the national health and nutrition examination survey reported that a recent year

Question

the national health and nutrition examination survey reported that a recent year, the mean serum cholesterol level for US adults was 202, with a standard deviation of 41 (the units are milligram per deciliter). A simple random sample of 110 adults are chosen.

A) what is the probability that the sample mean cholesterol level is greater than 207?

B) what is the probability that the sample mean cholesterol level is between 195 and 203?

C) would it be unusual for the sample mean to be less than 196? explain

Explanation / Answer

mean = 202

standard deviation = 41

sample size = 110

standard deviation of sample mean = 41/110 = 3.9

A)

z value for 207 is (207-202)/3.9 = 1.282, correspoding p value is 0.900079

P(X<207) = 0.900079

probability that the sample mean cholesterol level is greater than 207 =1- 0.900079 = 0.099921

B)

z value for 195 is (195-207)/3.9 = -3.0769 , p value is 0.001045

z value for 203 is (203-207)/3.9 = -1.025641 , p value is 0.15253

probability that the sample mean cholesterol level is between 195 and 203 is (0.15253-0.001045) i.e0.151485

c)

z value for 196 is (196-202)/3.9 = -1.5384 , p value is 0.061975

P(X<196) is 0.061975

which is very low

so  it is unusual for the sample mean to be less than 196

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