Ultimate Questions On Statistics! Advanced Placement Trivia Quiz

1) A large company grants its hourly employees pay raises after their six-month performance review. The amounts of the raises average $0.25 per hour, with a standard deviation of $0.05 per hour. Let Ŷ represent the average pay raise of 50 randomly selected hourly employees. Which calculation would give the approximate probability that Ŷ is at least $0.30 per hour?

A

B

C

D

E

Calculation B would give the approximate probability that Ŷ is at least $0.30 per hour. This is because Calculation B involves finding the z-score for the value $0.30 and then using the z-table to find the corresponding probability. The z-score is calculated by subtracting the mean ($0.25) from the desired value ($0.30) and dividing by the standard deviation ($0.05). By using Calculation B, we can determine the probability that Ŷ is at least $0.30 per hour.

Explanation

Calculation B would give the approximate probability that Ŷ is at least $0.30 per hour. This is because Calculation B involves finding the z-score for the value $0.30 and then using the z-table to find the corresponding probability. The z-score is calculated by subtracting the mean ($0.25) from the desired value ($0.30) and dividing by the standard deviation ($0.05). By using Calculation B, we can determine the probability that Ŷ is at least $0.30 per hour.

2) For a class project, Charlotte recorded the heights of all 28 students in her class and calculated several statistics. She then realized she made an error recording the height of the tallest person in the class. She correctly had him listed as the tallest, but needed to add two inches to his recorded height to correct it. Which of these measures of spread must remain unchanged?

Maximum

Standard deviation

Variance

Interquartile range

Range

The interquartile range is a measure of spread that is not affected by extreme values or outliers. It is calculated by finding the difference between the upper quartile and the lower quartile, which are not influenced by the tallest person's height or the two inches added to it. Therefore, the interquartile range remains unchanged in this scenario.

Explanation

The interquartile range is a measure of spread that is not affected by extreme values or outliers. It is calculated by finding the difference between the upper quartile and the lower quartile, which are not influenced by the tallest person's height or the two inches added to it. Therefore, the interquartile range remains unchanged in this scenario.

3) A wildlife study found a cheap and easy way to estimate the number of nesting sites of terns (a type of seabird) by monitoring the number of calls heard per minute. More calls happen when there are more birds. In fact, it turned out that the number of calls explained 71% of the variation in the abundance of nests between breeding sites. Which of the following statements is correct about the correlation between the number of calls and the abundance of nests?

The correlation coefficient is -0.71.

The correlation coefficient is 0.71.

The correlation coefficient is -0.84.

The correlation coefficient is 0.84.

The slope of the regression line is needed to determine the correlation.

The given information states that the number of calls explains 71% of the variation in the abundance of nests. This indicates a strong positive correlation between the number of calls and the abundance of nests. The correlation coefficient measures the strength and direction of the linear relationship between two variables, ranging from -1 to +1. Since the given information suggests a positive correlation, the correct answer is that the correlation coefficient is 0.84.

Explanation

The given information states that the number of calls explains 71% of the variation in the abundance of nests. This indicates a strong positive correlation between the number of calls and the abundance of nests. The correlation coefficient measures the strength and direction of the linear relationship between two variables, ranging from -1 to +1. Since the given information suggests a positive correlation, the correct answer is that the correlation coefficient is 0.84.

4) The back-to-back stemplot below shows the daily high temperatures for the month of October for two cities in different regions of the country. Which of the following statements is true?

The range of the October temperatures for Hendersonville is greater than the range of October temperatures for Sheboygan.

The median October temperature for Hendersonville is less than the median October temperature for Sheboygan.

The median of the October temperatures for Hendersonville is greater than the third quartile of the October temperatures for Sheboygan.

The mean of the October temperatures for Hendersonville is less than he mean of the October temperatures for Sheboygan.

In a stemplot, the median is the middle value when the data is arranged in ascending order. The third quartile is the value that separates the upper 75% of the data from the lower 25% of the data. Therefore, if the median of the October temperatures for Hendersonville is greater than the third quartile of the October temperatures for Sheboygan, it means that the temperatures in Hendersonville are generally higher than the temperatures in Sheboygan.

Explanation

In a stemplot, the median is the middle value when the data is arranged in ascending order. The third quartile is the value that separates the upper 75% of the data from the lower 25% of the data. Therefore, if the median of the October temperatures for Hendersonville is greater than the third quartile of the October temperatures for Sheboygan, it means that the temperatures in Hendersonville are generally higher than the temperatures in Sheboygan.

5) Students in a statistics class, as part of an activity, measured the length of their hair in inches. Summary statistics are shown below. Which of the following statements is true?

Seventy-five percent of the students had hair at least 18.5 inches long.

Fifty percent of the students had hair between 9.18 and 29 inches long.

Fifty percent of the students had hair between 2 and 18.5 inches long.

The distribution of hair lengths can be appropriately modeled by a normal distribution.

The distribution of hair lengths has at least one outlier.

The correct answer is "Fifty percent of the students had hair between 2 and 18.5 inches long." This statement is true because it falls within the range of the summary statistics provided. The summary statistics do not mention anything about the distribution being normal or having outliers, so these statements cannot be determined based on the given information.

Explanation

The correct answer is "Fifty percent of the students had hair between 2 and 18.5 inches long." This statement is true because it falls within the range of the summary statistics provided. The summary statistics do not mention anything about the distribution being normal or having outliers, so these statements cannot be determined based on the given information.