A recent study found that the life expectancy of a people living in Africa is normally distributed with an average of 53 years with a standard deviation of 7.5 years. If a person in Africa is selected at random, what is the probability that the person will die before the age of 65?

the probability that the person will die before the age of 65 is 0.9452.

= P(z <1.6 )

Now here we Using z table

So,  

= 0.9452

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P(X<65)=0.9452

Step-by-step explanation:

This is a normal distribution problem.

-Given the mean age is 53 years and the standard deviation is 75, the probability of dying before age 65 is calculated as:

#We check the value of z=1.60 on the z table:

Hence, the probability of dying before 65 is 0.9452

207/8 is the right

the answer is c.

step-by-step explanation: