Insurance companies track life expectancy information to assist in determining the
cost of life insurance policies. Last year the average life expectancy of all policyholders
was 77 years. ABI Insurance wants to determine if their clients now have a longer life
expectancy, on average, so they randomly sample some of their recently paid policies.
The ages of the clients in the sample are shown below. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | }
\hline 86 & 75 & 83 & 84 & 81 & 77 & 78 & 79 & 79 & 81 \
\hline 76 & 85 & 70 & 76 & 79 & 81 & 73 & 74 & 72 & 83 \
\hline
\end{array}$$
A. Based on the sample results, find the 90% confidence interval and interpret.
b. For more accurate cost determination, ABI Insurance wants to estimate the average life
expectancy to within one year with 95% confidence. How many randomly selected
recently paid policies would they need to sample?
c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7
years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret.

Question

Insurance companies track life expectancy information to assist in determining the
cost of life insurance policies. Last year the average life expectancy of all policyholders
was 77 years. ABI Insurance wants to determine if their clients now have a longer life
expectancy, on average, so they randomly sample some of their recently paid policies.
The ages of the clients in the sample are shown below. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | } 
\hline 86 & 75 & 83 & 84 & 81 & 77 & 78 & 79 & 79 & 81 \
\hline 76 & 85 & 70 & 76 & 79 & 81 & 73 & 74 & 72 & 83 \
\hline
\end{array}$$ 
A. Based on the sample results, find the 90% confidence interval and interpret.
b. For more accurate cost determination, ABI Insurance wants to estimate the average life
expectancy to within one year with 95% confidence. How many randomly selected
recently paid policies would they need to sample?
c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7
years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret.

Quizplus · Accepted Answer

The Answer of Insurance companies track life expectancy information to assist in determining the
cost of life insurance policies. Last year the average life expectancy of all policyholders
was 77 years. ABI Insurance wants to determine if their clients now have a longer life
expectancy, on average, so they randomly sample some of their recently paid policies.
The ages of the clients in the sample are shown below. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | } 
\hline 86 & 75 & 83 & 84 & 81 & 77 & 78 & 79 & 79 & 81 \
\hline 76 & 85 & 70 & 76 & 79 & 81 & 73 & 74 & 72 & 83 \
\hline
\end{array}$$ 
A. Based on the sample results, find the 90% confidence interval and interpret.
b. For more accurate cost determination, ABI Insurance wants to estimate the average life
expectancy to within one year with 95% confidence. How many randomly selected
recently paid policies would they need to sample?
c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7
years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret.

Insurance Companies Track Life Expectancy Information to Assist in Determining