Chapter 4.CR, Problem 1CR

Find (a) the mean, (b) the median, (c) the mode, and (d) the standard derivation of the following set of the raw material.

$581048106875$

To determine

(a)

To find:

The mean of the following set of raw material.

Answer to Problem 1CR

Solution:

The mean of the given set of raw material is $7.1$.

Explanation of Solution

Given:

The given following set of raw material is,

$581048106875$

Formula used:

The mean is the ratio of the sum of the data points to the number of the data points.

Consider the mean formula,

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Here, the mean is $\overline{x}$, the sum of the data points is ${\displaystyle \sum x}$ and the number of the data points $n$.

Calculation:

Consider the given set of raw materials,

$581048106875$

Consider the total sum of the raw materials,

$\begin{array}{rll}{\displaystyle \sum x}& =& 5+8+10+4+8+10+6+8+7+5\\ & =& 71\end{array}$

Consider the total number of data points,

$n=10$

Consider the mean formula,

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$ ……$\left(1\right)$

Substitute $10$ for $n$ and $71$ for ${\displaystyle \sum x}$ in equation $\left(1\right)$.

$\begin{array}{rll}\overline{x}& =& \frac{71}{10}\\ & =& 7.1\end{array}$

Thus, the mean of the given set of raw material is $7.1$.

Conclusion:

Thus, the mean of the given set of raw material is $7.1$.

To determine

(b)

To find:

The median of the following set of raw material.

Answer to Problem 1CR

Solution:

The median of the given data is $9$.

Explanation of Solution

Given:

The given following set of raw material is,

$581048106875$

Formula used:

The median is the middle value of a distribution of numbers.

Consider the median position formula,

$L=\frac{n+1}{2}\text{th}\text{value}$

Here, the median number position is $L$ and the number of the data points $n$.

Consider the median number formula,

$\text{M}=\frac{{\text{R}}_{\text{m1}}{\text{+R}}_{\text{m2}}}{\text{2}}$

Here, the median number is $\text{M}$, the first middle data is ${\text{R}}_{\text{m1}}$ and the second middle data is ${\text{R}}_{\text{m2}}$.

Calculation:

Consider the given set of raw materials,

$581048106875$

Consider the total number of data points,

$n=10$

Consider the mean formula,

$L=\frac{n+1}{2}\text{th}\text{value}$ ……$\left(1\right)$

Substitute $10$ for $n$ in equation $\left(1\right)$.

$\begin{array}{rll}L& =& \frac{10+1}{2}\\ & =& \frac{11}{2}\\ & =& \text{5}\text{.5th}\text{value}\end{array}$

Thus, the median of the given set of raw material is $\text{5}\text{.5th}\text{value}$.

Consider the fifth raw material,

${\text{R}}_{\text{m1}}=8$

Consider the sixth raw material,

${\text{R}}_{\text{m2}}=10$

Consider the median number formula,

$\text{M}=\frac{{\text{R}}_{\text{m1}}{\text{+R}}_{\text{m2}}}{\text{2}}$ ….. $\left(2\right)$

Substitute $8$ for ${\text{R}}_{\text{m1}}$ and $10$ for ${\text{R}}_{\text{m2}}$ in equation $\left(2\right)$.

$\begin{array}{rll}\text{M}& =& \frac{\text{8+10}}{\text{2}}\\ & =& \frac{18}{2}\\ & =& 9\end{array}$

Thus, the median of the given data is $9$.

Conclusion:

Thus, the median of the given data is $9$.

To determine

(c)

To find:

The mode of the following set of raw material.

Answer to Problem 1CR

Solution:

The mode of the given set of raw material is $8$.

Explanation of Solution

Given:

The given following set of raw material is,

$581048106875$

Concept:

The mode is the data points with highest frequency. Because it represents the most common number, the mode can be viewed as an average.

Approach:

Consider the given set of raw materials,

$581048106875$

The frequency of the number $10$ is $2$.

