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

94.52%

82.89%

94.95%

88.49%

Answer 1

To solve this problem, we need to calculate the z-score and then use a z-table to find the corresponding probability.

The z-score formula is given by:
z = (x - μ) / σ

where:
- x is the value we want to find the probability for (in this case, 65)
- μ is the mean (average) life expectancy in Africa (53)
- σ is the standard deviation (7.5)

Plugging in the values:

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

Next, we use a z-table or a calculator to find the cumulative probability for a z-score of 1.6. The cumulative probability (P) is the probability that a randomly selected person in Africa will die before the age of 65.

From the z-table or calculator, we find that the cumulative probability for a z-score of 1.6 is approximately 0.9452 or 94.52%.

Therefore, the correct answer is 94.52%.