Simple Random Sampling - Help!

Question

Suppose that, in fact, the blood cholesterol level of all men aged 20 to 34 follows the normal distribution with mean of u = 188 milligrams per deciliter (mg/dl) and standard deviation = 41 mg/dl.

1. Choose an SRS of 100 men from this population. What is the sampling distribution of x? What is the probability that x takes a value between 185 and 191 mg/dl? This is the probability that x estimates u within ± 3 mg/dl.
2. Choose an SRS of 1000 men from this population. Now what is the probability that x falls within ± 3 mg/dl of u? The larger sample is much more likely to give an accurate estimate of u.

Answer 1

This is actually pretty tough. It requires you to go from the population distribution to the distribution of sample means.

 

1) In this part, we're saying we have a random sample of 100 men. Let's say we took random samples of 100 men over and over and each time we took the mean cholesterol from those 100. We would end up with a bunch of means; if we took the average of all those means, we should end up with the same mean as the population (188). But what would the standard deviation of all those means be? Since this is a distribution of means, and not a distribution of individual people, the standard deviation will not be equal to 41 (since the mean is based on 100 people, it will be more accurate). The standard deviation of a distribution of means will be equal to the SD of the population divided by the square root of the sample size - in your case, this is 41/sqrt(100) = 4.1.

So our distribution of means has a mean of 188 and a standard deviation of 4.1. Now we want to know the probability of a mean falling between 185 and 191. Whenever we deal with probabilities in a distribution, it is much easier if we convert to z scores first. Any z-score can be found using the formula: z = (X - M)/SD where X is a score, M is the mean and SD is the standard deviation. In this case we want (185-188)/4.1 = -.7317 and (191 - 188)/4.1 = .7317.

Now, to answer the question, we need to find what percent of the distribution falls between the z-scores -.7317 and .7317. To do this, we need to use the normal distribution table (your teacher should have given you one or it should be in your textbook). Using the table, we see that approximately 53.56% of the distribution falls between those scores.

 

2) The process for this question is the same, but now we replace 100 with 1000. The standard deviation becomes 41/sqrt(1000) = 1.2965. Our z-scores become (185 - 188)/1.2965 = -2.3139 and (191 - 188)/1.2965 = 2.3139. Checking the normal distribution table, we see that approximately 97.93% of the distribution falls between those scores.