Question

A normal distribution of heights of adult women from a certain region has a mean of 62.7 inches and a standard deviation of 3 inches. Give the standard score and approximate percentiles for a woman with each of the following heights. a. 64.2 inches b. 61.2 inches c. 58.2 inches d. 67.2 inches Click the icon to view the standard scores and percentiles for a normal distribution. a. The standard score is and the percentile is (Type integers or decimals.)

Multiple question parts.

Answer 1

Here Heights of adults women from certain region normal distribution with mean of 62.7 inches and standard deviation 3.

$\mu = 62.7$ & $\sigma = 3$ inches

We know Standard score is also known as z-score and which is given by , Z score for X is $\frac{X-\mu}{\sigma}$

a. 64.2 inches

standard score = $\frac{X-\mu}{\sigma}$ =

64.2-62.7 3 =0.5

percentile = 69.15

The Standard score is 0.5 and the percentile is 69.15

b. 61.2 inches

standard score = $\frac{X-\mu}{\sigma}$ = $\frac{61.2-62.7}{3} = -0.5$

percentile = 30.85

The Standard score is -0.5 and the percentile is 30.85

c. 58.2 inches

standard score = $\frac{X-\mu}{\sigma}$ = $\frac{58.2-62.7}{3} = -1.5$

percentile = 6.68

The Standard score is -1.5 and the percentile is 6.68

d. 67.2 inches

standard score = $\frac{X-\mu}{\sigma}$ = $\frac{67.2-62.7}{3} = 1.5$

percentile = 93.32

The Standard score is 1.5 and the percentile is 93.32