Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000. a. If we select a random sample of 50 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 $112,000? d. What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

Answer 1

Answer:

a) $5,656.85

b) Bell-shaped(normally distributed).

c) 36.32% probability of selecting a sample with a mean of at least $112,000.

d) 96.16% probability of selecting a sample with a mean of more than $100,000.

e) 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.

Step-by-step explanation:

To solve this question, it is important to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size, of size at least 30, can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 110000, \sigma = 40000$

a. If we select a random sample of 50 households, what is the standard error of the mean?

This is the standard deviation of the sample, that is, s, when $n = 50$.

So

$s = \frac{\sigma}{\sqrt{n}} = \frac{40000}{\sqrt{50}} = 5656.85$

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

By the Central Limit Theorem, bell-shaped(normally distributed).

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

This is 1 subtracted by the pvalue of Z when X = 112000. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{112000 - 110000}{5656.85}$

$Z = 0.35$

$Z = 0.35$ has a pvalue of 0.6368

So 1-0.6368 = 0.3632 = 36.32% probability of selecting a sample with a mean of at least $112,000.

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

This is 1 subtracted by the pvalue of Z when X = 112000. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{100000 - 110000}{5656.85}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

So 1-0.0384 = 0.9616 = 96.16% probability of selecting a sample with a mean of more than $100,000.

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

This is the pvalue of Z when X = 112000 subtractex by the pvalue of Z when X = 100000.

So

X = 112000

$Z = \frac{X - \mu}{s}$

$Z = \frac{112000 - 110000}{5656.85}$

$Z = 0.35$

$Z = 0.35$ has a pvalue of 0.6368

X = 100000

$Z = \frac{X - \mu}{s}$

$Z = \frac{100000 - 110000}{5656.85}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

So 0.6368 - 0.0384 = 0.5984 = 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.