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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? 94.52% 82.89% 94.95% 88.49%

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? 94.52% 82.89% 94.95% 88.49%

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**Understanding Life Expectancy and Probability Calculations** 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 Choices:** - 94.52% - 82.89% - 94.95% - 88.49% **Explanation:** The problem examines statistical data about life expectancy and requires calculating the probability using the given parameters. Here, the average life expectancy is 53 years, with a standard deviation of 7.5 years. To solve it, you would typically use the Z-score formula and statistical tables or a calculator to find the correct probability for a person dying before the age of 65. **Steps to solve:** 1. Compute the Z-score, which measures how many standard deviations away 65 years is from the average (mean): - \( Z = \frac{X - \mu}{\sigma} \) - \( \mu \) = mean = 53 years - \( \sigma \) = standard deviation = 7.5 years - \( X \) = 65 years \( Z = \frac{65 - 53}{7.5} = \frac{12}{7.5} \approx 1.60 \) 2. Using a Z-table or standard normal distribution calculator, find the probability corresponding to \( Z = 1.60 \). 3. The Z-table gives the area to the left of the Z-value (the probability of living up to 65 years). Subtract this value from 1 to find the probability of dying before 65 years. In practical teaching, the correct answer will be provided by calculations and referencing the standard normal distribution tables or using statistical software. **Correct Answer:** 94.52%
Transcribed Image Text:**Understanding Life Expectancy and Probability Calculations** 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 Choices:** - 94.52% - 82.89% - 94.95% - 88.49% **Explanation:** The problem examines statistical data about life expectancy and requires calculating the probability using the given parameters. Here, the average life expectancy is 53 years, with a standard deviation of 7.5 years. To solve it, you would typically use the Z-score formula and statistical tables or a calculator to find the correct probability for a person dying before the age of 65. **Steps to solve:** 1. Compute the Z-score, which measures how many standard deviations away 65 years is from the average (mean): - \( Z = \frac{X - \mu}{\sigma} \) - \( \mu \) = mean = 53 years - \( \sigma \) = standard deviation = 7.5 years - \( X \) = 65 years \( Z = \frac{65 - 53}{7.5} = \frac{12}{7.5} \approx 1.60 \) 2. Using a Z-table or standard normal distribution calculator, find the probability corresponding to \( Z = 1.60 \). 3. The Z-table gives the area to the left of the Z-value (the probability of living up to 65 years). Subtract this value from 1 to find the probability of dying before 65 years. In practical teaching, the correct answer will be provided by calculations and referencing the standard normal distribution tables or using statistical software. **Correct Answer:** 94.52%
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