Question

A population has a mean of 75 and a standard deviation of 8. A random sample of 800 is selected. The expected value of LaTeX: \bar{x}x ¯ is
a.8

b.75

c.800

d.None of these alternatives is correct.

Answer 1

Answer:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we find the expected value for this statistic we got:

$E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n})= \frac{\sum_{i=1}^n E(X_i)}{n}$

And assuming that each observation have the same expected value $\mu$ we got:

$E(\bar X) = \frac{n \mu}{n}=\mu$

So then the expected value for the sample mean would be:

b.75

Step-by-step explanation:

For this case we have the information about the population:

$\mu = 75$ represent the mean

$\sigma = 8$ represent the deviation

We select a sample size of n = 800, a very large sample indeed. We know that the sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we find the expected value for this statistic we got:

$E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n})= \frac{\sum_{i=1}^n E(X_i)}{n}$

And assuming that each observation have the same expected value $\mu$ we got:

$E(\bar X) = \frac{n \mu}{n}=\mu$

So then the expected value for the sample mean would be:

b.75

Answer 2

Answer:

b.75

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem:

Mean of the population is 75.

By the Central Limit Theorem,

The mean of the sample, $\bar{x}$, is expected to be also 75.

So the correct answer is:

b.75