Question

watch your cholesterol: the mean serum cholesterol level for u.s. adults was 199, with a standard deviation of 39.8 (the units are milligrams per deciliter). a simple random sample of 107 adults is chosen. use excel. round the answers to at least four decimal places.
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part 1 of 3
(a) what is the probability that the sample mean cholesterol level is greater than 207?
the probability that the sample mean cholesterol level is greater than 207 is \boxed{}.

Explanation:

Step1: Identify distribution parameters

Population mean $\mu = 199$, population standard deviation $\sigma = 39.8$, sample size $n = 107$. The sampling distribution of the sample mean $\bar{X}$ has mean $\mu_{\bar{X}}=\mu = 199$ and standard deviation (standard error) $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{39.8}{\sqrt{107}}$.

Calculate $\sigma_{\bar{X}}$: $\sqrt{107}\approx10.3441$, so $\sigma_{\bar{X}}\approx\frac{39.8}{10.3441}\approx3.8476$.

Step2: Calculate z-score

For $\bar{x} = 207$, the z-score is $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{207 - 199}{3.8476}=\frac{8}{3.8476}\approx2.0792$.

Step3: Find probability using z-table or Excel

We need $P(\bar{X}>207)=P(Z > 2.0792)$. Using Excel, the formula is =1 - NORM.DIST(207, 199, 3.8476, TRUE) (or using z-score: =1 - NORM.S.DIST(2.0792, TRUE)).

Calculating NORM.S.DIST(2.0792, TRUE) gives the cumulative probability up to $z = 2.0792$. Let's compute it: $NORM.S.DIST(2.0792, TRUE)\approx0.9812$. So $1 - 0.9812 = 0.0188$.

Answer:

0.0188 (rounded to four decimal places)