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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?

User Gareth Latty
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1 Answer

12 votes
12 votes

Answer:

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

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 = (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 = (X - \mu)/(\sigma)


0.84 = (X - 68)/(4)


X - 68 = 4*0.84


X = 71

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

User Fbjorn
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