Question

The height of women ages 20-29 is normally distributed, with a mean of 64.6 inches. Assume o= 2.5 inches. Are you more likely to randomly select 1 woman with a height less than 66.2 inches or are you more likely to select a sample of 21 women with a mean height less than 66.2 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 66.2 inches? (Round to four decimal places as needed.) What is the probability of selecting a sample of 21 women with a mean height less than 66.2 inches? (Round to four decimal places as needed.) Are you more likely to randomly select 1 woman with a height less than 66.2 inches or are you more likely to select a sample of 21 women with a mean height less than 66.2 inches? Choose the correct answer below. O A. It is more likely to select 1 woman with a height less than 66.2 inches because the probability is higher. O B. It is more likely to select a sample of 21 women with a mean height less than 66.2 inches because the sample of 21 has a lower probability. O C. It is more likely to select 1 woman with a height less than 66.2 inches because the probability is lower. O D. It is more likely to select a sample of 21 women with a mean height less than 66.2 inches because the sample of 21 has a higher probability ? Click to select vour answer(s).

Answer 1

Solution :

Given that ,

mean = $\mu$ = 64.6

standard deviation = $\sigma$ = 2.5

a) P(x < 66.2) = P[(x - $\mu$ ) / $\sigma$ < (66.2 - 64.6) / 2.5]

= P(z < 0.64)

Using z table,

= 0.7389

b) n = 21

$\mu$_{$\bar x$} = $\mu$ = 64.6

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 2.5 / $\sqrt$ 21 = 0.546

P($\bar x$ < 66.2) = P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (66.2 - 64.6) / 0.546)

= P(z < 2.93)

Using z table

= 0.9983

c) correct option is = D.