Question

the height of women ages 20 - 29 is normally distributed, with a mean of 65 inches. assume $sigma = 2.8$ inches. are you more likely to randomly select 1 woman with a height less than 66.9 inches or are you more likely to select a sample of 25 women with a mean height less than 66.9 inches? explain. click the icon to view page 1 of the standard normal table. click the icon to view page 2 of the standard normal table. what is the probability of randomly selecting 1 woman with a height less than 66.9 inches? 0.7517 (round to four decimal places as needed.) what is the probability of selecting a sample of 25 women with a mean height less than 66.9 inches? (round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for single - woman case

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$. Given $\mu = 65$, $\sigma=2.8$, and $x = 66.9$. Then $z_1=\frac{66.9 - 65}{2.8}=\frac{1.9}{2.8}\approx0.68$. Looking up in the standard normal table, the probability $P(X\lt66.9)$ for a single woman is already given as $0.7517$.

Step2: Calculate z - score for sample - mean case

The formula for the z - score of the sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. Here, $\mu = 65$, $\sigma = 2.8$, $n = 25$, and $\bar{x}=66.9$. First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{2.8}{\sqrt{25}}=\frac{2.8}{5}=0.56$. Then $z_2=\frac{66.9 - 65}{0.56}=\frac{1.9}{0.56}\approx3.39$.

Step3: Find the probability for sample - mean case

Looking up $z = 3.39$ in the standard normal table, we find that $P(\bar{X}\lt66.9)$ corresponding to $z = 3.39$ is approximately $0.9996$.

Answer:

$0.9996$