Question

a company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. suppose the probability distribution of the lifetimes of the participants in the plan is approximately normal with mean equal to 66.3 years and standard deviation equal to 3.6 years. nine participants are randomly selected. what is the probability that the average age at death of these nine participants will exceed 68 years

Answer 1

Answer:

0.0778 = 7.78% probability that the average age at death of these nine participants will exceed 68 years

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Approximately normal with mean equal to 66.3 years and standard deviation equal to 3.6 years.

This means that $\mu = 66.3, \sigma = 3.6$

Sample of 9:

This means that $n = 9, s = \frac{3.6}{\sqrt{9}} = 1.2$

What is the probability that the average age at death of these nine participants will exceed 68 years?

This is 1 subtracted by the pvalue of Z when X = 68. So

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

By the Central Limit Theorem

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

$Z = \frac{68 - 66.3}{1.2}$

$Z = 1.42$

$Z = 1.42$ has a pvalue of 0.9222

1 - 0.9222 = 0.0778

0.0778 = 7.78% probability that the average age at death of these nine participants will exceed 68 years