Download Final Exam Review Problems

STAT 361 - Statistics for Management - Final Review
Final Exam Review Problems
Directions: Choose the most appropriate response for each question.
1. For an experiment, the items a machine produces are either defective or non-defective. It is known that ten percent
of the items produced by a machine are defective. Out of 15 items chosen at random,
What is the probability that less than 3 items will be defective? P (x < 3)
A. 0.1285 B. 0.816 C. 0.9445 D. 0.7432
2. An insurance company has determined that each week an average of nine claims are filed in their Atlanta branch.
The filing of a claim is believed to be random and the probability of filing a claim during the week is the same in
any week.
What is the probability that during the next week less than two claims will be filed?
A. 0.1171 B. 0.0011 C. 0.00 D. 0.0189
3. A major credit card company has determined that its customers charge an average of $280 per month on their
accounts with a standard deviation of $20.
What percentage of the customers charges between $240 and $285 per month?
A. 0.0987 B. 0.0228 C. 0.4772 D. 0.5759
4. Michael is running for president. The proportion of voters who favor Michael is 0.8. A simple random sample of
100 voters is taken.
What is the probability that the number of voters in the sample who favor Michael will be less than 0.75?
A. 0.49 B. 0.4938 C. 0.1587 D. 0.0062
5. If A and B are independent events with P(A) = 0.4 and P(B) = 0.6, then P(A ∩ B) =
A. 0.76 B. 1.00 C. 0.24 D. 0.20
6. If A and B are independent events with P(A) = 0.05 and P(B) = 0.65, then P(A⏐B) =
A. 0.05 B. 0.0325 C. 0.65 D. 0.8
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STAT 361 - Statistics for Management - Final Exam Review
7. In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 16 pieces of
carry-on luggage was weighed. The average weight was 20 pounds. Assume that we know the standard deviation
of the population to be 8 pounds.
a. Determine a 97% confidence interval estimate for the mean weight of the carry-on luggage.
b. Determine a 95% confidence interval estimate for the mean weight of the carry-on luggage.
8. A simple random sample of 36 items resulted in a sample mean of 40 and a standard deviation of 12. Construct a
95% confidence interval for the population mean.
9. From among 8 students how many committees consisting of 3 students can be selected?
10. Ten individuals are candidates for positions of president, vice president of an organization. How many possibilities
of selections exist?
11. A survey of a sample of business students resulted in the following information regarding the genders of the
individuals and their selected major.
Selected Major
Gender
Male
Management
40
Marketing
10
Others
30
Total
80
Female
30
20
70
120
Total
70
30
100
200
a.
b.
c.
d.
What is the probability of selecting an individual who is majoring in Marketing?
What is the probability of selecting an individual who is majoring in Management, given that
the person is female?
Given that a person is male, what is the probability that he is majoring in Management?
What is the probability of selecting a male individual?
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STAT 361 - Statistics for Management - Final Exam Review
12. For the following frequency distribution,
Class
45–47
48–50
51–53
54–56
57–59
a.
b.
Frequency
3
6
8
2
1
Compute the mean.
Compute the standard deviation. (Assume the data represent a population.)
13. Below you are given the ages of a sample of 10 college students who are enrolled in statistics.
20
a.
b.
c.
d.
e.
f.
g.
h.
18
20
22
18
20
22
Compute the mean.
Compute the variance.
Compute the standard deviation.
Compute the coefficient of variation.
Determine the 25th percentile.
Determine the median
Determine the 75th percentile.
Determine the range.
3
17
19
24
ID: A
Final Exam Review Problems
Answer Section
MULTIPLE CHOICE
1. ANS: B
0.2059 + 0.3432 + 0.2669 = 0.816
15!
0
15
P (x = 0) =
(0.1) (0.9) = 0.2059
0!14!
15!
P (x = 1) =
1!14!
15!
P (x = 2) =
2!13!
15!
P (x = 3) =
3!12!
(0.1) (0.9)
1
14
= 0.3432
(0.1) (0.9)
2
13
= 0.2669
3
12
= 0.1285
(0.1) (0.9)
2. ANS: B
Poisson Distribution
P ( X < 2 ) = P(X=1) + P (X=0) = 0.001 + 0.0001 = 0.0011
f (x) =
f (1) =
f (0) =
µx e
−µ
x!
µx e
−µ
x!
µx e
=
−µ
x!
=
9 1 e −9
1!
9 0 e −9
0!
= 0.0011
= 0.0001
1
ID: A
3. ANS: D
240 − 280 −40
Z1 =
=
= −2.00
20
20
Z2 =
285 − 280
20
=
5
20
0.4772 + 0.0987 = 0.5759
= 0.25
0.4772
-2.00
0.0987
0
0.25
z charge
2
ID: A
4. ANS: D
E(X) = P = 0.80.
p ÊÁË 1 − p ˆ˜¯
0.8 (0.2)
σ p8 =
=
=
n
100
p8 − µ p 0.70 − 0.80
Z1 =
=
= −2.50
0.04
σ p8
0.0016 = 0.04
P(P < 0.70 ) =0.5 - 0.4938 = 0.0062
0.4938
-2.50
z %favor
0
P = 0.80 (Favor Michael)
Sigma P-Hat: 0.04
Sample Size: N=100
5. ANS: C
6. ANS: A
3
ID: A
PROBLEM
7. ANS:
a.
b.
15.66 to 24.34
16.08 to 23.92
a. σ is known , Use Z = 2.17 for a 97% C.I.
ÊÁ σ ˆ˜
ÁÊ 8 ˜ˆ˜
Á x ˜˜
˜˜
˜˜ = 20 ± 2.17 ÁÁÁÁ
x ± Z ÁÁÁÁ
ÁÁ 16 ˜˜˜ = 20 ± 4.34
Á n ˜˜
Ë
¯
Ë
¯
15.66____24.34
0.97, 97% C.I.
-2.17
0
2.17
Mean = 20 (Pounds)
Sigma: 8
Sample Size: N=16
Sample Mean: 20
b. σ is known , Use Z = 1.96 for a 95% C.I.
ÊÁ σ ˆ˜
ÁÊ 8 ˜ˆ˜
Á x ˜˜
˜˜
˜˜ = 20 ± 1.96 ÁÁÁÁ
x ± Z ÁÁÁÁ
ÁÁ 16 ˜˜˜ = 20 ± 3.92
Á n ˜˜
Ë
¯
Ë
¯
16.08____23.92
4
z weight
ID: A
0.95, 95% C.I.
-1.96
0
1.96
Mean = 20 (Pounds)
Sigma: 8
Sample Size: N=16
Sample Mean: 20
8. ANS:
σ x is unknown
d. f. = n − 1 = 36 − 1 = 35
t(0.025, 95% C.I.) = 2.021
ÊÁ S ˆ˜
ÊÁ 12 ˆ˜
Á x ˜˜
˜˜
˜˜ = 40 ± 2.021 ÁÁÁÁ
˜˜ = 40 ± 4.042
x ± t ÁÁÁÁ
˜
Á
Á n˜
ÁË 36 ˜˜¯
Ë
¯
35.96____44.04
9. ANS:
56
10. ANS:
90
11. ANS:
a. 0.15
b. 0.25
c. 0.50
d. 0.40
12. ANS:
a.
b.
50.8
3.06
5
z weight
ID: A
13. ANS:
a.
b.
c.
d.
e.
f.
g.
h.
20
4.667
2.16
10.8%
18
20
22
7
6