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Normal distribution value?

Term Life Insurance & Whole Life Insurance Discussion

Normal distribution value?

Postby fernand » Wed Jul 31, 2013 4:32 pm

A life insurance company wants to estimate their 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 pass away?
fernand
 
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Normal distribution value?

Postby hymie » Wed Jul 31, 2013 4:33 pm

80% is represented by 0.8000 area under the standard normal curve
The z value which separates 0.8000 area from the rest (1.0000 - 0.8000 = 0.2000) is + 0.84 approximately
z = (X-Mean)/SD
X = Mean + (z*SD)
= 68 + (0.84*4)
= 71.36
80% of the plan participants will pass away by 71.36 years.
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Normal distribution value?

Postby aetos » Wed Jul 31, 2013 4:37 pm

Using a normal distribution table, you can see that .3 from the mean (meaning .5 + .3 = .8 or 80%) is .8425 standard deviation from the mean.

68 + .8425*4 = 71.37

80% are dead by the age of 71.37

This is probably the answer, considering how my work is usually impeccable.
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Normal distribution value?

Postby hwertun52 » Wed Jul 31, 2013 4:49 pm

Using a normal distribution table, you can see that .3 from the mean (meaning .5 + .3 = .8 or 80%) is .8425 standard deviation from the mean.

68 + .8425*4 = 71.37

80% are dead by the age of 71.37
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Joined: Tue Sep 18, 2012 11:19 am


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