The average score for games played in the NFL is 21.2 and the standard deviation is 9.3 points. 17 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of ? ~ N( 21.2 2.2556) O or b. What is the distribution of Σx? [x - N( 360.4 ✔ 38.344) or or c. P(> 19.1166) = 0.8212
The average score for games played in the NFL is 21.2 and the standard deviation is 9.3 points. 17 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of ? ~ N( 21.2 2.2556) O or b. What is the distribution of Σx? [x - N( 360.4 ✔ 38.344) or or c. P(> 19.1166) = 0.8212
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![# NFL Game Scores: Statistical Analysis
## Problem Description
The average score for games played in the NFL is 21.2 and the standard deviation is 9.3 points. We randomly select 17 games. All answers should be rounded to four decimal places where possible, and we assume a normal distribution.
## Solutions and Explanations
### a. Distribution of the Sample Mean
**Question:** What is the distribution of \(\bar{x}\)?
**Answer:** \(\bar{x} \sim N(21.2, \frac{9.3}{\sqrt{17}})\)
- Mean (\(\mu_{\bar{x}}\)) = 21.2
- Standard Error (\(\sigma_{\bar{x}}\)) = \( \frac{9.3}{\sqrt{17}} = 2.2556 \)
### b. Distribution of the Sum
**Question:** What is the distribution of \(\sum x\)?
To determine this, we multiply the sample mean and standard deviation by the sample size (17).
**Answer:** \(\sum x \sim N(360.4, \sqrt{17} \cdot 9.3)\)
- Mean (\(\mu_{\sum x}\)) = \( 17 \cdot 21.2 = 360.4 \)
- Standard Deviation (\(\sigma_{\sum x}\)) = \( \sqrt{17} \cdot 9.3 = 38.344 \)
### c. Probability Calculation
**Question:** What is the probability that the sample mean is greater than 19.1166?
**Calculation:**
\[ P(\bar{x} > 19.1166) \]
**Answer:** 0.8212
### d. 79th Percentile for the Mean Score
**Question:** Find the 79th percentile for the mean score for this sample size.
**Calculation:**
Using the Z-score table for the 79th percentile (\( z = 0.8133 \)):
\[ \mu_{\bar{x}} + (z \cdot \sigma_{\bar{x}}) \]
\[ 21.2 + (0.8133 \cdot 2.2556) = 23.018 \]
**Answer:** 23.018
### e. Probability of Sample Mean in Range
**Question:** What is the probability that the sample mean is between 19.8166](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb954a688-5e54-4f96-9877-503883322f61%2F7be8990b-5878-42fd-91b9-826a320b47fc%2Fuf5718n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# NFL Game Scores: Statistical Analysis
## Problem Description
The average score for games played in the NFL is 21.2 and the standard deviation is 9.3 points. We randomly select 17 games. All answers should be rounded to four decimal places where possible, and we assume a normal distribution.
## Solutions and Explanations
### a. Distribution of the Sample Mean
**Question:** What is the distribution of \(\bar{x}\)?
**Answer:** \(\bar{x} \sim N(21.2, \frac{9.3}{\sqrt{17}})\)
- Mean (\(\mu_{\bar{x}}\)) = 21.2
- Standard Error (\(\sigma_{\bar{x}}\)) = \( \frac{9.3}{\sqrt{17}} = 2.2556 \)
### b. Distribution of the Sum
**Question:** What is the distribution of \(\sum x\)?
To determine this, we multiply the sample mean and standard deviation by the sample size (17).
**Answer:** \(\sum x \sim N(360.4, \sqrt{17} \cdot 9.3)\)
- Mean (\(\mu_{\sum x}\)) = \( 17 \cdot 21.2 = 360.4 \)
- Standard Deviation (\(\sigma_{\sum x}\)) = \( \sqrt{17} \cdot 9.3 = 38.344 \)
### c. Probability Calculation
**Question:** What is the probability that the sample mean is greater than 19.1166?
**Calculation:**
\[ P(\bar{x} > 19.1166) \]
**Answer:** 0.8212
### d. 79th Percentile for the Mean Score
**Question:** Find the 79th percentile for the mean score for this sample size.
**Calculation:**
Using the Z-score table for the 79th percentile (\( z = 0.8133 \)):
\[ \mu_{\bar{x}} + (z \cdot \sigma_{\bar{x}}) \]
\[ 21.2 + (0.8133 \cdot 2.2556) = 23.018 \]
**Answer:** 23.018
### e. Probability of Sample Mean in Range
**Question:** What is the probability that the sample mean is between 19.8166
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