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Let's go through the steps to find the point estimate of the mean fan rating and the standard deviation for the given sample of NFL games ratings.
### Step-by-Step Solution:
#### Part (a): Developing a Point Estimate of the Mean Fan Rating
1. List the Ratings:
The fan ratings for the random sample of 12 games are:
[tex]\[ 57, 61, 86, 74, 72, 73, 20, 57, 80, 79, 83, 74 \][/tex]
2. Calculate the Mean:
To find the mean (average) rating, sum all the ratings and then divide by the number of ratings.
[tex]\[ \text{Mean} = \frac{57 + 61 + 86 + 74 + 72 + 73 + 20 + 57 + 80 + 79 + 83 + 74}{12} \][/tex]
[tex]\[ \text{Sum of Ratings} = 816 \][/tex]
[tex]\[ \text{Mean} = \frac{816}{12} = 68 \][/tex]
3. Conclusion:
The point estimate of the mean fan rating for the population of NFL games is 68.
#### Part (b): Developing a Point Estimate of the Standard Deviation
1. List the Ratings (from part (a)):
[tex]\[ 57, 61, 86, 74, 72, 73, 20, 57, 80, 79, 83, 74 \][/tex]
2. Calculate the Mean (already calculated as 68 from part (a)).
3. Calculate the Deviations from the Mean:
Subtract the mean from each rating to find the deviations.
[tex]\[ \begin{aligned} &57 - 68 = -11, \quad 61 - 68 = -7, \quad 86 - 68 = 18, \quad 74 - 68 = 6, \\ &72 - 68 = 4, \quad 73 - 68 = 5, \quad 20 - 68 = -48, \quad 57 - 68 = -11, \\ &80 - 68 = 12, \quad 79 - 68 = 11, \quad 83 - 68 = 15, \quad 74 - 68 = 6 \end{aligned} \][/tex]
4. Square the Deviations:
[tex]\[ \begin{aligned} &(-11)^2 = 121, \quad (-7)^2 = 49, \quad 18^2 = 324, \quad 6^2 = 36, \\ &4^2 = 16, \quad 5^2 = 25, \quad (-48)^2 = 2304, \quad (-11)^2 = 121, \\ &12^2 = 144, \quad 11^2 = 121, \quad 15^2 = 225, \quad 6^2 = 36 \end{aligned} \][/tex]
5. Sum the Squared Deviations:
[tex]\[ 121 + 49 + 324 + 36 + 16 + 25 + 2304 + 121 + 144 + 121 + 225 + 36 = 3522 \][/tex]
6. Calculate the Sample Variance:
Divide the sum of squared deviations by the sample size minus one (n-1), where n is the number of ratings.
[tex]\[ \text{Variance} = \frac{3522}{11} \approx 320.1818 \][/tex]
7. Calculate the Standard Deviation:
Take the square root of the variance to find the standard deviation.
[tex]\[ \text{Standard Deviation} = \sqrt{320.1818} \approx 17.8936 \][/tex]
8. Conclusion:
The point estimate of the standard deviation for the population of NFL games is approximately 17.8936 (to four decimal places).
### Final Answers:
- (a) The point estimate of the mean fan rating for the population of NFL games is 68.
- (b) The point estimate of the standard deviation for the population of NFL games is 17.8936.
### Step-by-Step Solution:
#### Part (a): Developing a Point Estimate of the Mean Fan Rating
1. List the Ratings:
The fan ratings for the random sample of 12 games are:
[tex]\[ 57, 61, 86, 74, 72, 73, 20, 57, 80, 79, 83, 74 \][/tex]
2. Calculate the Mean:
To find the mean (average) rating, sum all the ratings and then divide by the number of ratings.
[tex]\[ \text{Mean} = \frac{57 + 61 + 86 + 74 + 72 + 73 + 20 + 57 + 80 + 79 + 83 + 74}{12} \][/tex]
[tex]\[ \text{Sum of Ratings} = 816 \][/tex]
[tex]\[ \text{Mean} = \frac{816}{12} = 68 \][/tex]
3. Conclusion:
The point estimate of the mean fan rating for the population of NFL games is 68.
#### Part (b): Developing a Point Estimate of the Standard Deviation
1. List the Ratings (from part (a)):
[tex]\[ 57, 61, 86, 74, 72, 73, 20, 57, 80, 79, 83, 74 \][/tex]
2. Calculate the Mean (already calculated as 68 from part (a)).
3. Calculate the Deviations from the Mean:
Subtract the mean from each rating to find the deviations.
[tex]\[ \begin{aligned} &57 - 68 = -11, \quad 61 - 68 = -7, \quad 86 - 68 = 18, \quad 74 - 68 = 6, \\ &72 - 68 = 4, \quad 73 - 68 = 5, \quad 20 - 68 = -48, \quad 57 - 68 = -11, \\ &80 - 68 = 12, \quad 79 - 68 = 11, \quad 83 - 68 = 15, \quad 74 - 68 = 6 \end{aligned} \][/tex]
4. Square the Deviations:
[tex]\[ \begin{aligned} &(-11)^2 = 121, \quad (-7)^2 = 49, \quad 18^2 = 324, \quad 6^2 = 36, \\ &4^2 = 16, \quad 5^2 = 25, \quad (-48)^2 = 2304, \quad (-11)^2 = 121, \\ &12^2 = 144, \quad 11^2 = 121, \quad 15^2 = 225, \quad 6^2 = 36 \end{aligned} \][/tex]
5. Sum the Squared Deviations:
[tex]\[ 121 + 49 + 324 + 36 + 16 + 25 + 2304 + 121 + 144 + 121 + 225 + 36 = 3522 \][/tex]
6. Calculate the Sample Variance:
Divide the sum of squared deviations by the sample size minus one (n-1), where n is the number of ratings.
[tex]\[ \text{Variance} = \frac{3522}{11} \approx 320.1818 \][/tex]
7. Calculate the Standard Deviation:
Take the square root of the variance to find the standard deviation.
[tex]\[ \text{Standard Deviation} = \sqrt{320.1818} \approx 17.8936 \][/tex]
8. Conclusion:
The point estimate of the standard deviation for the population of NFL games is approximately 17.8936 (to four decimal places).
### Final Answers:
- (a) The point estimate of the mean fan rating for the population of NFL games is 68.
- (b) The point estimate of the standard deviation for the population of NFL games is 17.8936.
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