Chapter 2.1, Problem 32E

(a)

To determine

To find: the points where the mean and the median fall

Answer to Problem 32E

This then implies that the mean and median both have to occur at A.

Explanation of Solution

Given:

  

Calculation:

Mean is A, median is also A (the distribution is symmetric).

  

We note that the given graph is symmetric, because the peak is in the middleof the graph.

A symmetric distribution has the property that the mean and the median are both in the middle of the distribution and thus at the peak of the distribution.

This then implies that the mean and median both have to occur at A.

Conclusion:

This then implies that the mean and median both have to occur at A.

(a)

To determine

To find: the points where the mean and the median fall

Answer to Problem 32E

This then implies that the mean needs to be A, while the median has to be B.

Explanation of Solution

Given:

  

Calculation:

Mean A, median B (the left skew pulls the mean to the left).

  

We note that the given graph is skewed to the left, because the peak is to the right in the distribution and there is a tail of more unusual values to the left in the distribution.

The mean and median of a left-skewed distribution will both be to the left of the peak of the distribution.

The mean will be more affected by the more unusual values to the left in the distribution than the median, which causes the mean to be smaller than the median.

This then implies that the mean needs to be A, while the median has to be B.

Conclusion:

This then implies that the mean needs to be A, while the median has to be B.