Reconsider the prize wheels described in the Example 11.4 and in Exercise 11.4.15. The average distance the outcomes are from the expected value for the wheel from the example (wheel A) can be computed as follows: |0 − 2.33|(8/12) + |1 − 2.33|(3/12) + |25 − 2.33| (1/12) = $3.775 

a. Find the average distance the outcomes for wheel B are from the expected value of the wheel. 

b. We often think of standard deviation as approximately the average distance outcomes are away from their mean (or expected value). This tends to be a reasonable interpretation if a distribution is not highly skewed or has outliers. Compare the standard deviations and average distance from the mean for wheels A and B. Are the two standard deviations approximately the same as their respective average distances from the mean or not? Explain why or why not.

Data from Exercises 11.4.15.

The example in this section described a prize wheel divided into 12 equal-sized sectors where 8 are marked no prize, 3 are marked win $1, and 1 is marked win $25. We will call this wheel A. Suppose you had a similar wheel (which we’ll call wheel B) that had 12 sectors: 2 of the sectors marked no prize, 1 marked win $1, and 9 marked win $3.