Question

Weights of women are normally distributed with mean 143 pounds and standard deviation 29 pounds. If 36 women are randomly selected, what is the probability that their mean weight is between 120 and 160 pounds?

Answer 1

Answer on Question #48443 – Math – Statistics and Probability

Weights of women are normally distributed with mean $\mu = 143$ pounds and standard deviation $\sigma = 29$ pounds. If $n = 36$ women are randomly selected, what is the probability that their mean weight is between 120 and 160 pounds?

Solution

The mean weight is

$$\bar{X} = \frac{X_1 + X_2 + \cdots + X_n}{n},$$

The expectation of mean weight is

$$E(\bar{X}) = E(X_1) = \mu = 143,$$

The variance of mean weight is

$$Var(\bar{X}) = \frac{\sigma^2}{n} = \frac{29^2}{36}.$$

The probability that mean weight is between 120 and 160 pounds is

$$P(120 < \bar{X} < 160) = P(\bar{X} < 160) - P(\bar{X} < 120).$$ $$P(\bar{X} < 160) = P(z < z_1),$$

where

$$z_1 = \frac{160 - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{160 - 143}{\frac{29}{\sqrt{36}}} = 3.52.$$ $$P(\bar{X} < 120) = P(z < z_2),$$

where

$$z_1 = \frac{120 - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{120 - 143}{\frac{29}{\sqrt{36}}} = -4.76.$$

Thus

$$P(120 < \bar{X} < 160) = P(z < 3.52) - P(z < -4.76) = 0.99978 - 0.00001 = 0.99977.$$

Answer: 0.99977.

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