A population of values has a normal distribution with = 27. sample of size n = 171. What is the mean of the distribution of sample means? H₂ = What is the standard deviation of the distribution of sample m (Report answer accurate to 2 decimal places.)
A population of values has a normal distribution with = 27. sample of size n = 171. What is the mean of the distribution of sample means? H₂ = What is the standard deviation of the distribution of sample m (Report answer accurate to 2 decimal places.)
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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### Understanding Sample Means and Standard Deviation
A population of values has a normal distribution with a mean (μ) of 27.2 and a standard deviation (σ) of 23.1. You intend to draw a random sample of size \( n = 171 \).
#### Questions:
1. **What is the mean of the distribution of sample means?**
\(\mu_{\bar{x}} =\) [Enter your answer here]
2. **What is the standard deviation of the distribution of sample means?**
(Report answer accurate to 2 decimal places)
\(\sigma_{\bar{x}} =\) [Enter your answer here]
---
#### Explanation:
1. **Mean of the Distribution of Sample Means:**
The mean of the sample means (\(\mu_{\bar{x}}\)) is equal to the mean of the population (\(μ\)). Therefore:
\(\mu_{\bar{x}} = μ\)
2. **Standard Deviation of the Distribution of Sample Means:**
The standard deviation of the sample means (\(\sigma_{\bar{x}}\)) is given by the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\sigma\) is the standard deviation of the population.
- \(n\) is the sample size.
Plugging in the values:
\[
\sigma_{\bar{x}} = \frac{23.1}{\sqrt{171}}
\]
Calculate the value and round it to two decimal places for accuracy.
[Enter your calculated answer here]
---
In this exercise, understanding these concepts ensures that when dealing with normal distributions, the sample mean (\(\mu_{\bar{x}}\)) provides a reliable estimate of the population mean (μ), and the sample standard deviation (\(\sigma_{\bar{x}}\)) offers insights into the variability of the sample means around the population mean.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9dd31519-e039-435c-a120-3cd371187144%2Fa022a368-b872-44f4-a596-fb618f95ef03%2Fps1cgza_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Understanding Sample Means and Standard Deviation
A population of values has a normal distribution with a mean (μ) of 27.2 and a standard deviation (σ) of 23.1. You intend to draw a random sample of size \( n = 171 \).
#### Questions:
1. **What is the mean of the distribution of sample means?**
\(\mu_{\bar{x}} =\) [Enter your answer here]
2. **What is the standard deviation of the distribution of sample means?**
(Report answer accurate to 2 decimal places)
\(\sigma_{\bar{x}} =\) [Enter your answer here]
---
#### Explanation:
1. **Mean of the Distribution of Sample Means:**
The mean of the sample means (\(\mu_{\bar{x}}\)) is equal to the mean of the population (\(μ\)). Therefore:
\(\mu_{\bar{x}} = μ\)
2. **Standard Deviation of the Distribution of Sample Means:**
The standard deviation of the sample means (\(\sigma_{\bar{x}}\)) is given by the formula:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\sigma\) is the standard deviation of the population.
- \(n\) is the sample size.
Plugging in the values:
\[
\sigma_{\bar{x}} = \frac{23.1}{\sqrt{171}}
\]
Calculate the value and round it to two decimal places for accuracy.
[Enter your calculated answer here]
---
In this exercise, understanding these concepts ensures that when dealing with normal distributions, the sample mean (\(\mu_{\bar{x}}\)) provides a reliable estimate of the population mean (μ), and the sample standard deviation (\(\sigma_{\bar{x}}\)) offers insights into the variability of the sample means around the population mean.
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