The frequency of the number $8$ is $3$.

The frequency of the number $5$ is $2$.

The frequency of the number $4$ is $1$.

The frequency of the number $6$ is $1$.

The frequency of the number $7$ is $1$.

Therefore, the mode is $8$, because it has the highest frequency $3$.

Thus, the mode of the given set of raw material is $8$.

Conclusion:

Thus, the mode of the given set of raw material is $8$.

To determine

(d)

To find:

The standard derivation of the following set of raw material.

Answer to Problem 1CR

Solution:

The standard derivation of the given set of raw material is $7.1$.

Explanation of Solution

Given:

The given following set of raw material is,

$581048106875$

Formula used:

The mean is the ratio of the sum of the data points to the number of the data points.

The variance of the sample is found by dividing the sum of squares of the deviations by $n-1$.

The square root of the variance is called standard variance.

Consider the mean formula,

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Here, the mean is $\overline{x}$, the sum of the data points is ${\displaystyle \sum x}$ and the number of the data points $n$.

Consider the variance formula,

$s=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}$

Here, the variance is $s^2$, the mean is $\overline{x}$ and the number of the data points $n$.

Consider the standard formula,

$d=\sqrt{s}$

Here, the standard deviation is $d$.

Calculation:

Consider the given set of raw materials,

$x=581048106875$

Consider the total sum of the raw,

$\begin{array}{rll}{\displaystyle \sum x}& =& 5+8+10+4+8+10+6+8+7+5\\ & =& 71\end{array}$

Consider the total number of data points,

$n=10$

Consider the mean formula,

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$…$\left(1\right)$

Substitute $10$ for $n$ and $71$ for ${\displaystyle \sum x}$ in equation $\left(1\right)$.

$\begin{array}{rll}\overline{x}& =& \frac{71}{10}\\ & =& 7.1\end{array}$

Thus, the mean of the given set of raw material is $7.1$.

Consider sum of squares of the deviations,

$\begin{array}{rll}{\displaystyle \sum {\left(x-\overline{x}\right)}^2}& =& \left[\begin{array}{l}{\left(5-7.1\right)}^2+{\left(8-7.1\right)}^2+{\left(10-7.1\right)}^2+{\left(4-7.1\right)}^2+{\left(8-7.1\right)}^2\\ +{\left(10-7.1\right)}^2+{\left(6-7.1\right)}^2+{\left(8-7.1\right)}^2+{\left(7-7.1\right)}^2+{\left(5-7.1\right)}^2\end{array}\right]\\ & =& \left[\begin{array}{l}{\left(-2.1\right)}^2+{\left(0.9\right)}^2+{\left(2.9\right)}^2+{\left(-3.1\right)}^2+{\left(2.9\right)}^2\\ +{\left(2.9\right)}^2+{\left(-1.1\right)}^2+{\left(0.9\right)}^2+{\left(0.1\right)}^2+{\left(-2.1\right)}^2\end{array}\right]\\ & =& \left[\begin{array}{l}4.41+0.81+8.41+9.61+8.41\\ +8.41+1.21+0.81+0.01+4.41\end{array}\right]\\ & =& 46.5\end{array}$

Consider the variance formula,

$s=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}$…$\left(2\right)$

Substitute $46.5$ for ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ and $10$ for $n$ in equation $\left(2\right)$.

$\begin{array}{rll}s& =& \frac{46.5}{10-1}\\ & =& \frac{46.5}{9}\\ & =& 5.16667\end{array}$

Therefore, the variance is $5.16667$.

Consider the standard formula,

$d=\sqrt{s}$... $\left(3\right)$

Substitute $5.16667$ for $s$ in the above mentioned standard deviation formula.

$\begin{array}{rll}d& =& \sqrt{5.16667}\\ & =& 2.273030\end{array}$

Thus, the standard deviation is $2.273030$.

Conclusion:

Thus, the standard deviation is $2.273030$.