Chapter 14.3, Problem 10MCQ

To determine

To discuss: The variability of the data.

Answer to Problem 10MCQ

Terms	30
Median	84.5
$Q_1$	75
$Q_3$	90
Interquartile	15
Semi-interquartile	7.5
Outliers	One

Explanation of Solution

Given information:

  

Formula used:

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

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

Calculation:

Variability of the data can be studied by box-and-whisker plot.

It represent the median, quartiles, interquartile range , and extreme values in a set of the data.

Consider the data in ascending order,

  $\left\{\begin{array}{l}54,55,62,62,64,65,71,75,76,77,78,\\ 79,80,82,84,85,86,87,88,89,89,89,\\ 90,92,93,93,95,96,98,99\end{array}\right\}$

So, $n=30$ .

The median is $\mathrm{84.5.}$

The mode of is $89.$

The mean of scores are $70.$

For quartile points :-

Since the median is 84.5.

Lower half - $54,55,62,62,64,65,71,75,76,77,78,79,80,82,84.$ and

Upper half - $85,86,87,88,89,89,89,90,92,93,93,95,96,98,99.$

Median of lower half is −

  $\begin{array}{l}={\left(\frac{15+1}{2}\right)}^{th}\text{term}\\ ={\left(\frac{16}{2}\right)}^{th}\text{term}\\ ={\left(8\right)}^{th}\text{term}\\ =75\end{array}$

Median of upper half is −

  $\begin{array}{l}={\left(\frac{15+1}{2}\right)}^{th}\text{term}\\ ={\left(\frac{16}{2}\right)}^{th}\text{term}\\ ={\left(8\right)}^{th}\text{term}\\ =90\end{array}$

The quartile points are :-

  $\begin{array}{l}{Q}_1=75\\ Q_2=84.5\\ Q_3=90\end{array}$

For interquatrtile:-

The inter- quartile range is $Q_3-Q_1.$

  $\begin{array}{l}=90-75\\ =15.\end{array}$

For semi-interquartile :-

The semi-interquartile range -

  $\frac{\text{inter-quartile range}}{2}.$

  $\begin{array}{l}=\frac{15}{2}\\ =\mathrm{7.5.}\end{array}$

For outlier

  $\begin{array}{l}{Q}_1-1.5\left(7.5\right)=75-11.25=63.75\\ Q_3+1.5\left(7.5\right)=90+11.25=101.25\end{array}$

The lower extreme 54 and the upper extreme 99 are within he limits.

So, there will be only one outlier.

Therefore,

Terms	30
Median	84.5
$Q_1$	75
$Q_3$	90
Interquartile	15
Semi-interquartile	7.5
Outliers	One