Chapter 2.6, Problem 28E

To determine

To find: the value of d .

Answer to Problem 28E

The required value is $d=0.86$

Explanation of Solution

Given:

  $10.56;9.52;9.73;9.80;9.78;10.91$

Calculation:

Find the mean $\overline{x}$ of the data:

  $\begin{array}{l}\overline{x}=\frac{10.56+9.52+9.73+9.80+9.78+10.91}{6}\\ =\frac{60.3}{6}=10.05\end{array}$

The write the inequality:

  $\begin{array}{l}\left|x-\overline{x}\right|<d\\ \left|x-10.05\right|<d\end{array}$

Compute the absolute value differences and take as d the maximum of them:

  $\begin{array}{l}\left|10.56-10.05\right|=0.51\\ \left|9.52-10.05\right|=0.53\\ \left|9.73-10.05\right|=0.32\\ \left|9.80-10.05\right|=0.25\\ \left|9.78-10.05\right|=0.27\\ \left|10.91-10.05\right|=0.86\\ d=0.86\end{array}$

Conclusion:

Therefore, the required value is $d=0.86$