Chapter B.2, Problem 40E

To determine

To calculate:

Intervals containing the at least $\frac{3}{4}$ and $\frac{8}{9}$ .and changing of intervals when standard deviation is $225$ .

Answer to Problem 40E

When the Standard deviation decreased from $468$ to $225$ the interval get more near to the mean.

Explanation of Solution

Given information:

Standard deviation $225$

Calculation:

According to the given data, the average score on $437$ players who played in the $2013-2014$ .

Mean and Standard deviation is $563$ and $468$ .

Intervals containing the at least $\frac{3}{4}$ and $\frac{8}{9}$ .

  $\begin{array}{l}\frac{3}{4}\\ =0.75\end{array}$

Therefore, by using the Chebychev's Theorem, interval contain $75\%$ score which must lie within two standard deviation of the mean.

  $\begin{array}{l}=\left[mean\pm 2s\tan darddeviation\right]\\ =\left[563-2\left(468\right)+563-2\left(468\right)\right]\\ =\left[563-936+563-450\right]\\ =\left[-373,113\right]\end{array}$

Therefore, the interval contain at least $\frac{3}{4}$ score is $\left[-373,113\right]$

Intervals containing the at least $\frac{8}{9}$ .

  $\begin{array}{l}=\frac{8}{9}\\ =0.88\end{array}$

Therefore, by using the Chebychev's Theorem, interval contain $88.9$ score which must lie within two standard deviation of the mean.

  $\begin{array}{l}=\left[mean\pm 3s\tan darddeviation\right]\\ =\left[563-3\left(468\right)+563+3\left(468\right)\right]\\ =\left[563-1404+563-1404\right]\\ =\left[-841,1967\right]\end{array}$

Therefore, the interval contain at least $\frac{8}{9}$ score is $\left[-841,1967\right]$

Now consider the average score on the $563$ and standard deviation is $225$ .

Intervals containing the at least $\frac{3}{4}$

  $\begin{array}{l}\frac{3}{4}\\ =0.75\end{array}$

Therefore, by using the Chebychev's Theorem, interval contain $75\%$ score which must lie within two standard deviation of the mean.

  $\begin{array}{l}=\left[mean\pm 2s\tan darddeviation\right]\\ =\left[563-2\left(225\right)+563+2\left(225\right)\right]\\ =\left[563-450+563+450\right]\\ =\left[113,1013\right]\end{array}$

Therefore, the interval contain at least $\frac{3}{4}$ score is $\left[113,1013\right]$

Intervals containing the at least $\frac{8}{9}$ .

  $\begin{array}{l}=\frac{8}{9}\\ =0.88\end{array}$

Therefore, by using the Chebychev's Theorem, interval contain $88.9$ score which must lie within two standard deviation of the mean.

  $\begin{array}{l}=\left[mean\pm 3s\tan darddeviation\right]\\ =\left[563-3\left(225\right)+563+3\left(225\right)\right]\\ =\left[563-675+563+675\right]\\ =\left[-112,1238\right]\end{array}$

Therefore, the interval contain at least $\frac{8}{9}$ score is $\left[-112,1238\right]$

Hence, when the Standard deviation decreased from $468$ to $225$ the interval get more near to the mean.