A normal distributed population has the following parameters,

Question

µ = 74.7 and σ = 91.6. If a random sample of size n= 139 is selected,

a. What is the mean of the distribution of sample means?

b. What is the standard deviation of the distribution of sample means?

Answer 1

Hi Bailey,

You were given that the population was normally distributed. Any time you have this, you can apply the Central Limit Theorem, which states that if the population is normally distributed, the sample mean x-bar, which part a asks you to compute is: N(mu, sigma/sqrt(n)).

Let’s break that down

N=Normal distribution, just an indicator, not a variable here

mu=population mean=approximate sample mean

sigma=population standard deviation

n=sample size

Recall that sigma/sqrt(n) was the equation for standard error; that is equivalent to sample standard deviation, which this problem asked you to find.

Thus, for our problem:

a) x-bar=mu=74.7

b) N(mu, sigma/sqrt n)

We now need to compute standard error (sample std. dev.) by using sigma/sqrt(n):

mu=74.7

sigma=91.6

n=139

SE=91.6/sqrt(139

SE=7.77

Did we hit the jackpot at the casino? I would commit that N(mu, sigma/sqrt(n)) and its use for normal populations to memory. I hope this helps.

Answer 2

The mean of the sampling distribution is the mean of the parent distribution, µ. The notation for a sample mean is x-bar. You may hear it called an "unbiased estimator" of the population mean.

The standard deviation of the distribution of sample means is so important for inferential statistics that it has a special name, the standard error. The standard error is the standard deviation of the population divided by the square root of the sample size:

S.E. = sigma/√n = 91.6/√139

Note that the standard error depends only on the sample size, not the size of the population.