
(0.7)2
+ e−0.7
+ e−0.7
1!
0! 0
2!
1
2
(0.7)
(0.7)
(0.7)
= e−0.7
+
+
0!
1!
2!
= e−0.7
= e−0.7 [1 + 0.7 + 0.245] = e−0.7 (1.945) = 0.966
c. between one and three, inclusive.
P (1 ≤ X ≤ 3) = P (X = 1) + P (X = 3) + P (X = 3)
(0.7)1
(0.7)2
(0.7)3
+ e−0.7
+ e−0.7
2!
1! 1
3!
2
3
(0.7)
(0.7)
(0.7)
= e−0.7
+
+
1!
2!
3!
= e−0.7
= e−0.7 [0.7 + 0.245 + 0.057] = e−0.7 (1.002) = 0.498
d. Find and interpret the mean of the random variable X.
Mean = µ = λ = 0.7
On average 0.7 wars begin in a calendar year.
e. Determine the standard deviation of X.
√
Standard Deviation = σ =
5
λ=
√
0.7 = 0.837