Question

The American Council of Life Insurers provides information about life insurance in force per covered family in the Life Insurers Fact Book. Assume that standard deviation of life insurance in force is \(\(50,900\).

a. Determine the probability that the sampling error made in estimating the population mean life insurance in force by that of a sample of \(500\) covered families will be \(\)2000\) or less.

b. Must you assume that life-insurance amounts are normally distributed in order to answer part (a)? What if the sample size is \(20\) instead of \(500\)?

c. Repeat part (a) for a sample size of \(5000\).

Short answer

Part a. The probability that the sampling error made in estimation of population mean life insurance in force will be \($2000\) or less is \(0.6212\).

Part b. If the sample size were \(20\) (a small sample) instead of \(500\), then it would be necessary to assume normal distribution.

Part c. The probability that the sampling error made in estimation of population mean life insurance in force will be \($2000\) or less is \(0.9946\).

Step-by-step solution

Part a. Step 1. Given information

The standard deviation of life insurance in force is \($50,000\)

Sample size is \(500\).

Part a. Step 2. Calculation

As per the given information,

\(n=500, \mu_{\bar{x}}=\mu, \sigma=50900\)

\(\mu_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{50900}{\sqrt{500}}=2276.32\)

The probability will be given as below -

\(P(\mu-2000\leq \bar{x}\leq \mu+2000)\)

\(z\)-score computation:

\(\bar{x}=\mu-2000\rightarrow z=\frac{(\mu-2000)-\mu}{22.76.32}=-0.88\)

Area less than \((z=-0.88)\) is \(0.1894\)

\(\bar{x}=\mu-2000\rightarrow z=\frac{(\mu+2000)-\mu}{22.76.32}=0.88\)

Area less than \((z=0.88)\) is \(0.8106\).

Therefore, total area \(=0.8106-0.1894=0.6212\)

Hence, the probability that the sampling error made in estimation of population mean life insurance in force will be \($2000\) or less is \(0.6212\).

Part b. Step 1. Calculation

Sample size in part (a) is given as \(500\). This is sufficiently large sample. Therefore, it is not necessary to assume that life-insurance population is normally distributed. Since sample size is large, therefore \(\bar{x}\) is approximately normally distributed regardless of distribution of the population of life insurance amounts.

If the sample size were \(20\) (a small sample) instead of \(500\), then it would be necessary to assume normal distribution.

Part c. Step 1. Calculation

As per the given information,

\(n=5000, \mu_{\bar{x}}=\mu, \sigma=50900\)

\(\mu_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{50900}{\sqrt{5000}}=719.83\)

The probability will be given as below -

\(P(\mu-2000\leq \bar{x}\leq \mu+2000)\)

\(z\)-score computation:

\(\bar{x}=\mu-2000\rightarrow z=\frac{(\mu-2000)-\mu}{719.83}=-2.78\)

Area less than \((z=-2.78)\) is \(0.0027\)

\(\bar{x}=\mu-2000\rightarrow z=\frac{(\mu+2000)-\mu}{719.83}=2.78\)

Area less than \((z=2.78)\) is \(0.9973\).

Therefore, total area\(=0.9973-0.0027=0.9946\)

Hence, the probability that the sampling error made in estimation of population mean life insurance in force will be \($2000\) or less is \(0.9946\).