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2: (2>60) (2480) 5. A The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. a. What proportion of this population is between 60 and 80? 2 = -2.5 = 6.21 2= 60-70 4 2-80-70 4 2= = 2.5=6.21 71.5-45 4 ماما - 49379 1-,99379 -3 O Stry b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75? 2= 68.5-45 5.87 4 -4 = 6.21% +6.21= 12.2400 is between 60-80 Years old Po

2: (2>60) (2480) 5. A The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. a. What proportion of this population is between 60 and 80? 2 = -2.5 = 6.21 2= 60-70 4 2-80-70 4 2= = 2.5=6.21 71.5-45 4 ماما - 49379 1-,99379 -3 O Stry b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75? 2= 68.5-45 5.87 4 -4 = 6.21% +6.21= 12.2400 is between 60-80 Years old Po

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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## Problem 5: Statistical Analysis of a Retirement Community The problem involves analyzing the ages of thousands of residents in a retirement community, which are normally distributed with a mean of 70 years and a standard deviation of 4 years. ### a. Proportion of Population Between 60 and 80 To find the proportion of the population between ages 60 and 80: 1. Calculate the Z-scores for ages 60 and 80. - \( Z = \frac{60 - 70}{4} = -2.5 \) - \( Z = \frac{80 - 70}{4} = 2.5 \) 2. Using the Z-table, determine the proportion for each Z-score. - For \( Z = -2.5 \), approximately \( 0.0062 \) (or 0.62%) - For \( Z = 2.5 \), approximately \( 0.9938 \) (or 99.38%) 3. The proportion of residents between ages 60 and 80 is: - \( 0.9938 - 0.0062 = 0.9876 \) (or 98.76%) ### b. Probability of Sample Mean Age Given that a sample of 45 residents is chosen at random: 1. Calculate the Z-scores for the sample mean ages 68.5 and 71.75. - \( Z = \frac{68.5 - 70}{\frac{4}{\sqrt{45}}} \approx -2.24 \) - \( Z = \frac{71.75 - 70}{\frac{4}{\sqrt{45}}} \approx 2.92 \) The probability that the sample mean is between 68.5 and 71.75 can be determined similarly using the Z-table values for each Z-score. ### c. 95% Confidence Interval for the Sample Means The task is to find the limits within which 95% of the possible sample mean values lie. This typically corresponds to Z-scores of -1.96 and 1.96 for a normal distribution. The interval can be calculated using these Z-scores and the standard error of the mean. Note: Handwritten corrections and calculations in red mark the numerical answers, but further numerical accuracy might require precise Z-table lookups and calculations.
Transcribed Image Text:## Problem 5: Statistical Analysis of a Retirement Community The problem involves analyzing the ages of thousands of residents in a retirement community, which are normally distributed with a mean of 70 years and a standard deviation of 4 years. ### a. Proportion of Population Between 60 and 80 To find the proportion of the population between ages 60 and 80: 1. Calculate the Z-scores for ages 60 and 80. - \( Z = \frac{60 - 70}{4} = -2.5 \) - \( Z = \frac{80 - 70}{4} = 2.5 \) 2. Using the Z-table, determine the proportion for each Z-score. - For \( Z = -2.5 \), approximately \( 0.0062 \) (or 0.62%) - For \( Z = 2.5 \), approximately \( 0.9938 \) (or 99.38%) 3. The proportion of residents between ages 60 and 80 is: - \( 0.9938 - 0.0062 = 0.9876 \) (or 98.76%) ### b. Probability of Sample Mean Age Given that a sample of 45 residents is chosen at random: 1. Calculate the Z-scores for the sample mean ages 68.5 and 71.75. - \( Z = \frac{68.5 - 70}{\frac{4}{\sqrt{45}}} \approx -2.24 \) - \( Z = \frac{71.75 - 70}{\frac{4}{\sqrt{45}}} \approx 2.92 \) The probability that the sample mean is between 68.5 and 71.75 can be determined similarly using the Z-table values for each Z-score. ### c. 95% Confidence Interval for the Sample Means The task is to find the limits within which 95% of the possible sample mean values lie. This typically corresponds to Z-scores of -1.96 and 1.96 for a normal distribution. The interval can be calculated using these Z-scores and the standard error of the mean. Note: Handwritten corrections and calculations in red mark the numerical answers, but further numerical accuracy might require precise Z-table lookups and calculations.
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