Question
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?
Answers
GPT 3.5
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.
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.