South Dakota State University
STAT 281
Test 2 Practice Problems
Chapter 5 5.1 The article “Scrambled Statistics: What Are the Chances of Finding Multi-Yolk Eggs?” (Significance [August 2016]: 11) gives the probability of a double-yolk egg as .001.
A. Give a relative frequency interpretation of the given probability.
Answer: In the long run, about 1 out of one tho
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Test 2 Practice Problems
Chapter 5 5.1 The article “Scrambled Statistics: What Are the Chances of Finding Multi-Yolk Eggs?” (Significance [August 2016]: 11) gives the probability of a double-yolk egg as .001.
A. Give a relative frequency interpretation of the given probability.
Answer: In the long run, about 1 out of one thousand eggs will contain a double yolk.
B. If 5000 eggs were randomly selected, about how many double-yolk eggs would you expect to
find?
.001 x 5000 = 5 double-yolk eggs
5.2 An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is .86. Give a relative frequency interpretation of this probability.
Answer: In the long run, 86% of the time this particular flight that flies between Phoenix and
Atlanta will arrive on time. (As you gathered the data on this flight over a long period of time, the percent of times that the flight arrives on time would approach 86%).
5.3 For a monthly subscription fee, a video download site allows people to download and watch up to five movies per month. Based on past download histories, the following table gives the estimated probabilities that a randomly selected subscriber will download 0, 1, 2, 3, 4, or 5 movies in a particular month.
If a subscriber is selected at random, what is the estimated probability that this subscriber downloads
|
Number of Downloads
|
0
|
1
|
2
|
3
|
4
|
5
|
|
Estimated Probability
|
.03
|
.45
|
.25
|
.10
|
.10
|
.07
|
a. three or fewer movies?
Answer: .83
b. at most three movies?
Answer: .83
c. four or more movies?
Answer: .17
d. zero or one movie?
Answer: .48
e. more than one movie?
Answer: .52
5.11 Phoenix is a hub for a large airline. Suppose that on a particular day, 8000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1800 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 480 passengers missed their connecting flight. Of these 480 passengers, 75 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8000 passengers. Compute the following probabilities:
a. the probability that the selected passenger had Phoenix as a final destination.
Answer: .225
b. the probability that the selected passenger did not have Phoenix as a final destination.
Answer: .775
c. the probability that the selected passenger was connecting and missed the connecting flight.
Answer: .06
d. the probability that the selected passenger was a connecting passenger and did not miss
the connecting flight.
Answer: .715
e. the probability that the selected passenger either had Phoenix as a final destination or was
delayed overnight in Phoenix.
Answer: .234
f. An independent customer satisfaction survey is planned. Fifty passengers selected at
random from the 8000 passengers who arrived in Phoenix on the day described above will be
contacted for the survey. The airline knows that the survey results will not be favorable if too
many people who were delayed overnight are included. Write a few sentences explaining whether
or not you think the airline should be worried, using relevant probabilities to support your
answer.
Answer: The airline should not be particularly worried. The probability of selecting one person who was delayed overnight is approximately 75/8000 = .0094, and the probability of selecting additional people who were delayed overnight decreases from .0094 (after selecting one, you’d be left with 74 people, etc.).
Extra problem 1: A single card is randomly drawn from a deck of 52 cards. (You can leave the probabilities in A and B as fractions)
A. Find the probability that it is an ace.
There are four aces (one of each of the four suits). So the probability is 4/52.
B. Find the probability that it is a number less than 5 (not including the ace)
These would be numbers 2, 3, 4, each in four suits. So 12/52 or 3/13.
Extra problem 2: Extra problem 2: Someone is checking the sex of three newborn puppies. Assume male and female are equally likely. What is the probability that there are two males and one female in the litter? (As we check the sex of the three puppies, a male followed by a female is different than a female followed by a male)
The sample space is {MMM, FFF, MMF, MFF, MFM, FFM, FMM, FMF}. You might have been able to predict that it contains 8 possibilities, because 2x2x2=8.
There are 3 cases where there are 2 males and a female. Thus the probability is 3/8.
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