Question

assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. if 75 women are randomly selected, find the probability that they have a mean height between 63 and 65 inches. a. 0.3071 b. 0.9811 c. 0.2119 d. 0.0188

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 2.5$ (population standard deviation) and $n = 75$ (sample size).
$$\sigma_{\bar{x}}=\frac{2.5}{\sqrt{75}}\approx\frac{2.5}{8.6603}\approx0.2887$$

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\mu = 63.6$ (population mean).
For $\bar{x}=63$:
$$z_1=\frac{63 - 63.6}{0.2887}=\frac{- 0.6}{0.2887}\approx - 2.08$$
For $\bar{x}=65$:
$$z_2=\frac{65 - 63.6}{0.2887}=\frac{1.4}{0.2887}\approx4.85$$

Step3: Find the probabilities

Using the standard normal distribution table (or a calculator with a normal - distribution function), $P(Z\lt - 2.08)=0.0188$ and $P(Z\lt4.85)\approx1$ (since for large $z$ - values, the probability approaches 1 in the standard normal distribution).
The probability $P(63\lt\bar{X}\lt65)=P(Z\lt4.85)-P(Z\lt - 2.08)$
$$P(63\lt\bar{X}\lt65)=1 - 0.0188=0.9812\approx0.9811$$

Answer:

B. 0.9811