Question

The feet of the average adult woman are 24.6 cm long, and foot lengths are normally distributed. If 16% of adult women have feet that are shorter than 22 cm, approximately what percent of adult women have feet longer than 27.2 cm?

Answer 1

Answer:

Approximately 16% of adult women have feet longer than 27.2 cm.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The feet of the average adult woman are 24.6 cm long

This means that $\mu = 24.6$

16% of adult women have feet that are shorter than 22 cm

This means that when $X = 22$, Z has a p-value of 0.16, so when $X = 22, Z = -1$. We use this to find $\sigma$

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

$-1 = \frac{22 - 24.6}{\sigma}$

$-\sigma = -2.6$

$\sigma = 2.6$

Approximately what percent of adult women have feet longer than 27.2 cm?

The proportion is 1 subtracted by the p-value of Z when X = 27.2. So

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

$Z = \frac{27.2 - 24.6}{2.6}$

$Z = 1$

$Z = 1$ has a p-value of 0.84.

1 - 0.84 = 0.16

0.16*100% = 16%.

Approximately 16% of adult women have feet longer than 27.2 cm.