The height of women ages 20-29 is normally distributed with a mean of 64 9 inche
ID: 3316658 • Letter: T
Question
The height of women ages 20-29 is normally distributed with a mean of 64 9 inches Assume = 2 8 inches Are you more likely to height less than 66 9 inches or are you more likely to select a sample of 22 women with a mean height less than 66 9 inches? Explain y select 1 woman with a EE Click the icon to vww 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? (Round to four decimal places as needed ) What is the probability of selecting a sample of 22 women with a mean height less than 66 9 inches? (Round to four decimal places as needed ) 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 22 women with a mean height less than 66 9 inches? Choose the correct answer below 0 A. It is more likely to select 1 woman with a height less than 66.9 inches because the probability is hgher O B. It is more likely to select a sample of 22 women with a mean height less than 66.9 inches because the sample of 22 has a higher probability O C. It is more likely to select a sample of 22 women with a mean height less than 66 9 inches because the sample of 22 has a lower probability D. It is more likely to select 1 woman with a height less than 66 9 inches because the probability is lowerExplanation / Answer
Here mean = 64.9 inches
Standard deviaation = 2.8 iches
(a) Pr(X < 66.9 inches) = NORM(X < 66.9 ; 64.9 ; 2.8)
Z = (66.9 - 64.9)/2.8 = 0.7143
Pr(X < 66.9 inches) = Pr(Z < 0.7143) = 0.7625
(b) Here let say sample mean = x
standard error of sample mean = /sqrt(n) = 2.8/ sqrt(22) = 0.597
Pr(x < 66.9 ; 64.9; 0.597) = Pr(Z)
Z = (66.9 - 64.9)/ 0.597 = 3.35
Pr(x < 66.9 ; 64.9; 0.597) = Pr(Z)
Pr(x < 66.9 ; 64.9; 0.597) = 0.9996
(c)Option B is correct. Sample mean of 22 is more likely to select.
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