The average score for games played in the NFL is 20.9 and the standard deviation is 9.3 points. 18 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of x? x ~ N( 20.9 9.3 X) 0 b. What is the distribution of Σx? Σ - N( X 376.2, 1556.8 x) o c. P( 23.112) = 0.8438✔ 0 < d. Find the 60th percentile for the mean score for this sample size. 21.469 x e. P(21.512 < x < 22.196) = 0.1121 OF f. Q1 for the distribution = 19.4204 X g. Pa x > 423.216) = 0.1170 h. For part c) and e), is the assumption of normal necessary? No Yes می

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### Statistics in NFL Game Scores

The average score for games played in the NFL is known to be 20.9, with a standard deviation of 9.3 points. In a statistical analysis, 18 games are randomly selected. All answers are rounded to four decimal places, and a normal distribution is assumed.

#### Questions and Solutions

**a. What is the distribution of \(\bar{x}\)?**

\[
\bar{x} \sim N(20.9, \frac{9.3}{\sqrt{18}})
\]

**Solution:**

\[
\bar{x} \sim N(20.9, 2.1925) \quad \textcolor{green}{\checkmark}
\]

**b. What is the distribution of \(\sum x\)?**

\[
\sum x \sim N(18 \times 20.9, 18 \times 9.3^2)
\]

**Solution:**

\[
\sum x \sim N(376.2, 1556.76) \quad \textcolor{green}{\checkmark}
\]

**c. \(\text{P}(\bar{x} < 23.112) = ?\)**

**Solution:**

\[
\text{P}(\bar{x} < 23.112) = 0.8438 \quad \textcolor{green}{\checkmark}
\]

**d. Find the 60th percentile for the mean score for this sample size.**

**Solution:**

\[
60\text{th percentile} = 21.1658 \quad \textcolor{red}{\times}
\]

**e. \(\text{P}(21.512 < \bar{x} < 22.196) = ?\)**

**Solution:**

\[
\text{P}(21.512 < \bar{x} < 22.196) = 0.1121 \quad \textcolor{green}{\checkmark}
\]

**f. Q1 for the \(\bar{x}\) distribution.**

**Solution:**

\[
Q1 = 19.4204 \quad \textcolor{red}{\times}
\]

**g. \(\text{P}(\sum x > 423.216) = ?\)**

**Solution:**

\[
\text{P}(\sum
Transcribed Image Text:### Statistics in NFL Game Scores The average score for games played in the NFL is known to be 20.9, with a standard deviation of 9.3 points. In a statistical analysis, 18 games are randomly selected. All answers are rounded to four decimal places, and a normal distribution is assumed. #### Questions and Solutions **a. What is the distribution of \(\bar{x}\)?** \[ \bar{x} \sim N(20.9, \frac{9.3}{\sqrt{18}}) \] **Solution:** \[ \bar{x} \sim N(20.9, 2.1925) \quad \textcolor{green}{\checkmark} \] **b. What is the distribution of \(\sum x\)?** \[ \sum x \sim N(18 \times 20.9, 18 \times 9.3^2) \] **Solution:** \[ \sum x \sim N(376.2, 1556.76) \quad \textcolor{green}{\checkmark} \] **c. \(\text{P}(\bar{x} < 23.112) = ?\)** **Solution:** \[ \text{P}(\bar{x} < 23.112) = 0.8438 \quad \textcolor{green}{\checkmark} \] **d. Find the 60th percentile for the mean score for this sample size.** **Solution:** \[ 60\text{th percentile} = 21.1658 \quad \textcolor{red}{\times} \] **e. \(\text{P}(21.512 < \bar{x} < 22.196) = ?\)** **Solution:** \[ \text{P}(21.512 < \bar{x} < 22.196) = 0.1121 \quad \textcolor{green}{\checkmark} \] **f. Q1 for the \(\bar{x}\) distribution.** **Solution:** \[ Q1 = 19.4204 \quad \textcolor{red}{\times} \] **g. \(\text{P}(\sum x > 423.216) = ?\)** **Solution:** \[ \text{P}(\sum
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