A random sample of n = 64 scores is obtained from a normal population with μ = 30 and o= 10. What is the probability that the sample mean will be greater than M = 31?
A random sample of n = 64 scores is obtained from a normal population with μ = 30 and o= 10. What is the probability that the sample mean will be greater than M = 31?
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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![**Calculating the Probability of the Sample Mean Being Greater Than a Given Value**
Consider a scenario where a random sample of size \( n = 64 \) scores is obtained from a normal population with:
- Population mean \( \mu = 30 \)
- Population standard deviation \( \sigma = 10 \)
We are interested in determining the probability that the sample mean \( \bar{X} \) will be greater than \( M = 31 \).
**Solution Steps:**
1. **Identify the Known Values:**
- Sample size, \( n = 64 \)
- Population mean, \( \mu = 30 \)
- Population standard deviation, \( \sigma = 10 \)
- Value to compare with the sample mean, \( M = 31 \)
2. **Calculate the Standard Error of the Mean (SEM):**
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{64}} = \frac{10}{8} = 1.25
\]
3. **Convert the Problem to a Standard Normal Distribution (Z-Score):**
\[
Z = \frac{M - \mu}{\text{SEM}} = \frac{31 - 30}{1.25} = \frac{1}{1.25} = 0.8
\]
4. **Find the Probability Using the Standard Normal Distribution:**
- Use a Z-table or standard normal distribution calculator to find the probability corresponding to \( Z = 0.8 \).
5. **Determine the Probability:**
- The standard normal table indicates that the cumulative probability associated with \( Z = 0.8 \) is approximately 0.7881.
- Thus, the probability that the sample mean is greater than 31 is:
\[
P(\bar{X} > 31) = 1 - P(Z \leq 0.8) = 1 - 0.7881 = 0.2119
\]
Hence, the probability that the sample mean will be greater than 31 is approximately 0.2119, or 21.19%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f48232e-2a28-4b7d-af98-51d5c912a267%2F49198419-35c0-4e91-aea1-03730a6e82c2%2Fvi28dhn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating the Probability of the Sample Mean Being Greater Than a Given Value**
Consider a scenario where a random sample of size \( n = 64 \) scores is obtained from a normal population with:
- Population mean \( \mu = 30 \)
- Population standard deviation \( \sigma = 10 \)
We are interested in determining the probability that the sample mean \( \bar{X} \) will be greater than \( M = 31 \).
**Solution Steps:**
1. **Identify the Known Values:**
- Sample size, \( n = 64 \)
- Population mean, \( \mu = 30 \)
- Population standard deviation, \( \sigma = 10 \)
- Value to compare with the sample mean, \( M = 31 \)
2. **Calculate the Standard Error of the Mean (SEM):**
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{64}} = \frac{10}{8} = 1.25
\]
3. **Convert the Problem to a Standard Normal Distribution (Z-Score):**
\[
Z = \frac{M - \mu}{\text{SEM}} = \frac{31 - 30}{1.25} = \frac{1}{1.25} = 0.8
\]
4. **Find the Probability Using the Standard Normal Distribution:**
- Use a Z-table or standard normal distribution calculator to find the probability corresponding to \( Z = 0.8 \).
5. **Determine the Probability:**
- The standard normal table indicates that the cumulative probability associated with \( Z = 0.8 \) is approximately 0.7881.
- Thus, the probability that the sample mean is greater than 31 is:
\[
P(\bar{X} > 31) = 1 - P(Z \leq 0.8) = 1 - 0.7881 = 0.2119
\]
Hence, the probability that the sample mean will be greater than 31 is approximately 0.2119, or 21.19%.
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