Chapter 13, Problem 9PS

To determine

The intervals containing at least $\frac{3}{4}$ and $\frac{8}{9}$ of the scores and then to find the way in which the intervals would change after altering standard deviation to $16$ .

Answer to Problem 9PS

For portion $\frac{3}{4}$ ,

interval is $[179,291]$ .

For portion $\frac{8}{9}$ ,

interval is $[151,319]$ .

If standard deviation was $16$ ,

For portion $\frac{3}{4}$ ,

interval would be $[203,267]$ .

For portion $\frac{8}{9}$ ,

interval would be $[187,283]$ .

Explanation of Solution

Given information: Mean is $235$ , standard deviation is $28$ .

Calculation:

Chebychev’s theorem states that the portion of any data set lying within $k$ standard deviations of the mean is at least $1-\frac{1}{k^2}$

For portion $\frac{3}{4}$ , using Chebychev’s theorem,

  $1-\frac{1}{k^2}=\frac{3}{4}$

  $\begin{array}{l}1-\frac{1}{k^2}=\frac{3}{4}\\ \\ \frac{1}{k^2}=\frac{1}{4}\\ \\ k=2\end{array}$

Interval is $x\pm \sigma k=235\pm (28)(2)$

  $\Rightarrow$ Interval is $[179,291]$ .

For portion $\frac{8}{9}$ , using Chebychev’s theorem,

  $1-\frac{1}{k^2}=\frac{8}{9}$

  $\begin{array}{l}1-\frac{1}{k^2}=\frac{8}{9}\\ \\ \frac{1}{k^2}=\frac{1}{9}\\ \\ k=3\end{array}$

Interval is $x\pm \sigma k=235\pm (28)(3)$

  $\Rightarrow$ Interval is $[151,319]$ .

If standard deviation was $16$ ,

For portion $\frac{3}{4}$ ,

Interval would be $x\pm \sigma k=235\pm (16)(2)$

  $\Rightarrow$ Interval would be $[203,267]$ .

For portion $\frac{8}{9}$ ,

Interval would be $x\pm \sigma k=235\pm (16)(3)$

  $\Rightarrow$ Interval would be $[187,283]$ .