Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.

a. If we select a random sample of 58 households, what is the standard error of the mean?
b. What is the expected shape of the distribution of the sample mean?
c. What is the likelihood of selecting a sample with a mean of at least $124,000?
d. What is the likelihood of selecting a sample with a mean of more than $112,000?

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

We know the following info:

[tex]\mu = 120000 , \sigma =38000[/tex]

Part a

We select a sample of 58>30 households. From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And the standard error is given by:

[tex] SE= \frac{38000}{\sqrt{58}}= 4989.644[/tex]

Part b

Normal and with the following parameters:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Part c

We can calculate the z score for 124000 and we got:

[tex] z = \frac{124000-120000}{\frac{38000}{\sqrt{58}}}= 0.802[/tex]

So we want this probability:

[tex] P(\bar X >124000) = P(z>0.802) [/tex]

And using the complement rule and the normal standard distribution or excel we got:

[tex] P(\bar X >124000) = P(z>0.802)=1-P(z<0.802) = 1-0.789=0.211[/tex]

Part d

We can calculate the z score for 124000 and we got:

[tex] z = \frac{112000-120000}{\frac{38000}{\sqrt{58}}}= -1.603[/tex]

So we want this probability:

[tex] P(\bar X >112000) = P(z>-1.603) [/tex]

And using the complement rule and the normal standard distribution or excel we got:

[tex] P(\bar X >112000) = P(z>-1.603)=1-P(z<-1.603) = 1-0.0545=0.9455[/tex]