Question

the height of women ages 20 - 29 is normally distributed, with a mean of 63.6 inches. assume \\( \sigma = 2.8 \\) inches. are you more likely to randomly select 1 woman with a height less than 66 inches or are you more likely to select a sample of 14 women with a mean height less than 66 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.
are you more likely to randomly select 1 woman with a height less than 66 inches or are you more likely to select a sample of 14 women with a mean height less than 66 inches? choose the correct answer below.
a. it is more likely to select a sample of 14 women with a mean height less than 66 inches because the sample of 14 has a lower probability.
b. it is more likely to select 1 woman with a height less than 66 inches because the probability is higher.
c. it is more likely to select a sample of 14 women with a mean height less than 66 inches because the sample of 14 has a higher probability.

Explanation:

Step1: Calculate z - score for single - woman case

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
Given \(\mu = 63.6\), \(\sigma=2.8\), and \(x = 66\).

$$z_1=\frac{66 - 63.6}{2.8}=\frac{2.4}{2.8}\approx0.86$$

Using the standard normal table, \(P(X\lt66)=P(Z\lt0.86)\). Looking up in the standard normal table, \(P(Z\lt0.86)=0.8051\)

Step2: Calculate z - score for sample - mean case

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
Given \(n = 14\), \(\mu = 63.6\), \(\sigma=2.8\), and \(\bar{x}=66\)
\(\frac{\sigma}{\sqrt{n}}=\frac{2.8}{\sqrt{14}}\approx0.75\)

$$z_2=\frac{66 - 63.6}{0.75}=\frac{2.4}{0.75}=3.2$$

Using the standard normal table, \(P(\bar{X}\lt66)=P(Z\lt3.2)\). Looking up in the standard normal table, \(P(Z\lt3.2)=0.9993\)

Answer:

C. It is more likely to select a sample of 14 women with a mean height less than 66 inches because the sample of 14 has a higher probability.