Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
The number of electrical outages in a city varies from day to day. Assume that the number of electrical
outages (x) in the city has the following probability distribution.
x
f (x)
0
0.8
1
0.15
2
0.04
3
0.01
1) The mean and the standard deviation for the number of electrical outages (respectively) are
a. 0 and 0.8
b. 2.6 and 5.77
c. 3 and 0.01
d. 0.26 and 0.577
Answer: D
x
0
1
2
3
Totals
f (x)
0.8
0.15
0.04
0.01
xf (x)
( x − u ) 2 f ( x)
0*0.8=0
1*.15=0.15
2*.04=0.08
3*.01=0.03
(0-0.26)^2=0.05408
(1-0.26)^2=0.08214
(2-0.04)^2=0.121104
(3-0.01)^2=0.075076
Mean = ∑ xf (x) = 0.26
Standard deviation=
∑ ( x − u)
2
Variance = ∑ ( x − µ ) 2 f ( x) = 0.3324
f ( x) = 0.3324 = 0.576542
2) Four percent of the customers of a mortgage company default on their payments. A sample of five
customers is selected. What is the probability that exactly two customers in the sample will default on their
payments?
a. 0.7408
b. 0.2592
c. 0.0142
d. 0.9588
Answer: C
n
n!
5!
f ( x) =   =
p x (1 − p) n − x =
(0.04) 2 (0.96) 3 = 10 * 0.04 3 * 0.96 3 = 0.01416
x
x
!
(
n
−
x
)!
2
!
3
!
 
3) Twenty percent of the students in a class of 100 are planning to go to graduate school. The mean and
standard deviation of this binomial distribution is
a. 4 and 20
b. 2 and 16
c. 16 and 2
d. 20 and 4
e) 4 and 17
Answer: D
E ( x ) = np = 0.2 * 100 = 20
σ = np (1 − p ) = ( 0.2)(100)( 0.8) = 4
4) A production process produces 2% defective parts. A sample of five parts from the production process is
selected. What is the probability that the sample contains exactly two defective parts?
a. 0.02
b. 0.10
1
Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
c. 0.0038
d. 0.0004
e) none of the above
Answer: C
n
n!
5!
f ( x) =   =
p x (1 − p) n − x =
(0.02) 2 (0.98) 8 = 10 * 0.004 * 0.9604 = 0.003842
x
x
!
(
n
−
x
)!
2
!
3
!
 
5) X is a random variable with the probability function: f(X) = X/6
The expected value of X is
a) 0.333
b) 2.333
c) 0.500
d) 2.000
Answer: B
x
1
2
3
Totals
xf (x)
( x − u ) 2 f ( x)
1*.0.1667=0.1667
2*0.3333=0.6666
3*0.150 =0.150
(1-0.26)^2=0.08214
(2-0.04)^2=0.121104
(3-0.01)^2=0.075076
f (x)
1/6=0.1667
2/6=0.333
3/6=0.50
for X = 1, 2 or 3
Mean = E ( x) = ∑ xf ( x) = 0.1667 + 0.6666 + 0.150 = 2.3333
Number of fatal accidents
Number of days
0
45
1
75
2
120
3
45
4
15
6) The table above shows the police record of a metropolitan area kept over the past 300 days show the
number of fatal accidents. What is the probability that in a given day there will less than 3 accidents?
a) 0.5
b) 120
c) 0.2
d) 0.8
Answer: D
f (x)
x
# of fatal accidents
0
45
45/300 =0.15
1
75
75/300 =0.25
2
120
120/300 =0.4
3
45
45/300 =0.15
4
15
15/300 =0.05
Using the computed relative frequencies, the cumulative relative frequencies for less than 3 accidents
is p ( x < 3) = p ( x = 0) + p( x = 1) + p( x = 2) = 0.15 = 0.25 + 0.40 = 0.80
Number of fatal accidents
0
1
2
3
4
Number of days
45
75
120
45
15
2
Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
7) The table above shows the police record of a metropolitan area kept over the past 300 days show the
number of fatal accidents. What is the probability that in a given day there will be no accidents?
