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.
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 0.0188.
part: 1 / 3
part 2 of 3
(b) what is the probability that the sample mean cholesterol level is between 187 and 193?
the probability that the sample mean cholesterol level is between 187 and 193 is .

Explanation:

Step1: Identify the distribution

The population mean \(\mu = 199\), population standard deviation \(\sigma = 39.8\), sample size \(n = 107\). The sampling distribution of the sample mean \(\bar{X}\) is approximately normal with mean \(\mu_{\bar{X}}=\mu = 199\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{39.8}{\sqrt{107}}\approx\frac{39.8}{10.344}\approx3.847\).

Step2: Calculate z - scores

For \(x = 187\):
The z - score formula is \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}\)
\(z_1=\frac{187 - 199}{3.847}=\frac{- 12}{3.847}\approx - 3.12\)
For \(x = 193\):
\(z_2=\frac{193 - 199}{3.847}=\frac{-6}{3.847}\approx - 1.56\)

Step3: Find probabilities using Excel

We want \(P(187<\bar{X}<193)=P(-3.12 < Z<-1.56)\)
In Excel, we can use the formula \(=NORM.DIST(193,199,3.847,TRUE)-NORM.DIST(187,199,3.847,TRUE)\)
First, \(NORM.DIST(193,199,3.847,TRUE)\) gives the cumulative probability for \(x = 193\)
\(NORM.DIST(193,199,3.847,TRUE)\approx0.0594\)
\(NORM.DIST(187,199,3.847,TRUE)\approx0.0009\)
Then \(P(187<\bar{X}<193)=0.0594 - 0.0009 = 0.0585\) (or we can use z - scores: \(P(-3.12<Z < - 1.56)=P(Z < - 1.56)-P(Z < - 3.12)\)
From standard normal table or Excel, \(P(Z < - 1.56)\approx0.0594\), \(P(Z < - 3.12)\approx0.0009\), so the difference is \(0.0594 - 0.0009=0.0585\))

Answer:

\(0.0585\)