Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $165,000. This distribution follows the normal distribution with a standard deviation of $40,000.

If we select a random sample of 50 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)

What is the expected shape of the distribution of the sample mean?

What is the likelihood of selecting a sample with a mean of at least $167,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

What is the likelihood of selecting a sample with a mean of more than $155,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Find the likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

We have given,

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $165,000. This distribution follows the normal distribution with a standard deviation of $40,000.

a)  the standard error of the mean

$=\frac{\sigma }{\sqrt{n}}$

$=\frac{40000 }{\sqrt{50}}$

=5656.85

$\approx 5657$

b) the expected shape of the distribution of the sample mean = Normal.

c)  the likelihood of selecting a sample with a mean of at least $167,000

$=P\left [ \bar{x}\geq 167000 \right ]$

$=P\left [ Z\geq \frac{167000-165000}{\frac{40000}{\sqrt{50}}} \right ]$

$=P\left [ Z\geq 0.35 \right ]$

$=1-P\left [ Z< 0.35 \right ]$

=1-0.6368..................................................by normal probability table.

=0.3632

d)  the likelihood of selecting a sample with a mean of more than $155,000

$=P\left [ \bar{x}> \frac{155000-165000}{\frac{40000}{\sqrt{50}}} \right ]$

=P[Z>-1.77]

$=1-P[Z\leq -1.77]$

=1-0.0384.........................by using normal probability table.

=0.9616

e) the likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000.

$=P\left [ \frac{155000-165000}{\frac{40000}{\sqrt{50}}}<\bar{x}< \frac{167000-165000}{\frac{40000}{\sqrt{50}}} \right ]$

=P[-1.77<Z<0.35]

$=P[Z\leq 0.35]-P[Z\leq -1.77]$

=0.6368-0.0384.................by using normal probability table.

=0.5984