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
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0dca65e3-5d84-435b-a40e-c7b10d92b1e9%2F238f95cb-cd93-4c8c-93d9-e8b2a5e905b7%2F9t3iym7_processed.jpeg&w=3840&q=75)
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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