Chapter 0, Problem 29PO

To determine

To find :The mean, median, mode, range, and standard deviation, and to identify any outliers.

Answer to Problem 29PO

Mean = $36$

Median = $\text{33}\text{.5}$

Mode = $35$

Range = $\text{59}$

Standard deviation = $18$

Outlier = $\text{75}$

Explanation of Solution

Given:

The number of students present at $8$ student council meetings are $23,45,16,75,32,35,28,35$

Formula used:

The mean is calculated using the formula,

  $\overline{X}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$ $\left(1\right)$

Where, $X_i$ represents the individual data point,

  ${\displaystyle \sum _{i=1}^n{X}_i}$ is the sum of all data points

  $n$ is the number of data points, in this problem $n=8$

Median $\left(M\right)$ can be calculated using the formula,

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

The mode is the data point whichoccurs more number of times in the given data set.

  $\text{Range = }\left(\text{maximum value in the data set}\right)\text{ –}\left(\text{ minimum value in the data set}\right)$  $\left(3\right)$

Standard deviation $\left(\text{s }\right)$ can be calculated using the equation,

  $\text{s = }\sqrt{\frac{{\displaystyle \sum _{i=1}^2{\left(X_i-\overline{X}\right)}^2}}{n-1}}$  $\left(4\right)$

Interquartile range $\left(IQR\right)$ = upper quartile − lower quartile

  $IQR$ = Third quartile $\left(Q_3\right)$ − first quartile $\left(Q_1\right)$  $\left(5\right)$

First quartile, $Q_1=\frac{1}{4}{\left(n+1\right)}^{th}\text{ term}$  $\left(6\right)$

Third quartile, $Q_3=\frac{3}{4}{\left(n+1\right)}^{th}\text{ term}$  $\left(7\right)$

Calculation:

The data can be write in the ascending order as,

  $16,23,28,32,35,35,45,75$

Mean value of the given set of data is,

  $\overline{X}=\frac{23+45+16+75+32+35+28+35}{8}$

  $\text{=}\frac{289}{8}=36.12\approx 36$

Median can be obtained by substituting the numerical values in equation $\left(2\right)$ ,

  $M={\left(\frac{8+1}{2}\right)}^{th}\text{ value}$

  $\text{=}{\left(\frac{9}{2}\right)}^{th}\text{ value}$

  $\text{= 4}{\text{.5}}^{th}\text{ value}$

  $\text{= 33}\text{.5}$

  $35$ appears two times in the data set hence it is the mode of the given data set.

Range can be calculated by substituting the maximum value = $75$ and minimum value = $16$ in the equation $\left(3\right)$ ,

  $\text{Range = 75 - 16 = 59}$

Standard deviation can be calculated using the equation $\left(4\right)$ ,

  $\text{s =}\sqrt{\frac{{\left(16-36\right)}^2\text{+ }{\left(23-36\right)}^2+{\left(28-36\right)}^2+{\left(32-36\right)}^2+{\left(35-36\right)}^2+{\left(35-36\right)}^2+{\left(45-36\right)}^2+{\left(75-36\right)}^2}{8-1}}$

  $=\sqrt{\frac{400+169+64+16+1+1+81+1521}{7}}$

  $=\sqrt{\frac{2253}{7}}$

  $=\sqrt{321.85}$

  $=17.84\approx 18$

To find an outlier:

First and third quartile can be obtained by substituting the numerical values in equations $\left(6\right)$ and $\left(7\right)$ ,

  $Q_1=\frac{1}{4}{\left(8+1\right)}^{th}\text{ term}$

  $\text{=}\frac{1}{4}{\left(9\right)}^{th}\text{ term = 2}{\text{.25}}^{th}\text{ term = 24}$

  $Q_3=\frac{3}{4}{\left(8+1\right)}^{th}\text{ term}$

  $\text{=}\frac{3}{4}{\left(9\right)}^{th}\text{ term = 6}{\text{.75}}^{th}\text{ term = 42}$

Substituting the values of $Q_1$ and $Q_3$ in equation $\left(5\right)$ ,

  $IQR=42-24=18$

  $IQR\times 1.5=18\times 1.5=27$

Addition of third quartile to this number gives,

  $27+42=69$

Since $\text{75}$ is greater than $\text{69}$ , hence $\text{75}$ is an outlier.

Conclusion:

Mean = $36$

Median = $\text{33}\text{.5}$

Mode = $35$

Range = $\text{59}$

Standard deviation = $18$

Outlier = $\text{75}$