Chapter MPS, Problem 12.5MPS

a.

To determine

Find the mean for $2006$ data.

Answer to Problem 12.5MPS

  $36.333$

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

Consider the table data and calculate mean,

  $Mean\left(\mu \right)=36.333$

Now standard deviation is,

  $\begin{array}{l}Mean\left(\overline{x}\right)=\frac{{\displaystyle \sum x}}{n}\\ Mean\left(\overline{x}\right)=\frac{\begin{array}{l}35+24+37+28+38+35\\ 38+44+44+17+54+42\end{array}}{12}\\ =\frac{463}{12}\end{array}$

Hence, the mean is, $36.333$ .

b.

To determine

Find the mean for $2007$ data.

Answer to Problem 12.5MPS

  $32.583$

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

Consider the table data and calculate mean,

  $\begin{array}{l}Mean\left(\overline{x}\right)=\frac{{\displaystyle \sum x}}{n}\\ Mean\left(\overline{x}\right)=\frac{\begin{array}{l}38+20+33+58+30+23\\ 48+24+37+38+28+14\end{array}}{12}\\ =\frac{391}{12}\\ Mean=32.583\end{array}$

Hence, the mean is, $32.583$ .

c.

To determine

Find the standard deviation for $2006$ data.

Answer to Problem 12.5MPS

  $9.432$

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

Consider the table data and calculate mean,

  $Mean\left(\mu \right)=36.333$

Now standard deviation is,

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum _{k=1}^{12}\left(x_k-\mu \right)}}{n}}\\ =\sqrt{\frac{\begin{array}{l}{\left(38-36.333\right)}^2+{\left(24-36.333\right)}^2+{\left(37-36.333\right)}^2+{\left(28-36.333\right)}^2+\\ {\left(38-36.333\right)}^2+{\left(35-36.333\right)}^2+{\left(38-36.333\right)}^2+{\left(44-36.333\right)}^2+\\ {\left(44-36.333\right)}^2+{\left(17-36.333\right)}^2+{\left(54-36.333\right)}^2+{\left(42-36.333\right)}^2\end{array}}{12}}\\ =\sqrt{\frac{1067.668668}{12}}\\ =\sqrt{88.97238900}\\ =9.432517638\end{array}$

Hence, the standard deviation is, $9.432$ .

d.

To determine

Find the standard deviation for $2007$ data.

Answer to Problem 12.5MPS

  $11.758$

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

Consider the table data and calculate mean,

  $Mean\left(\mu \right)=32.583$

Now calculate standard deviation,

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum _{k=1}^{12}\left(x_k-\mu \right)}}{n}}\\ =\sqrt{\frac{\begin{array}{l}{\left(38-32.583\right)}^2+{\left(20-32.583\right)}^2+{\left(33-32.583\right)}^2+{\left(58-32.583\right)}^2+\\ {\left(30-32.583\right)}^2+{\left(23-32.583\right)}^2+{\left(48-32.583\right)}^2+{\left(24-32.583\right)}^2+\\ {\left(37-32.583\right)}^2+{\left(38-32.583\right)}^2+{\left(28-32.583\right)}^2+{\left(14-32.583\right)}^2\end{array}}{12}}\\ =\sqrt{\frac{1658.916668}{12}}\\ =\sqrt{138.2430557}\\ =11.75768071\end{array}$

Hence, the standard deviation is, $11.758$ .

e.

To determine

Compare $2006and2007$ standard deviations.

Answer to Problem 12.5MPS

  $2007$ Standard deviation is greater than $2006$ .

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

  $2007$ Standard deviation is greater than $2006$ .

Hence, $2007$ standard deviation is greater than $2006$ .

f.

To determine

Conclude the $2006and2007$ seasons.

Answer to Problem 12.5MPS

The mean of season $2006$ is stronger is equal to $36.333$ .

Explanation of Solution

Given information:

Table shows the Ohio State university football score,

  $\begin{array}{l}2006Season\\ \begin{array}{cccc}35& 24& 37& 28\\ 38& 35& 38& 44\\ 44& 17& 54& 42\end{array}\end{array}$    $\begin{array}{l}2007Season\\ \begin{array}{cccc}38& 20& 33& 58\\ 30& 23& 48& 24\\ 37& 38& 28& 14\end{array}\end{array}$

Calculation:

The mean of season $2006$ is stronger is equal to $36.333$ .

Hence, the mean of season $2006$ is stronger.