Question

According to a recent poll, 29% of adults in a certain area have high levels of cholesterol. They report that such elevated levels "could be financially devastating to the regions healthcare system" and are a major concern to health insurance providers. According to recent studies, cholesterol levels in healthy adults from the area average about 205 mg/dL, with a standard deviation of about 30 mg/dL, and are roughly Normally distributed. If the cholesterol levels of a sample of 42 healthy adults from the region is taken, answer parts (a) through (d).

(a) What is the probability that the mean cholesterol level of the sample will be no more than 205? P(y s 205)-(Round to four decimal places as needed.)
(b) What is the probability that the mean cholesterol level of the sample will be between 200 and 210? P(200 <-<210)= □ (Round to four decimal places as needed.)
(c) What is the probability that the mean cholesterol level of the sample will be less than 195? P(y <195)(Round to four decimal places as needed.)
(d) What is the probability that the mean cholesterol level of the sample will be greater than 217? P(y > 217)Round to four decimal places as needed.)

Answer 1

a) 0.5 = 50% probability that the mean cholesterol level of the sample will be no more than 205

b) 0.7198 = 71.98% probability that the mean cholesterol level of the sample will be between 200 and 210

c) 0.0154 = 1.54% probability that the mean cholesterol level of the sample will be less than 195

d) 0.0048 = 0.48% probability that the mean cholesterol level of the sample will be greater than 217

Explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 205, \sigma = 30, n = 42, s = (30)/(√(42)) = 4.6291`

(a) What is the probability that the mean cholesterol level of the sample will be no more than 205?

This is the pvalue of Z when X = 205.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (205 - 205)/(4.6191)`

`Z = 0` has a pvalue of 0.5.

0.5 = 50% probability that the mean cholesterol level of the sample will be no more than 205

needed.)

(b) What is the probability that the mean cholesterol level of the sample will be between 200 and 210?

pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 200.

X = 210

`Z = (X - \mu)/(s)`

`Z = (210 - 205)/(4.6191)`

`Z = 1.08`

`Z = 1.08` has a pvalue of 0.8599

X = 200

`Z = (X - \mu)/(s)`

`Z = (210 - 205)/(4.6191)`

`Z = -1.08`

`Z = -1.08` has a pvalue of 0.1401

0.8599 - 0.1401 = 0.7198

0.7198 = 71.98% probability that the mean cholesterol level of the sample will be between 200 and 210

(c) What is the probability that the mean cholesterol level of the sample will be less than 195?

This is the pvalue of Z when X = 195.

`Z = (X - \mu)/(s)`

`Z = (195 - 205)/(4.6191)`

`Z = -2.16`

`Z = -2.16` has a pvalue of 0.0154

0.0154 = 1.54% probability that the mean cholesterol level of the sample will be less than 195

(d) What is the probability that the mean cholesterol level of the sample will be greater than 217?

This is 1 subtracted by the pvalue of Z when X = 217.

`Z = (X - \mu)/(s)`

`Z = (217 - 205)/(4.6191)`

`Z = 2.59`

`Z = 2.59` has a pvalue of 0.9952

1 - 0.9952 = 0.0048

0.0048 = 0.48% probability that the mean cholesterol level of the sample will be greater than 217