Chapter 2.2, Problem 2.28E

(a)

To determine

To Calculate: the median and explain the meaning of this means in this setting.

Answer to Problem 2.28E

  ${\text{M=Q}}_2=87.75$

Explanation of Solution

Given:

93 93 87.5 91 94.5 72 96 95 93.5 93.5 73 82 45 88 80

86 85.5 87.5 81 78 86 89 92 91 98 85 82.5 88 94.5 43

Formula used:

  ${\text{Q}}_2=2\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}$

Calculation:

Order the number in ascending order

43 45 72 73 78 80 81 82 82.5 85 85.5 86 86 87.5 87.5

88 88 88 89 91 91 92 93 93.5 93.5 94.5 94.5 95 96 98

  $\begin{array}{c}{\text{Q}}_2=2\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}\\ =2\times \left(\frac{\text{30+1}}{4}\right)\text{th postion}\\ \text{=15}\text{.5}\\ \text{=15th positon + 0}\text{.5}\left(16th\text{ - 15th}\right)\\ =87.5+0.5\left(88-87.5\right)\\ =87.75\\ {\text{M=Q}}_2=87.75\end{array}$

I

The median or 2^nd Quartile means 50% of the data fall less than a certain value and that percentage value is known as 50^th percentile or median here 87.75 is the median which implies that 50% of the data fall less than 87.5 and 50% of the data fall greater than 87.5

(b)

To determine

To Calculate: the quartiles and IQR and interpret meaning of IQR in this setting.

Answer to Problem 2.28E

  $\begin{array}{c}{\text{Q}}_1=81.75\\ {\text{Q}}_3=93.125\\ \text{IQR}=11.375\end{array}$

Explanation of Solution

Given:

43 45 72 73 78 80 81 82 82.5 85 85.5 86 86 87.5 87.5

88 88 88 89 91 91 92 93 93.5 93.5 94.5 94.5 95 96 98

Formula used:

  $\begin{array}{c}{\text{Q}}_1=1\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}\\ {\text{Q}}_3=3\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}\\ {\text{IQR=Q}}_3-{\text{Q}}_1\end{array}$

Calculation:

  $\begin{array}{c}{\text{Q}}_1=1\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}\\ =1\times \left(\frac{\text{30+1}}{4}\right)\text{th postion}\\ \text{=7}\text{.75}\\ \text{=7th position +0}\text{.75}\left(8th-7th\right)\\ =81+0.75\left(82-81\right)\\ =81.75\end{array}$

  $\begin{array}{c}{\text{Q}}_3=3\times \left(\frac{\text{N+1}}{4}\right)\text{th postion}\\ =3\times \left(\frac{\text{30+1}}{4}\right)\text{th postion}\\ \text{=3}\times \text{7}\text{.75}\\ \text{=23}\text{.75}\\ \text{=23rd position+0}\text{.25}\left(24th-23rd\right)\\ =93+0.25\left(93.5-93\right)\\ =93+0.125\\ =93.125\end{array}$

  $\begin{array}{c}{\text{IQR=Q}}_3-{\text{Q}}_1\\ =93.125-81.75\\ =11.375\end{array}$

IQR stands for inter quartile range. It is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile and the quartile. Here 11.375 data value represent that spread of the middle 50% of the data.

(c)

To determine

To Explain: that is there outlier and justifies the answer.

Answer to Problem 2.28E

43 and 45 are extremely high outliers.

Explanation of Solution

Given:

43 45 72 73 78 80 81 82 82.5 85 85.5 86 86 87.5 87.5

88 88 88 89 91 91 92 93 93.5 93.5 94.5 94.5 95 96 98

Formula used:

For greater than ${\text{Q}}_3+1.5\left(\text{IQR}\right)$ and

For lesser than ${\text{Q}}_1-1.5\left(\text{IQR}\right)$

Calculation:

  $\begin{array}{c}{\text{Q}}_3+1.5\left(\text{IQR}\right)=93.125+1.5\left(11.375\right)\\ =110.1875\\ {\text{Q}}_1-1.5\left(\text{IQR}\right)=81.75-1.5\left(11.375\right)\\ =64.6875\end{array}$

An outlier is an extremely high or extremely low value in the data set. It can be identify an outlier if it is greater than ${\text{Q}}_3+1.5\left(\text{IQR}\right)$ and lesser then ${\text{Q}}_1-1.5\left(\text{IQR}\right)$ and there are two extremely high outliers 43 and 45.

(d)

To determine

To construct: the box plot on the basis of given data.

Explanation of Solution

Given:

93 93 87.5 91 94.5 72 96 95 93.5 93.5 73 82 45 88 80

86 85.5 87.5 81 78 86 89 92 91 98 85 82.5 88 94.5 43

Graph:

  

By seeing the box plot it is clearly observed that there are 2 high outliers which are 43 and 45 and the 1^st quartile and 3^rd quartile are about 81 and 93 respectively and the median is about 88 which are looking forth.