91Ó°ÊÓ

Range = largest value - Smallest value

=760 – 135

= 625

$$\begin{gathered} \mathbf{StandardDeviation}\mathbf{=}\frac{\mathbf{Range}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{625}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{156}\mathbf{.}\mathbf{25} \end{gathered}$$

Therefore, the estimated s is 156.25.

Based on the Chebyshev and Empirical rule, we know that most measurements will fall in the 2 and 3 standard deviation range from the mean. Therefore, the standard deviation should lie within that range.

We also know that the standard deviation cannot be more than ¼th of the range. Therefore, to estimate the standard deviation, we use$\mathrm{s}=\frac{\mathrm{Range}}{4}$formula.

The standard deviation as 25 can't be feasible because the standard deviation would lie in the 104.16 to 156.25 range.

We found the lower range by applying the same procedure in step 1 except for the denominator. We will change it to 6 because most observations will lie in the 3 standard deviation range.

$$\begin{gathered} \mathbf{Lower}\mathbf{}\mathbf{range}\mathbf{=}\frac{\mathbf{Range}}{\mathbf{6}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{625}}{\mathbf{6}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{104}\mathbf{.}\mathbf{16} \end{gathered}$$