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 100 women are randomly selected, find the probability that they have a mean height greater than 63.0 inches.
a. 0.8989
b. 0.2881
c. 0.0082
d. 0.9918

Explanation:

Step1: Calculate the standard error

The formula for the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 2.5 \) (population standard deviation) and \( n = 100 \) (sample size).

$$ \sigma_{\bar{x}}=\frac{2.5}{\sqrt{100}}=\frac{2.5}{10}=0.25 $$

Step2: Calculate the z - score

The formula for the z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \), where \( \bar{x} = 63.0 \) (sample mean), \( \mu = 63.6 \) (population mean), and \( \sigma_{\bar{x}}=0.25 \).

$$ z=\frac{63.0 - 63.6}{0.25}=\frac{- 0.6}{0.25}=-2.4 $$

Step3: Find the probability

We want \( P(\bar{X}>63.0) \), which is equivalent to \( P(Z>-2.4) \).
Since \( P(Z > z)=1 - P(Z\leq z) \), and from the standard normal table \( P(Z\leq - 2.4)=0.0082 \).

$$ P(Z>-2.4)=1 - 0.0082 = 0.9918 $$

Answer:

D. \(0.9918\)