a) 0.00
b) 0.15
c) 1.00
d) 0.85
Answer: B
f (x)
x
#
0
45
45/300 =0.15
1
75
75/300 =0.25
2
120
120/300 =0.4
3
45
45/300 =0.15
4
15
15/300 =0.05
N = 300
p ( x = 0) = 0.15
Number of new clients
Probability f (x)
0
0.05
1
0.10
2
0.15
3
0.35
4
0.20
5
0.10
6
0.05
8) The table above is for Roth is a computer-consulting firm. The number of new clients that they have
obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that
is shown above. The expected number of new clients per month is:
a) 21
b) 6
c) 0
d) 3.05
Answer: D
Number of clients
xf (x)
Probability f (x)
0
0.05
0*0.05=0
1
0.10
1*0.10=0.10
2
0.15
2*0.15=0.30
3
0.30
3*0.30=1.05
4
0.20
4*0.20=1.80
5
0.10
5*0.10=0.50
6
0.05
6*0.05=0.30
E ( x) =
Number of new clients Probability
0
0.05
1
0.10
2
0.15
3
0.35
4
0.20
5
0.10
6
0.05
f (x)
3
∑ xf ( x) =0 + 0.10 + 0.30 + 1.05 + 0.80 + 0.50 + 0.30 = 3.05
Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
9) The table above is for Roth is a computer-consulting firm. The number of new clients that they have
obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that
is shown above. The standard deviation is:
a. 2.047
b.3.05
c. 21
d. 1.431
Answer: D
Number of Probability
( x − u ) 2 * f ( x)
clients
f (x)
xf (x)
0
1
2
3
4
5
6
0.05
0.10
0.15
0.30
0.20
0.10
0.05
0*0.05=0
1*0.10=0.10
2*0.15=0.30
3*0.30=1.05
4*0.20=1.80
5*0.10=0.50
6*0.05=0.30
E ( x) =
∑ xf ( x) =3.05
(0-3.05)^2*.05=0.465
(1-3.05)^2*.10=0.420
(2-3.05)^2*.15=0.115
(3-3.05)^2*.30=0.0009
(4-3.05)^2*.20=0.181
(5-3.05)^2*.10=0.380
(6-3.05)^2*.05=0.435
Variance=
( x − µ ) 2 f ( x) = 2.0475
St. Dev.=
Number of goals
0
1
2
3
4
Probability
0.05
0.15
0.35
0.30
0.15
∑
∑ ( x − u)
2
f ( x) = 2.0475 = 1.431
f (x)
10) The table above shows the probability distribution for the number of gals the lions soccer team makes
per game. What is the expected number of goals per game?
a) 2.35
b) 0
c) 1
d) 2, since it has the highest probability
e) None of the above
Answer: A
Number of goals
xf (x)
Probability f (x)
0
0.05
0*0.05=0
1
0.15
1*0.15=0.15
2
0.35
2*0.35=0.70
3
0.30
3*0.30=0.90
4
0.15
4*0.15=0.60
E ( x) =
Daily Sales (In $1,000)
40
50
60
70
Probability f (x)
0.1
0.4
0.3
0.2
4
∑ xf ( x) =0 + 0.15 + 0.70 + 0.90 + 0.60 = 2.35
Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
11) The table above shows the probability distribution for the daily sales at Michael’s Co. The probability of
having sales of at least $50,000 is:
a) 0.90
b) 0.5
c) 0.30
d) 0.10
e) None of the above
Answer: A
Using the relative frequencies given on the table, we need to calculate the cumulative relative
frequencies for sales greater than or equal to $50,000.
p ( x ≥ $50,000) = p ( x = $50,000) + p ( x = $60,000) + p ( x = $70,000) = 0.4 + 0.3 + 0.2 = 0.90
12) Bob would rather risk getting a parking ticket than pay the parking fee at a campus parking facility. Bob
parks at the facility Monday through Friday. The probability of getting a ticket on any given weekday is 15%.
