Question

A life insurance company wants to estimate its annual payouts. Assume that the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 4 years. By what age have 80% of the plan participants passed away?

Answer 1

Answer:

By 71 years of age 80% of the plan participants have passed away.

Step-by-step explanation:

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 68 years and a standard deviation of 4 years.

This means that $\mu = 68, \sigma = 4$

By what age have 80% of the plan participants passed away?

By the 80th percentile of ages, which is X when Z has a p-value of 0.8, so X when Z = 0.84.

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

$0.84 = \frac{X - 68}{4}$

$X - 68 = 4*0.84$

$X = 71$

By 71 years of age 80% of the plan participants have passed away.