# Question #1cd46

Sep 27, 2017

Boundaries = 124.5 and 131.5
Midpoint= 128
Width (range) = 6

#### Explanation:

The boundaries are found using these formulas:

${B}_{U} = {L}_{U} + \frac{1}{2} \left(M\right)$ and
${B}_{L} = {L}_{L} - \frac{1}{2} \left(M\right)$,

where ${B}_{U}$ is the upper boundary, ${B}_{L}$ is the lower boundary, ${L}_{U}$ is the upper limit of the data, ${L}_{L}$ is the lower limit of the data, and $M$ is the unit of measurement.

Firstly we find the unit of measurement. From 125-131 we seem to be counting in ones, thus our unit of measurement is $M = 1$. Next we find ${L}_{U} \mathmr{and} {L}_{L}$. These are simply the highest and lowest numbers, so we get ${L}_{U} = 131 \mathmr{and} {L}_{L} = 125$. We now plug these numbers into the equation to get:

${B}_{U} = 131 + \frac{1}{2} \left(1\right) \text{ } {B}_{L} = 125 - \frac{1}{2} \left(1\right)$
${B}_{U} = 131 + \frac{1}{2} \text{ } {B}_{L} = 125 - \frac{1}{2}$
${B}_{U} = 131.5 \text{ } {B}_{L} = 124.5$

Therefore the boundaries of 125-131 are 124.5 and 131.5.

I invite you too lok at Gamaliel B's answer for more details.

The width (or range) is found by subtracting the highest value from the lowest value, which looks like this:
$131 - 125 = 6$, therefore 6 is the width (or range).

The midpoint $p$ is found like this:

$p = \frac{{L}_{L} + {L}_{U}}{2}$

where $p$ is the midpoint, and ${L}_{L}$ & ${L}_{U}$ are the lower and upper limits of the data, respectively. (The midpoint is just the average of the lowest and highest values.) We know that ${L}_{L} = 125$ and ${L}_{U} = 131.$ Plug these into the formula to get

$p = \frac{125 + 131}{2}$
$p = \frac{256}{2}$
$p = 128 ,$ the midpoint.

I hope I helped!