91Ó°ÊÓ

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

Sample size is \(500\).

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 \(0.6212\).

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.

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 \(0.9946\).