What is the probability of a male living more than 86 years in the U.S.?

AI Thread Summary
The discussion focuses on calculating the probability of a male living more than 86 years in the U.S., given a normal distribution with a mean of 80 years and a standard deviation of 5 years. The z-score is calculated as 1.2, which corresponds to a cumulative probability of approximately 0.8849. By subtracting this value from 1, the probability of living beyond 86 years is found to be 0.1151. The calculation appears to be correct, and participants are confirming the accuracy of the solution. This analysis highlights the application of normal distribution in determining life expectancy probabilities.
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Homework Statement


Assume the life expectancy of U.S. males is normally distributed with a mean of 80 years and a standard deviation of 5 years. What is the probability that a randomly selected male lives more than 86 years?


Homework Equations


(work shown below)


The Attempt at a Solution



z=(86-80)/(5) = 6/5 = 1.2
P=1 - .8849 = 0.1151
 
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Looks fine. Just checking if you did it right?
 
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