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

Answer:

a) 6.68% of heights less than 150 centimeters

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

Step-by-step 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:

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

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 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

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

$Z = \frac{150 - 162}{8}$

$Z = -1.5$

$Z = -1.5$ 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

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

$Z = \frac{180 - 162}{8}$

$Z = 2.25$

$Z = 2.25$ has a pvalue of 0.9878

X = 160

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

$Z = \frac{160 - 162}{8}$

$Z = -0.25$

$Z = -0.25$ has a pvalue of 0.4013

0.9878 - 0.4013 = 0.5865

58.65% of heights between 160 centimeters and 180 centimeters