A simple random sample of size nequals10 is obtained from a population with muequals63 and sigmaequals18. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean? Assuming that this condition is​ true, describe the sampling distribution of x overbar.

Respuesta :

Answer:

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.

Step-by-step explanation:

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

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

In this problem

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

If it is:

[tex]\mu = 63, \sigma = 18, n = 10, s = \frac{18}{\sqrt{10}} = 5.69[/tex]

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.