The average score for games played in the NFL is 21.2 and the standard deviation is 9 points. 48 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.
a. What is the distribution of [tex]$\bar{x} \sim N($[/tex] [tex]$\square$[/tex] , [tex]$\square$[/tex] )
b. What is the distribution of [tex]$\qquad x ? \sum x \sim N($[/tex] [tex]$\square$[/tex] , [tex]$\square$[/tex] )
c. [tex]$P (\bar{x}\ \textgreater \ 19.6515)= \square$[/tex]
d. Find the 68th percentile for the mean score for this sample size. [tex]$\square$[/tex]
e. [tex]$P (19.5515\ \textless \ \bar{x}\ \textless \ 22.5495)= \square$[/tex]
f. Q1 for the [tex]$\bar{x}$[/tex] distribution [tex]$= \square$[/tex]
g. [tex]$P \left(\sum x\ \textgreater \ 996.072\right)= \square$[/tex]
h. For part c) and e), is the assumption of normal necessary? Yes No
Answer (1)
The distribution of the sample mean is x ˉ ∼ N ( 21.2 , 1.6875 ) .
The distribution of the sum of the scores is ∑ x ∼ N ( 1017.6 , 3888 ) .
Calculate 19.6515) = 0.8834"> P ( x ˉ > 19.6515 ) = 0.8834 .
The 68th percentile for the sample mean is 21.8076 .
Calculate P ( 19.5515 < x ˉ < 22.5495 ) = 0.7482 .
The first quartile for the x ˉ distribution is 20.3238 .
Calculate 996.072) = 0.6350"> P ( ∑ x > 996.072 ) = 0.6350 .
The assumption of normality is not necessary for parts c) and e). N o
Explanation
Understand the problem and provided data We are given that the average score for games played in the NFL is μ = 21.2 and the standard deviation is σ = 9 points. We are also given that 48 games are randomly selected, so n = 48 . We are asked to find several probabilities and percentiles related to the sample mean and the sum of the scores. We will assume a normal distribution.
Find the distribution of the sample mean a. The distribution of the sample mean x ˉ is normal with mean μ x ˉ = μ = 21.2 and standard deviation σ x ˉ = n σ = 48 9 ≈ 1.2990 . Thus, x ˉ ∼ N ( 21.2 , ( 1.2990 ) 2 ) . Therefore, x ˉ ∼ N ( 21.2 , 1.6875 ) .
Find the distribution of the sum of the scores b. The distribution of the sum of the scores ∑ x is normal with mean μ ∑ x = n μ = 48 ( 21.2 ) = 1017.6 and standard deviation σ ∑ x = σ n = 9 48 ≈ 62.3538 . Thus, ∑ x ∼ N ( 1017.6 , ( 62.3538 ) 2 ) . Therefore, ∑ x ∼ N ( 1017.6 , 3888 ) .
Calculate 19.6515)"> P ( x ˉ > 19.6515 ) c. To calculate 19.6515)"> P ( x ˉ > 19.6515 ) , standardize x ˉ by calculating the z-score: z = σ x ˉ x ˉ − μ x ˉ = 9/ 48 19.6515 − 21.2 = 1.2990 − 1.5485 ≈ − 1.1920 . Then, 19.6515) = P(Z > -1.1920) = 1 - P(Z < -1.1920) = 1 - 0.1166 = 0.8834"> P ( x ˉ > 19.6515 ) = P ( Z > − 1.1920 ) = 1 − P ( Z < − 1.1920 ) = 1 − 0.1166 = 0.8834 .
Find the 68th percentile for the sample mean d. The 68th percentile corresponds to the value of x ˉ such that P ( x ˉ < x 0.68 ) = 0.68 . Find the z-score corresponding to the 68th percentile, z 0.68 ≈ 0.4677 , from a standard normal table or calculator. Then, use the formula x 0.68 = μ x ˉ + z 0.68 σ x ˉ = 21.2 + 0.4677 ⋅ 48 9 = 21.2 + 0.4677 ⋅ 1.2990 ≈ 21.2 + 0.6076 = 21.8076 .
Calculate P ( 19.5515 < x ˉ < 22.5495 ) e. To calculate P ( 19.5515 < x ˉ < 22.5495 ) , standardize both values: z 1 = 9/ 48 19.5515 − 21.2 = 1.2990 − 1.6485 ≈ − 1.2691 and z 2 = 9/ 48 22.5495 − 21.2 = 1.2990 1.3495 ≈ 1.0389 . Then, P ( − 1.2691 < Z < 1.0389 ) = P ( Z < 1.0389 ) − P ( Z < − 1.2691 ) = 0.8504 − 0.1022 = 0.7482 .
Find the first quartile (Q1) for the x ˉ distribution f. The first quartile (Q1) corresponds to the 25th percentile, so P ( x ˉ < Q 1 ) = 0.25 . Find the z-score corresponding to the 25th percentile, z 0.25 ≈ − 0.6745 , from a standard normal table or calculator. Then, use the formula Q 1 = μ x ˉ + z 0.25 σ x ˉ = 21.2 + ( − 0.6745 ) ⋅ 48 9 = 21.2 − 0.6745 ⋅ 1.2990 ≈ 21.2 − 0.8762 = 20.3238 .
Calculate 996.072)"> P ( ∑ x > 996.072 ) g. To calculate 996.072)"> P ( ∑ x > 996.072 ) , standardize ∑ x by calculating the z-score: z = σ ∑ x ∑ x − μ ∑ x = 9 48 996.072 − 48 ( 21.2 ) = 62.3538 996.072 − 1017.6 = 62.3538 − 21.528 ≈ − 0.3453 . Then, 996.072) = P(Z > -0.3453) = 1 - P(Z < -0.3453) = 1 - 0.3650 = 0.6350"> P ( ∑ x > 996.072 ) = P ( Z > − 0.3453 ) = 1 − P ( Z < − 0.3453 ) = 1 − 0.3650 = 0.6350 .
Is the assumption of normal necessary for parts c) and e? h. Since the sample size is 30"> n = 48 > 30 , the Central Limit Theorem applies. Therefore, the assumption of normality is not strictly necessary for parts c) and e).
Examples
Understanding the distribution of average scores and probabilities can be useful in sports analysis. For example, a coach might want to know the probability that their team's average score over a certain number of games will exceed a certain threshold. This can help them assess their team's performance and make strategic decisions. Also, knowing the distribution of the sum of scores can help in predicting the total points scored in a season, which is relevant for setting betting odds or evaluating team consistency. These statistical insights provide a quantitative way to understand and predict team performance.
