Question

Heights of Women. Heights of adult women are distributed normally with a mean of 162 centimeters and a standard deviation of 8 centimeters. Use the Table B.3 Areas under the Normal Curve (page 519 of the textbook) to find the indicated quantities: a) The percentage of heights less than 150 centimeters b) The percentage of heights between 160 centimeters and 180 centimeters

Answer 1

a) 6.68% of heights less than 150 centimeters

b) 58.65% of heights between 160 centimeters and 180 centimeters

Explanation:

When the distribution is normal, we use 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 question, we have that:

`\mu = 162, \sigma = 8`

a) The percentage of heights less than 150 centimeters

We have to find the pvalue of Z when X = 150. So

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

`Z = (150 - 162)/(8)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

6.68% of heights less than 150 centimeters

b) The percentage of heights between 160 centimeters and 180 centimeters

We have to find the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 160.

X = 180

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

`Z = (180 - 162)/(8)`

`Z = 2.25`

`Z = 2.25` has a pvalue of 0.9878

X = 160

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

`Z = (160 - 162)/(8)`

`Z = -0.25`

`Z = -0.25` has a pvalue of 0.4013

0.9878 - 0.4013 = 0.5865

58.65% of heights between 160 centimeters and 180 centimeters