
(a)
To find: the points where the
(a)

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 find: the points where the mean and the median fall
(a)

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.
Chapter 2 Solutions
The Practice of Statistics for AP - 4th Edition
Additional Math Textbook Solutions
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
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University Calculus: Early Transcendentals (4th Edition)
A First Course in Probability (10th Edition)
Calculus: Early Transcendentals (2nd Edition)
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