The population mean and standard deviation are given below. Find the indicated probability and determine whether a sample mean in the given range below would be considered unusual. If convenient, use technology to find the probability. For a sample of n = 31, find the probability of a sample mean being less than 12,750 or greater than 12,753 when u = 12,750 and o= 1.6. ..... For the given sample, the probability of a sample mean being less than 12,750 or greater than 12,753 is. (Round to four decimal places as needed.)
The population mean and standard deviation are given below. Find the indicated probability and determine whether a sample mean in the given range below would be considered unusual. If convenient, use technology to find the probability. For a sample of n = 31, find the probability of a sample mean being less than 12,750 or greater than 12,753 when u = 12,750 and o= 1.6. ..... For the given sample, the probability of a sample mean being less than 12,750 or greater than 12,753 is. (Round to four decimal places as needed.)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Educational Content: Determining Probability of Sample Means**
In this exercise, we are given the population mean (\( \mu \)) and the standard deviation (\( \sigma \)) to find the probability of certain sample means. We will determine if these sample means would be considered unusual.
**Problem Statement:**
The population parameters are:
- Mean (\( \mu \)) = 12,750
- Standard deviation (\( \sigma \)) = 1.6
For a sample size of \( n = 31 \), we are tasked with finding the probability of a sample mean being either:
1. Less than 12,750
2. Greater than 12,753
**Steps to Solve:**
1. **Calculate the Standard Error (SE):**
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{1.6}{\sqrt{31}}
\]
2. **Find the Z-scores:**
- For \( x = 12,750 \)
\[
Z = \frac{x - \mu}{SE}
\]
- For \( x = 12,753 \)
\[
Z = \frac{x - \mu}{SE}
\]
3. **Determine the Probability:**
- Use the Z-score table or technology to find the probabilities corresponding to the calculated Z-scores.
4. **Evaluation:**
- Compare the probabilities to typical cutoff values (e.g., 0.05) to determine if the sample mean is unusual.
**Note:**
After calculations, ensure that probabilities are rounded to four decimal places as needed.
This approach uses the Central Limit Theorem, which allows us to assume a normal distribution of the sample means when the sample size is large (n > 30).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdabeff17-6e30-4e7b-b604-e664e91c4527%2F510d9473-6f2c-4a90-9c83-fff6afbf1acc%2Fzr78rhi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Determining Probability of Sample Means**
In this exercise, we are given the population mean (\( \mu \)) and the standard deviation (\( \sigma \)) to find the probability of certain sample means. We will determine if these sample means would be considered unusual.
**Problem Statement:**
The population parameters are:
- Mean (\( \mu \)) = 12,750
- Standard deviation (\( \sigma \)) = 1.6
For a sample size of \( n = 31 \), we are tasked with finding the probability of a sample mean being either:
1. Less than 12,750
2. Greater than 12,753
**Steps to Solve:**
1. **Calculate the Standard Error (SE):**
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{1.6}{\sqrt{31}}
\]
2. **Find the Z-scores:**
- For \( x = 12,750 \)
\[
Z = \frac{x - \mu}{SE}
\]
- For \( x = 12,753 \)
\[
Z = \frac{x - \mu}{SE}
\]
3. **Determine the Probability:**
- Use the Z-score table or technology to find the probabilities corresponding to the calculated Z-scores.
4. **Evaluation:**
- Compare the probabilities to typical cutoff values (e.g., 0.05) to determine if the sample mean is unusual.
**Note:**
After calculations, ensure that probabilities are rounded to four decimal places as needed.
This approach uses the Central Limit Theorem, which allows us to assume a normal distribution of the sample means when the sample size is large (n > 30).
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