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Chapter 6: Problem 52

College women have heights with the following distribution (inches): \(N(65,2.5)\). a. Find the height at the 75 th percentile. b. Find the height at the 25 th percentile. c. Find the interquartile range for heights. d. Is the interquartile range larger or smaller than the standard deviation? Explain.

Short Answer

Expert verified
a) The height at the 75th percentile is approximately 66.675 inches. b) The height at the 25th percentile is approximately 63.325 inches. c) The interquartile range for heights is approximately 3.35 inches. d) The interquartile range is larger than the standard deviation.

Step by step solution

01

Find the height at the 75th Percentile

The 75th percentile, also known as the third quartile (Q3), is calculated using the Z-Score for 0.75 which is 0.67. Hence the calculation would be: Mean+(Z-score*Standard Deviation). Plugging in the given values, it turns out to be \(65+(0.67*2.5)= 66.675\) inches. The height at the 75th percentile is approximately 66.675 inches.
02

Find the height at the 25th Percentile

The 25th percentile, also known as the first quartile (Q1), is calculated using the Z-Score for 0.25 which is -0.67. So, the calculation would be: Mean+(Z-score*Standard Deviation). By substituting the given values, it results in \(65+(-0.67*2.5)= 63.325\) inches. The height at the 25th percentile is approximately 63.325 inches.
03

Find the Interquartile Range for Heights

Interquartile Range (IQR) is the range between the first quartile and the third quartile. Its calculation would be Q3 - Q1. Substituting the calculated values from the previous steps, it becomes \(66.675 - 63.325 = 3.35\) inches. The interquartile range for heights is approximately 3.35 inches.
04

Compare the Interquartile Range with the Standard Deviation

We know the standard deviation for the height is 2.5 inches while the interquartile range is 3.35 inches. Since 3.35 is greater than 2.5, the interquartile range is larger than the standard deviation.

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