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Chapter 6: Problem 54

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
Height at the 75th Percentile = \(68.675\) inches. Height at the 25th Percentile = \(61.325\) inches. Interquartile Range = \(68.675 - 61.325 = 7.35\) inches. The interquartile range is larger than the standard deviation.

Step by step solution

01

Calculate Height at the 75th Percentile

A percentile indicates the relative standing of a data point in a data set. 75th percentile would mean the point below which 75% of the observations fall. To find the 75th percentile, first find the z-score corresponding to 0.75 by referring to a standard z-table. The z-score is 0.67 approximately. Then apply the z-score formula \(Z = (X - 渭) / 蟽\), where \(X\) is the score, \(渭\) is the mean, and \(蟽\) is standard deviation. Rearranging for \(X\), we get \(X = Z * 蟽 + 渭 = 0.67 * 2.5 + 65\).
02

Calculate Height at the 25th Percentile

Similarly find the z-score corresponding to 0.25 from the z-table, which is approximately -0.67. Apply the formula \(X = Z * 蟽 + 渭 = -0.67 * 2.5 + 65\).
03

Calculate the Interquartile Range

Interquartile range (IQR) is the range between the first quartile (25th percentile) and the third quartile (75th percentile). Thus, IQR is the difference between the height calculated in step 1 and step 2.
04

Compare the IQR with the standard deviation

The final task is to compare the calculated IQR with the standard deviation given, to ascertain whether IQR is larger or smaller. If the IQR is greater than 2.5, it is larger, if less, it is smaller, and if equal, they are the same.

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