Statistics questions

Stats Midterm
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Question 1
Marks: 10
Jeannie is an experienced business traveler, often traveling back and forth from San
Francisco to the East Coast several times per month. To catch her flights from San Francisco
she leaves her office one hour before her flight leaves. Her travel time from her office to the
departing gate at the San Francisco airport, including the time to park and go through
security screening, is normally distributed with a mean of 46 minutes and a standard
deviation of 5 minutes. What is the probability that Jeannie will miss her flight because her
total time for catching her plane exceeds one hour? Round your final answer to 4 decimal
places.
Choose one answer.
A. 0.0017
B. 0.0026
C. 0.0059
D. 0.0137
E. None of the above
Answer:
We have µ = 46 and σ = 5
The respective Z-score with X = 60 is
Z = (X - µ)/σ
= (60 – 46)/5
= 2.8
Using Z-tables, the probability is
P [Z > 2.8] =1 – 0.9974
= 0.0026
Question 2
Marks: 10
Use the same information about Jeannie in the above problem. Jeannie is known to be
somewhat lax about getting to the airport in time to catch her flight. Suppose that she
decides to leave her office so that she has a 96% chance of catching her flight;
consequently there is a 4% chance that she will miss her flight. How many minutes before
the flight leaves should she leave her office? Round your final answer to the nearest minute,
then input that whole number below.
Answer:
We have µ = 46 and σ = 5
The respective Z-score with p < 0.96 is 1.75
Z = (X - µ)/σ
1.75 = (X – 46)/5
1.75 x 5 = X – 46
X – 46 = 8.75
X = 54.75 ~ 55
Question 3
Marks: 10
In a busy coffee shop, which is a member of an international chain of coffee shops, 40% of
customers order a pastry in addition to their drink. If 20 customers were selected at random
during one business day, what is the probability that at least 11 of them did not order a
pastry with their drink. Round your answer to 4 decimal places.
Choose one answer.
A. 0.1275
B. 0.3763
C. 0.5956
D. 0.7553
E. None of the above
Answer:
We have p = 0.40 and n = 20
Since np = 20 x 0.40 = 8 and n (1 – p) = 20 x 0.60 = 12 both are greater 5, we use normal
approximation to binomial.
Mean, µ = np = 20 x 0.40 = 8
Standard deviation, σ = √ np (1 – p) = √20 x 0.40 x 0.60 = 2.19
P [X ≥ 11] = P [X > 10.5]
The respective Z-score with X = 10.5 is
Z = (X - µ)/σ
= (10.5 – 8)/2.19
= 1.14
Using Z-tables, the probability is
P [Z > 1.14] = 1 – 0.8725 = 0.1275
Question 4
Marks: 10
It has been conjectured by the U.S. Census Bureau that "approximately 60% of foreign-
born people who live in the U.S. are not naturalized citizens". Suppose that in a national
random sample of 70 foreign-born people who live in the U.S. that exactly 32 of them are
not naturalized citizens. Select the best answer below.
Choose one answer.
A. The percentage of foreign-born people who live in the U.S. and who are not
naturalized citizens is overstated by the U.S. Census Bureau.
B. The percentage of foreign-born people who live in the U.S. and who are not
naturalized citizens is not overstated by the U.S. Census Bureau.
C. Not enough information to answer the question.
D. None of these.
Answer:
Here n=70, p=60%=0.6 and x=32
P(x=32) =70c32 *(0.6)^32 *(0.4)^(70-32)
=0.005=0.5%
0.5%<60%
Question 5
Marks: 10
A civil service exam yields scores with a mean of 81 and a standard deviation of 5.5. Using
Chebyshev's Theorem what can we say about the percentage of scores that are above 92?
Choose one answer.
A. At most 12.5% of the scores are above 92.
B. At most 25% of the scores are above 92.
C. At least 75% of the scores are above 92.
D. At least 25% of the scores are above 92.
E. None of the above
Answer:
k = (92 – 81)/5.5 = 2
1/k2 = ¼ = 0.25
At least 75% is within ± 2 standard deviation and 25% outside ±2 standard deviation.
Question 6
Marks: 10
A civil service exam yields scores which are normally distributed with a mean of 81 and a
standard deviation of 5.5. If the civil service wishes to set a cut-off score on the exam so
that 15% of the test takers fail the exam, what should the cut-off score be? Remember to
round your z-value to 2 decimal places.
Choose one answer.
A. 75.28
B. 86.72
C. 60.24
D. 64.56
E. None of the above
Answer:
We have µ= 81 and σ = 5.5
The respective Z-score with p < 0.15 is -1.04 is
Z = (X - µ)/σ
-1.04 = (X – 81)/5.5
-1.04 x 5.5 = X – 81...