Question

The heights (measured in inches) of men aged 20 to 29 follow approximately the normal distribution with mean 69.3 and standard deviation 2.8. Between what two values does that middle 92% of all heights fall? (Please give responses to 3 decimal places)

Answer 1

Answer:

The middle 92% of all heights fall between 64.4 inches and 74.2 inches.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 69.3, \sigma = 2.8$

Between what two values does that middle 92% of all heights fall?

The middle 92% falls from X when Z has a pvalue of 0.5 - 0.92/2 = 0.04 to X when Z has a pvalue of 0.5 + 0.92/2 = 0.96. So from the 4th percentile to the 96th percentile.

4th percentile

X when $Z = -1.75$

$Z = \frac{X - \mu}{\sigma}$

$-1.75 = \frac{X - 69.3}{2.8}$

$X - 69.3 = -1.75*2.8$

$X = 64.4$

96th percentile

X when $Z = 1.75$

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{X - 69.3}{2.8}$

$X - 69.3 = 1.75*2.8$

$X = 74.2$

The middle 92% of all heights fall between 64.4 inches and 74.2 inches.