**Transcription for Educational Website**---**Problem 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?**\[Z = \frac{X - \mu}{\sigma}\]- For age 60: \[ Z = \frac{60 - 70}{4} = -2.5 \approx -6.21 \]- For age 80: \[ Z = \frac{80 - 70}{4} = 2.5 \approx 6.21 \]Calculation: 1 - (Probability of Z < -6.21 + Probability of Z > 6.21) = 0.99379Proportion = 6.21% + 6.21% = 12.24%**Conclusion:** 12.24% of the population is between 60 and 80 years old.---**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?**\[Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}\]- For age 68.5: \[ Z = \frac{68.5 - 70}{4/\sqrt{45}} = -5.87 \]- For age 71.75: \[ Z = \frac{71.75 - 70}{4/\sqrt{45}} = 6.6 \]Diagrams and detailed calculations show that this configuration results in a probability of a sample mean being in this range, with students marked -3 for incorrect graph plotting and -4 for calculation errors. **Note:** Exact Z scores, such as -6.21, indicate more precise computations and rounding in context.

Given,Mean(μ)=70standard deviation(σ)=4Sample size(n)=45

Answered: 2= (>>60) (z<80) 5. A The ages of the…

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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