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miv72 [106K]
3 years ago
11

Susan is 66 inches tall. The average height of women is 63.7 inches with a standard deviation of 1.75. How does Susan compare to

the mean (how many SD above or below the mean is she)? Write an example of the average distance that the individual values of a problem vary from the mean in context.
Advanced Placement (AP)
1 answer:
Assoli18 [71]3 years ago
8 0

Answer:

Kindly check explanation

Explanation:

Given that :

Mean height of women = 63.7 inches

Susan's height = 66 inches standard deviation = 1.75 inches

Number of standard deviations above or below the mean:

Zscore = (Susan's height - mean height) standard deviation

Zscore = (66 - 63.7) / 1.75

Zscore = 2.3 / 1.75

Zscore = 1.314

Meaning Susan's height is 1.314 standard deviations above the mean height of women.

To evaluate the average distance by which individual values vary from the mean. Use the relation:

[The value (X) - mean value (m)] / standard deviation

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