Answer the questions based on a five-day period. What is the probability that Bob will get a ticket during the
week (Monday-Friday)?
a) 0.1500
b) 0.0783
c) 0.0058
d) 0.3915
e) None of the above
Answer: D
n = 5, p = 0.15
f ( x) = n p p (1 − p ) n − p
The probability that bob will get late during the week is: f ( x = 1) = 5(0.15)1 (0.85) 4 = 0.3915
13) Suppose you are receiving a shipment of a product in lots of 5,000. You randomly select 25 units from
each lot and accept the shipment if the number of defective units is less than 3. Find the probability of
accepting a lot, if the actual fraction of defectives in the lot is 0.01 (1%).
a) 0.9980
b) 0.5371
c) 0.0982
d) 0.0090
e) None of the above
Answer: A
n = 25, p = 0.01
f ( x) = n x p (1 − p ) n − x
f ( x = 0) = 25(0.01)0 (0.99) 25 = 0.7778
f ( x = 1) = 25(0.01)1 (0.99) 24 = 0.1964
f ( x = 2) = 25(0.01) 2 (0.99) 23 = 0.0238
The probability that less than 3 items will be defective is
p ( x < 3) = p ( x = 0) + p ( x = 1) + p ( x = 2) = 0.7778 + 0.1964 + 0.0238 = 0.9998
14) Bob would rather risk getting a parking ticket than pay the parking fee at a campus parking facility. Bob
parks at the facility Monday through Friday. The probability of getting a ticket on any given weekday is 15%.
Answer the questions based on a five-day period.
16) What is the probability that Bob will get at least one ticket during the week?
a) 0.5563
b) 0.1648
5
Economics 270: Applied Business Statistics For Economics & Business (Summer 2012)
Department of Economics
Indiana University-Purdue University @ Indianapolis (IUPUI)
Lecturer: Baiyee-Mbi Agbor-Baiyee, PhD.
Answer Key: Quiz2-Chapter5: Discrete Probability Distribution
c) 0.0266
d) 0.0135
e) None of the above
Answer: A
n = 5, p = 0.15
f ( x) = n x p (1 − p ) n − x
f (x = 1) = 5(0.15)1 (0.85) 4 = 0.3915
f (x = 2) = 5(0.15) 2 (0.85)3 = 0.1382
f (x = 3) = 5(0.15)3 (0.85) 2 = 0.0244
f (x = 4) = 5(0.15) 4 (0.85)1 = 0.0022
f (x = 5) = 5(0.15)5 (0.85)0 = 0.0001
The probability that Bob will get at least 1 ticket during the week
is: f ( x ≥ 1) = p ( x = 1) + p ( x = 2) + p ( x = 3) + p( x = 4) + p( x = 5) = 0.3915 + .0.1382 + 0.0244 + 0.0022 + 0.0001 = 0.5564
The easier way to solve this particular problem is to define the problem f ( x ≥ 1) = 1 − p( x = 0)
Where p( x = 0) = 5(0.15)0 (0.85)5 = 0.4437 and therefore f ( x ≥ 1) = 1 − 0.4437 = 0.5563
15) The E270 common departmental final has 30 multiple questions. Each question has 5 choices. If you
guessed the answers to all the questions, how many questions should you expect to answer correctly?
a) 5
b) 6
c) 7
d) 14
e) None of the above
Answer: B
The number of correct guesses, x, has a binomial distribution with n = 30, p = 0.2
That is because there are 5 answers in each question, the probability to guess each question
correctly is 1/5=20%.
Therefore the expected number of questions that can be guessed correctly is:
E ( x) = 0.2 * 30 = 6
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