Chapter 1.3, Problem 89E

(a)

To determine

To find: the interquartile range (IQR).

Answer to Problem 89E

The interquartile range is 14

Explanation of Solution

Given:

S No	Quiz grades
1	74
2	75
3	76
4	78
5	80
6	82
7	84
8	86
9	87
10	90
11	91
12	93
13	96
14	98

Calculation:

The inter-quartile range is calculated as

  $\begin{array}{l}{Q}_1=\frac{N+1}{4}\text{th}\text{item}\\ Q_1=\frac{14+1}{4}\text{th}\text{item}\\ Q_1=\frac{15}{4}\text{th}\text{item}\\ Q_1=3.75\text{th}\text{item}\\ \text{=}\frac{1}{4}x3rd\text{item}+\frac{3}{4}x4\text{th}\text{item}\\ Q_1=\frac{1}{4}x76+\frac{3}{4}x78\\ =77.5\\ Q_3=\frac{3\left(N+1\right)}{4}\text{th}\text{item}\\ Q_3=\frac{3\left(14+1\right)}{4}\text{th}\text{item}\\ Q_3=\frac{45}{4}\text{th}\text{item}\\ Q_3=11.25\text{th}\text{item}=\frac{3}{4}x11\text{th}\text{item}+\frac{1}{4}x12\text{th}\text{item}\\ Q_3=\frac{3}{4}x91+\frac{1}{4}x93\\ =91.5\\ IQR=Q_3-Q_1\\ =91.5-77.5\\ =14\end{array}$

Conclusion:

The interquartile range is 14

(b)

To determine

whether there are any outliers.

Answer to Problem 89E

All the items are within the range thus there are no outliers.

Explanation of Solution

Calculation:

From Joe’s first 14 quiz grades in a marking period data we can say that there are no outliers in the data. We call an observation as an outlier if it falls more than (1.5) IQR above the third quartile or below the first quartile.

  $\begin{array}{l}IQR\times 1.5\\ 14\times 1.5=21\\ Q_1-21=77.5-21\\ =56.5\\ Q_3-21=91.5+21\\ =112.5\end{array}$

All the items are within the range thus there are no outliers.

Conclusion:

Thus there are no outliers.