Distribution of sample means

Question

For a normal population with a mean of 100 and a standard deviation of 20, for a sample of n=25, what is the probability that the sample mean will be within 5 points of the population mean?

Answer 1

Hi! First off, since we know that the underlying population has a normal distribution (with mean 100 and standard deviation 20), we can use z-scores to figure this out. If we didn't know the underlying distribution, we might have to think about using t-statistics, since we have an n < 30 (remember that the Central Limit Theorem only kicks in around n = 30). But since we know that it's normal, here's how we'll solve:

1. Standard error. Remember that the standard error is the standard deviation divided by the square root of n. Since the SD is 20 and n is 25, we have SE = 20/sqrt(25) = 20/5 = 4.
2. Identifying the probabilities we care about. The two z scores we care about are the ones corresponding to 95 and105 (remember that the population mean is 100 and we want to know the probability of a value falling within five points of the mean). At the end of the day, we want to know P(95 < x < 105), which we know can also be written as P(x < 105) - P(x < 95). All we need are the two z-scores that will help us solve this.
3. Calculating z-scores. Remember that we get a z-score by taking the value that we care about, subtracting the mean, and dividing by the standard error: z = (x-xbar)/SE. So let's do that:

For 95: (95 - 100)/4 = -5/4

For 105: (105-100)/4 = 5/4

1. Calculating the probabilities. Now we can either look these z-scores up on z tables, or we can use a statistical software package. I like to use the pnorm() command in R as follows:

P(x < 105) = P(z < 5/4) = pnorm(5/4) = 0.8944

P(x < 95) = P(z < -5/4) = pnorm(-5/4) = 0.1056

1. Solve the problem. Now, because of step 2, we know that all we need to do is subtract the two probabilities we solved for in Step 4:

P(95 < x < 105) = P(x < 105) - P(x < 95) = 0.8944 - 0.1056 = about 0.7887.

Hope this helps and good luck!