Chapter 4, Problem 4.70P

To determine

The mean median and standard deviation.

Answer to Problem 4.70P

The values of mean, median and standard deviation are $0.6733$ , $0.685$ and $0.0411$ .

Explanation of Solution

Given:

The number of sample subsets are $6$ .

The number of measured values in each sample subset are $4$ .

Formula used:

The expression foraverage of each subset of samplesis given as,

  $\overline{x}=\frac{x_1+x_2+x_3+x_4}{4}$

Here, $x_1$ , $x_2$ , $x_3$ and $x_4$ are the variables of subset.

The expression formean or average of averages is given as,

  $\overline{\overline{x}}=\frac{{\overline{x}}_1+{\overline{x}}_2+{\overline{x}}_3+{\overline{x}}_4+{\overline{x}}_5+{\overline{x}}_6}{6}$

The expression formedian is given as,

  $\text{Median}=\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}$

Here, $n$ is the total number of averages.

The expression forstandard deviation is given as,

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{\overline{x}}\right)}^2+{\left(x_2-\overline{x}\right)}^2+{\left(x_3-\overline{\overline{x}}\right)}^2+\mathrm{...}+{\left(x_n-\overline{\overline{x}}\right)}^2}{N-1}}$

Here, $N$ is the number of measurement and $x_n$ is the measured value of each part.

Calculation:

The average of the first subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_1=\frac{0.65+0.75+0.67+0.65}{4}\\ {\overline{x}}_1=\frac{2.72}{4}\\ {\overline{x}}_1=0.68\end{array}$

The average of the second subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_2=\frac{0.69+0.73+0.70+0.68}{4}\\ {\overline{x}}_2=\frac{2.8}{4}\\ {\overline{x}}_2=0.7\end{array}$

The average of the third subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_3=\frac{0.65+0.68+0.65+0.61}{4}\\ {\overline{x}}_3=\frac{2.59}{4}\\ {\overline{x}}_3=0.6475\end{array}$

The average of the fourth subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_4=\frac{0.64+0.65+0.60+0.60}{4}\\ {\overline{x}}_4=\frac{2.49}{4}\\ {\overline{x}}_4=0.6225\end{array}$

The average of the fifth subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_5=\frac{0.68+0.72+0.70+0.66}{4}\\ {\overline{x}}_5=\frac{2.76}{4}\\ {\overline{x}}_5=0.69\end{array}$

The average of the sixth subset can be calculated as,

  $\begin{array}{l}{\overline{x}}_6=\frac{0.70+0.74+0.65+0.71}{4}\\ {\overline{x}}_6=\frac{2.8}{4}\\ {\overline{x}}_6=0.7\end{array}$

The mean can be calculated as,

  $\begin{array}{l}\overline{\overline{x}}=\frac{{\overline{x}}_1+{\overline{x}}_2+{\overline{x}}_3+{\overline{x}}_4+{\overline{x}}_5+{\overline{x}}_6}{6}\\ \overline{\overline{x}}=\frac{0.68+0.7+0.6475+0.6225+0.69+0.7}{6}\\ \overline{\overline{x}}=\frac{4.04}{6}\\ \overline{\overline{x}}=0.6733\end{array}$

For the calculation of median arrange the data in increasing order,

  $0.6225,0.6425,0.68,0.69,0.7,0.7$ .

The median can be calculated as,

  $\begin{array}{l}\text{Median}=\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}\\ \text{Median}=\frac{3^{rd}\text{term}+4^{th}\text{term}}{2}\\ \text{Median}=\frac{0.68+0.69}{2}\\ \text{Median}=0.685\end{array}$

Standard deviation can be calculated as,

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{\overline{x}}\right)}^2+{\left(x_2-\overline{x}\right)}^2+{\left(x_3-\overline{\overline{x}}\right)}^2+\mathrm{...}+{\left(x_{24}-\overline{\overline{x}}\right)}^2}{24-1}}\\ \sigma =\left\{\sqrt{\frac{\begin{array}{l}{\left(0.65-0.6733\right)}^2+{\left(0.75-0.6733\right)}^2+{\left(0.67-0.6733\right)}^2\\ +{\left(0.65-0.6733\right)}^2+{\left(0.69-0.6733\right)}^2\mathrm{...}+{\left(71-0.6733\right)}^2\end{array}}{23}}\right\}\\ \sigma =0.0411\end{array}$

Conclusion:

Therefore, the values of mean, median and standard deviation are $0.6733$ , $0.685$ and $0.0411$ .