Question

the average score for games played in the nfl is 22.1 and the standard deviation is 9.3 points. 6 games are randomly selected. round all answers to 4 decimal places where possible and assume a normal distribution.

a. what is the distribution of \\( \overline{x} \\)? \\( \overline{x} \sim n(22.1,9.3000) \\)
b. what is the distribution of \\( \sum x \\)? \\( \sum x \sim n(22.1,3.7970) \\)
c. \\( p(\overline{x}>23.105)=0.3955 \\)
d. find the 67th percentile for the mean score for this sample size. 23.7707
e. \\( p(17.405<\overline{x}<25.0984)=0.6757 \\)
f. q1 for the \\( \overline{x} \\) distribution = 19.5384
g. \\( p(\sum x>151.23)=0.2077 \\)
h. for part c) and e), is the assumption of normal necessary? no yes

Explanation:

Step1: Calculate the mean and standard deviation for \(\sum x\)

The mean of \(\sum x\) is \(n\mu\), where \(n = 6\) and \(\mu=22.1\). So, \(\mu_{\sum x}=n\mu = 6\times22.1=132.6\).
The standard deviation of \(\sum x\) is \(\sigma_{\sum x}=\sqrt{n}\sigma\), where \(n = 6\) and \(\sigma = 9.3\). So, \(\sigma_{\sum x}=\sqrt{6}\times9.3\approx22.77\).

Step2: Calculate \(P(17.405<\overline{x}<25.0984)\)

First, find the \(z -\)scores.
The \(z -\)score formula is \(z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
For \(\overline{x}_1 = 17.405\), \(z_1=\frac{17.405 - 22.1}{\frac{9.3}{\sqrt{6}}}=\frac{- 4.695}{\frac{9.3}{\sqrt{6}}}\approx - 1.25\).
For \(\overline{x}_2 = 25.0984\), \(z_2=\frac{25.0984 - 22.1}{\frac{9.3}{\sqrt{6}}}=\frac{2.9984}{\frac{9.3}{\sqrt{6}}}\approx0.80\).
Then, \(P(17.405<\overline{x}<25.0984)=P(-1.25<Z<0.80)\).
Using the standard normal table, \(P(-1.25 < Z<0.80)=\Phi(0.80)-\Phi(- 1.25)\).
Since \(\Phi(-z)=1 - \Phi(z)\), \(\Phi(-1.25)=1-\Phi(1.25)\).
From the standard - normal table, \(\Phi(0.80)=0.7881\), \(\Phi(1.25)=0.8944\).
So, \(P(-1.25 < Z<0.80)=0.7881-(1 - 0.8944)=0.6825\).

Answer:

a. \(\overline{x}\sim N(22.1,\frac{9.3}{\sqrt{6}})\approx N(22.1,3.7970)\)
b. \(\sum x\sim N(132.6,22.77)\)
c. \(P(\overline{x}>23.105)\approx0.3955\)
d. The 67th percentile for the mean score: \(z\) - value for the 67th percentile is \(z\approx0.44\) (from the standard - normal table). Using \(z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), \(0.44=\frac{\overline{x}-22.1}{\frac{9.3}{\sqrt{6}}}\), \(\overline{x}=22.1 + 0.44\times\frac{9.3}{\sqrt{6}}\approx23.7702\)
e. \(P(17.405<\overline{x}<25.0984)\approx0.6825\)
f. \(Q1\) (25th percentile): \(z\) - value for the 25th percentile is \(z\approx - 0.67\). Using \(z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), \(-0.67=\frac{\overline{x}-22.1}{\frac{9.3}{\sqrt{6}}}\), \(\overline{x}=22.1-0.67\times\frac{9.3}{\sqrt{6}}\approx19.5392\)
g. For \(\sum x\), \(\mu_{\sum x}=132.6\), \(\sigma_{\sum x}=\sqrt{6}\times9.3\approx22.77\). \(z=\frac{151.23 - 132.6}{22.77}=\frac{18.63}{22.77}\approx0.82\). \(P(\sum x>151.23)=1 - P(\sum x\leqslant151.23)=1-\Phi(0.82)\approx0.2067\)
h. Yes