A recent study found that the average life expectancy of a person living in Africa is 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?

Question

Nadia David

Mathematics

1 January, 21:42

A recent study found that the average life expectancy of a person living in Africa is 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?

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Answers (1)

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Javon Vaughn · Answer 1

0.945200708

Step-by-step explanation:

In this question, we are asked to calculate the probability that a person selected at random would live less than 65.

To solve this problem, what we need to do is to first calculate the z-score.

Mathematically, z = (x-mean) / standard deviation

According to the question, the mean is 53 years and the standard deviation is 7.5

Substituting these values in the z-score equation:

Z-score = (65-53) / 7.5 = 12/7.5 = 1.6

We proceed to use the z-score table to obtain the probability. Using the table, z score of 1.6 corresponds to a probability of

0.945200708