A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 204 with a standard deviation of 37. A random sample of 116 adults was Chosen. Find the 33rd percentile of the sample mean.
A) 205.1
B) 202.5
C) 187.7
D) 203.9

Question

Quizplus · Accepted Answer

Since the sample size is greater than 30, we can use the Central Limit Theorem to find the distribution of the sample mean. The z-score corresponding to the 33rd percentile is -0.44 (found using a standard normal distribution table).

Using the formula for the standard error of the mean:

SE = standard deviation / √n 
SE = 37 / √116 
SE = 3.44

We can now use the formula for the confidence interval of the sample mean:

Sample mean ± (z-score)(SE)

Plugging in the values:

Sample mean ± (-0.44)(3.44) 
204 ± (-1.51) 
202.49

Therefore, the 33rd percentile of the sample mean is approximately 202.5.

A Survey Reported That in a Recent Year, the Mean