LAST AP STATS REVIEW

A manager of a local fast food restaurant wants to determine the average amount of time it takes from when a customer enters the drive-through line until they receive their order, regardless of the time of the day and the day of the week. That is, he would like to know the true average drive-through wait time for all his customers. Which of the following methods would be most likely to estimate the desired parameter with low bias and low variability?

• Select a random sample of 100 customers over a period of one month, determine each selected customer’s wait time, then compute the average wait time for all 100 customers.

• Select a random sample of 100 customers over a weekend, determine each selected customer’s wait time, then compute the average wait time for all 500 customers.

• Select a random sample of 500 customers on a randomly selected weekday, determine each selected customer’s wait time, then compute the average wait time for all 500 customers.

• Select a random sample of 500 customers over a weekend, determine each selected customer’s wait time, then compute the average wait time for all 500 customers.

• Select a random sample of 500 customers over a period of one month, determine each selected customer’s wait time, then compute the average wait time for all 500 customers.

Answer explanation

The correct answer is (E). In order to estimate the desired parameter with low bias, the manager should use a random sample over a long period of time. Limiting the data collection to just one weekend or one randomly selected weekday might introduce bias. Weekends may be busier than normal and may have longer wait times. Additionally, a larger sample size will produce an estimate that has lower variability than an estimate coming from a smaller sample size.

Sam is thinking about switching cell phone providers. Sam is particularly interested in making sure that he has good signal strength when he is at work. He asks each of his 30 coworkers what cell phone provider they use and if they have poor signal strength, moderate signal strength, or good signal strength at work. Here is a segmented bar chart of the results.

Which of the following statements is supported by the data?

• All three providers are equally good choices because all three bars extend to 1, or 100%.

• Over 50% of his coworkers do not have cell service while at work.

• Provider B has much better signal strength than Provider A at his workplace.

• Provider A has reports of the greatest signal strength.

• Provider C has reports of the greatest strength of service.

Answer explanation

The correct answer is (D). Those who use Provider A have the greatest proportion of users that have good service at Sam’s workplace, therefore the data seem to indicate that Provider A receives the greatest strength of service. 

Which of the following distributions will best be modeled by a normal distribution?

• A bimodal distribution with a peak at 3, a peak at 10, and a standard deviation of 5.

• A uniform distribution with a mean of 5 for which 20% of the observations are between 1 and 3.

• A binomial distribution with n=150 and p=0.5.

• A geometric distribution with p=0.9

• A chi-square distribution with 2 degrees of freedom.

Answer explanation

The correct answer is (C). A binomial distribution with n = 150 and p = 0.5 can be modeled by a normal distribution because np = 150(0.5) = 75 ≥ 10 and n(1 – p) = 150(0.5) ≥ 10. 

A college professor recorded the speed with which students ran out the door after class after taking the final exam (using a radar gun) for each of the last 4 semesters. The histogram below shows the distribution of speed for 168 students. The maximum value is 18 mph.

The value, 18 mph, was actually supposed to be 8 mph and this error is corrected. Which of the following is true?

• The mean will change the same amount as than the median, and the range will change the same amount as the SD.

• The mean will change more than the median, and the range will change more than the SD.

• The median will change more than the mean, and the range will change more than the SD.

• The mean will change more than the median, and the SD will change more than the range.

• The median will change more than the mean, and the SD will change more than the range.

Answer explanation

The correct answer is (B). The correction will decrease the mean because 8 is less than 18 and will decrease the standard deviation because the value, 8, is closer to the mean than the value 18.

Suppose that the McAllister's purchased 15 plane tickets for an upcoming family trip to Paris. Each ticket independently has a 0.01 probability of being accidentally thrown away. What is the probability that exactly one of the tickets is accidentally thrown away?

• (1/15)(0.01)

• (0.01)(1)+(0.99)(14)

Answer explanation

A random variable X has the following distribution:

What is the expected value of X?

• 0.1

• 0.2

• 0.4

• 0.6

• 0.8

Answer explanation

The correct answer is (D). The probabilities must add to 1, so 2c + 4c + 0.3 + 0.1 = 1.

Therefore 6c = 0.6 and c = 0.1.

The expected value of X is (–2)(0.2) + (0)(0.4) + (2)(0.3) + (4)(0.1) = 0.6.

There are forty multiple-choice questions on this exam, each having answer choices A, B, C, D, or E. Only one answer choice per question is correct. Suppose a student randomly guesses their answer choice to each question, and their guesses from question to question are independent. Which of the following is the probability that the student guesses at least 12 questions correctly on this portion of the exam?

• 0.0238

• 0.0442

• 0.0875

• 0.9125

• 0.9806

Answer explanation

Researchers are using a new fitness watch with heart monitoring capabilities to study the human heart. In particular, they are examining the relationship between the presence of arrhythmias, or irregular heartbeats, and stress. A random sample of 100 women between the ages of 30 and 50 were selected for a study. At the beginning of the study, each woman completed a questionnaire that classified them as having high or low stress.

Each woman was also given a fitness watch to wear for 1 week. The watches revealed to the researchers whether each woman had a heart arrhythmia. The following table gives the number of women that fall into each category.

• There is evidence that high stress causes heart arrhythmias, and the conclusion can be generalized to all women aged 30 to 50.

• There is evidence that high stress causes heart arrhythmias, and the conclusion can be generalized to all women aged 30 to 50 who wear a fitness watch.

• There is evidence that high stress causes heart arrhythmias, and the conclusion can be generalized to all women similar to those in the study.

• Although cause-and-effect cannot be established, there is an association between stress and the presence of heart arrhythmias for the population of all women aged 30 to 50.

• Although cause-and-effect cannot be established, there is an association between stress and the presence of heart arrhythmias for the population of all women.

Answer explanation

The correct answer is (D). Cause and effect cannot be established in this study because there was no random assignment of the women to treatments. However, because knowledge of a woman’s stress level can help predict whether or not she has an arrhythmia, there is an association between stress and the presence of heart arrhythmias for the population of all women aged 30 to 50. Note that while we can generalize this conclusion to all women aged 30 to 50, we cannot generalize this all women in general because the study only used a random sample of women that were aged 30 to 50.

Have you ever run into someone you know when you were far away from home? A random sample of 1000 adults were asked this question. Here are their responses:

Based on the responses, which of the following is a 90 percent confidence interval for the proportion of all adults who would respond that they have run into someone they know when they were far away from home?

• 0.801 ± 0.0325

Answer explanation

The United States Postal Service delivers many packages every day. A postal carrier recorded the number of packages he delivered each day for 92 days. Each day had a unique number of packages delivered (no doubles). Here are the results:

Mean: 110

Standard Deviation: 24.25

Q₁: 75

Median: 98

Q₃: 122

How many days did he deliver fewer than 75 packages?

• 7

• 23

• 25

• 35

• 46

Answer explanation

The correct answer is (B). According to the results displayed, the first quartile is 75 packages, meaning that 25% of the days he delivered fewer than 75 packages.

He recorded data for a period of 92 days so he delivered fewer than 75 packages for (92)(0.25) = 23 of the days for which he recorded data.