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A population of values has a normal distribution with μ = 133 and o= 48.9. You intend to draw a random sample of size n = 18. Find the probability that a single randomly selected value is greater than 123.8. P(X> 123.8)=5746 Find the probability that a sample of size n = 18 is randomly selected with a mean greater than 123.8. P(M> 123.8) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z- scores rounded to 3 decimal places are accepted. Question Help: Video Post to forum Submit Question

A population of values has a normal distribution with μ = 133 and o= 48.9. You intend to draw a random sample of size n = 18. Find the probability that a single randomly selected value is greater than 123.8. P(X> 123.8)=5746 Find the probability that a sample of size n = 18 is randomly selected with a mean greater than 123.8. P(M> 123.8) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z- scores rounded to 3 decimal places are accepted. Question Help: Video Post to forum Submit Question

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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### Probability and Statistics Exercise **Problem Statement:** A population of values has a normal distribution with \(\mu = 133\) and \(\sigma = 48.9\). You intend to draw a random sample of size \(n = 18\). **Questions:** 1. **Find the probability that a single randomly selected value is greater than 123.8.** \( P(X > 123.8) = 0.5746 \) 2. **Find the probability that a sample of size \(n = 18\) is randomly selected with a mean greater than 123.8.** \( P(M > 123.8) = \) **Instructions:** - Enter your answers as numbers accurate to 4 decimal places. - Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. **Answer Input:** - Probability 1: Already provided as 0.5746. - Probability 2: To be calculated. **Question Help:** - [Video](#) - [Post to forum](#) \[ \text{Submit Question} \] --- **Explanation of Calculation:** For Question 1, the provided probability is \( P(X > 123.8) = 0.5746 \), indicating this is a calculated value based on the given normal distribution parameters. For Question 2, you need to adjust for the sample size \(n = 18\), and calculate the probability that the sample mean is greater than 123.8. The formula to find this involves the standard error of the mean, calculated as follows: \[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{48.9}{\sqrt{18}} \] Then use the z-score formula for the sample mean to find the probability: \[ z = \frac{M - \mu}{SE} \] Where \(M\) is the sample mean. Ensure your final answer is rounded to 4 decimal places.
Transcribed Image Text:### Probability and Statistics Exercise **Problem Statement:** A population of values has a normal distribution with \(\mu = 133\) and \(\sigma = 48.9\). You intend to draw a random sample of size \(n = 18\). **Questions:** 1. **Find the probability that a single randomly selected value is greater than 123.8.** \( P(X > 123.8) = 0.5746 \) 2. **Find the probability that a sample of size \(n = 18\) is randomly selected with a mean greater than 123.8.** \( P(M > 123.8) = \) **Instructions:** - Enter your answers as numbers accurate to 4 decimal places. - Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. **Answer Input:** - Probability 1: Already provided as 0.5746. - Probability 2: To be calculated. **Question Help:** - [Video](#) - [Post to forum](#) \[ \text{Submit Question} \] --- **Explanation of Calculation:** For Question 1, the provided probability is \( P(X > 123.8) = 0.5746 \), indicating this is a calculated value based on the given normal distribution parameters. For Question 2, you need to adjust for the sample size \(n = 18\), and calculate the probability that the sample mean is greater than 123.8. The formula to find this involves the standard error of the mean, calculated as follows: \[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{48.9}{\sqrt{18}} \] Then use the z-score formula for the sample mean to find the probability: \[ z = \frac{M - \mu}{SE} \] Where \(M\) is the sample mean. Ensure your final answer is rounded to 4 decimal places.
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