Chapter 19, Problem 3P

For Problem 19.1, using Equations (19.1) and (19.6), calculate the mean and standard deviation of the class scores.

To determine

Calculate the mean and standard deviation of the class scores.

Answer to Problem 3P

The mean and standard deviation of class scores are 73.2 and 12.5 respectively.

Explanation of Solution

Given data:

The required data are given below with reference to the problem 19.1 in the textbook.

Number of students, $n=30$.

Students test scores are 57, 94, 81, 77, 66, 97, 62, 86, 75, 87, 91, 78, 61, 82, 74, 72, 70, 88, 66, 75, 55, 66, 58, 73, 79, 51, 63, 77, 52, 84.

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$\overline{x}=\frac{x_1+x_2+x_3+\mathrm{............}+x_{n-1}+x_n}{n}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$ (1)

Here,

$\overline{x}$ is the mean,

$x_i$ is the data points,

$n$ is the number of data points.

From equation 19.6 in the textbook, the formula to find standard deviation is,

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

Calculation:

Substitute all the value of student scores for $x_i$ up to the range $n$, and $30$ for $n$ in equation (1) to calculate mean $\left(\overline{x}\right)$,

$\begin{array}{c}\overline{x}=\frac{\begin{array}{l}57+94+81+77+66+97+62+86+75+87+91+78+61+82+74+72+\\ 70+88+66+75+55+66+58+73+79+51+63+77+52+84\end{array}}{30}\\ =\frac{2197}{30}\\ \overline{x}=73.2\end{array}$

Find the terms ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$ in the equation (2) as by substituting all the value of student scores for $x_i$ up to the range $n$, $73.2$ for $\overline{x}$ as,

${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}=\left(\begin{array}{l}{\left(57-73.2\right)}^2+{\left(94-73.2\right)}^2+{\left(81-73.2\right)}^2+{\left(77-73.2\right)}^2+{\left(66-73.2\right)}^2+\\ {\left(97-73.2\right)}^2+{\left(62-73.2\right)}^2+{\left(86-73.2\right)}^2+{\left(75-73.2\right)}^2+{\left(87-73.2\right)}^2+\\ {\left(91-73.2\right)}^2+{\left(78-73.2\right)}^2+{\left(61-73.2\right)}^2+{\left(82-73.2\right)}^2+{\left(74-73.2\right)}^2+\\ {\left(72-73.2\right)}^2+{\left(70-73.2\right)}^2+{\left(88-73.2\right)}^2+{\left(66-73.2\right)}^2+{\left(75-73.2\right)}^2+\\ {\left(55-73.2\right)}^2+{\left(66-73.2\right)}^2+{\left(58-73.2\right)}^2+{\left(73-73.2\right)}^2+{\left(79-73.2\right)}^2+\\ {\left(51-73.2\right)}^2+{\left(63-73.2\right)}^2+{\left(77-73.2\right)}^2+{\left(52-73.2\right)}^2+{\left(84-73.2\right)}^2\end{array}\right)$

${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}=4531.25$

Substitute all the value $4531.25$ for ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$ and $30$ for $n$ in equation (2) to find standard deviation $\left(s\right)$,

$\begin{array}{l}s=\sqrt{\frac{4531.25}{30-1}}\\ s=\sqrt{\frac{4531.25}{29}}\\ s=\sqrt{156.25}\\ s=12.5\end{array}$

Therefore, the mean and standard deviation of class scores are 73.2 and 12.5 respectively.

Conclusion:

Thus, the mean and standard deviation of class scores are 73.2 and 12.5 respectively.