Question

Consider population of 14 year old males & suppose that cholesterol levels are normally distributed with m= 170 mg/dl and Stnd dev= 30 mg/dl.

(A) Taking samples of size ten what should we expect the mean to be?
(B) Taking samples of size ten what should we expect the value of standarad deviation of our sample distribution to be?
(C) We select a random sample of 10 males from the population fInd the probability there average cholesterol level exceeds 180 mg/dl?

Answer 1

A) The expected mean of the sample distribution is 170 mg/dL. B) The expected standard deviation of the sample distribution is 9.49 mg/dL. C) The probability of the average cholesterol level exceeding 180 mg/dL can be found using the z-score.

Step-by-step explanation:

A) When taking samples of size ten from a population, the expected mean of the sample distribution can be found using the formula:

Sample Mean = Population Mean

In this case, the population mean is given as 170 mg/dl. So, the expected mean of the sample distribution would also be 170 mg/dl.

B) When taking samples of size ten from a population, the expected standard deviation of the sample distribution can be found using the formula:

Sample Standard Deviation = Population Standard Deviation / sqrt(Sample Size)

In this case, the population standard deviation is given as 30 mg/dl and the sample size is ten. So, the expected standard deviation of the sample distribution would be 30 / sqrt(10) ≈ 9.49 mg/dl.

C) To find the probability that the average cholesterol level exceeds 180 mg/dl in a random sample of 10 males, we need to calculate the z-score for this value using the formula:

z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(Sample Size))

Then, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. The probability can be interpreted as the likelihood of obtaining a sample average cholesterol level greater than 180 mg/dl.