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?

Answer 1

To solve this problem, we will use z-score formula:

z = (x - μ) / σ

where z is the z-score, x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

First, we need to calculate the z-score for the value 65:

z = (65 - 53) / 7.5
z = 12 / 7.5
z = 1.6

Now, we need to find the probability of the z-score being less than 1.6. We can look up the z-score in the standard normal distribution table or use a calculator.

The probability corresponding to a z-score of 1.6 is approximately 0.9452.

Therefore, the probability that a person in Africa will die before the age of 65 is approximately 0.9452.