Question

The heights (measured in inches) of men aged 20 to 29 follow approximately the normal distribution with mean 69.3 and standard deviation 2.7. Between what two values does the middle 91% of all heights fall? (Please give responses to at least one decimal place)

Answer 1

The middle 91% of all heights fall between 64.7 inches and 73.9 inches.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 69.3, \sigma = 2.7`

Between what two values does the middle 91% of all heights fall?

From X when Z has a pvalue of 0.5 - 0.91/2 = 0.045 to X when Z has a pvalue of 0.5 + 0.91/2 = 0.955.

Lower bound

X when Z has a pvalue of 0.045. So X when Z = -1.695.

`Z = (X - \mu)/(\sigma)`

`-1.695 = (X - 69.3)/(2.7)`

`X - 69.3 = -1.695*2.7`

`X = 64.7`

Upper bound

X when Z has a pvalue of 0.955. So X when Z = 1.695.

`Z = (X - \mu)/(\sigma)`

`1.695 = (X - 69.3)/(2.7)`

`X - 69.3 = 1.695*2.7`

`X = 73.9`

The middle 91% of all heights fall between 64.7 inches and 73.9 inches.