Question

the height of women ages 20 - 29 is normally distributed, with a mean of 63.6 inches. assume σ = 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. what is the probability of randomly selecting 1 woman with a height less than 66 inches? (round to four decimal places as needed.) what is the probability of selecting a sample of 14 women with a mean height less than 66 inches? (round to four decimal places as needed.)

Explanation:

Step1: Calculate \( z \)-score for one - woman case

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

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

Using the standard - normal table, \( P(X\lt66)=P(Z\lt0.86) \).
From the standard - normal table, \( P(Z\lt0.86)=0.8051 \).

Step2: Calculate \( z \)-score for the sample - mean case

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

$$ z=\frac{66 - 63.6}{\frac{2.8}{\sqrt{14}}}=\frac{2.4}{\frac{2.8}{3.742}}\approx\frac{2.4}{0.748}\approx3.21 $$

Using the standard - normal table, \( P(\bar{X}\lt66)=P(Z\lt3.21) \).
From the standard - normal table, \( P(Z\lt3.21)=0.9993 \).

Answer:

The probability of randomly selecting 1 woman with a height less than 66 inches is \( 0.8051 \).
The probability of selecting a sample of 14 women with a mean height less than 66 inches is \( 0.9993 \).
Since \( 0.9993>0.8051 \), you are more likely to select a sample of 14 women with a mean height less than 66 inches.