Question

42, 36, 45, 38, 51, 47, 36, 49 1. What is the range of the data? A. 8 G C. 15 B. 13 D. 25 2. What is the IQR of the data? A. 9 C. 11 D. 14 B. 10

Answer 1

Arrange the given data in ascending order.

`36,36,38,42,45,47,49,51`

The range is difference between highest and lowest value.

Determine the range for the data.

`\begin{gathered} R=51-36 \\ =15 \end{gathered}`

So range is 15.

The lower half of the data is 36,36,38 and 42 and upper half f the data is 45,47,49 and 51.

Determine the Quartile 1 and Quartile 3 for the data.

`\begin{gathered} Q_1=(36+38)/(2) \\ =37 \end{gathered}`
`\begin{gathered} Q_3=(47+49)/(2) \\ =48 \end{gathered}`

Determine the Inter Quartile range (IQR) for the data.

`\begin{gathered} \text{IQR}=Q_3-Q_1 \\ =48-37 \\ =11 \end{gathered}`

So IQR is 11.

Determine the mean value of the data.

`\begin{gathered} \text{Mean}=(42+36+45+38+51+47+36+49)/(8) \\ =(344)/(8) \\ =43 \end{gathered}`

Determine the absolute deviation of the data.

`\begin{gathered} |43-36|=|7| \\ =7 \end{gathered}`
`\begin{gathered} |43-36|=|7| \\ =7 \end{gathered}`
`\begin{gathered} |43-38|=|5| \\ =5 \end{gathered}`
`\begin{gathered} |43-42|=|1| \\ =1 \end{gathered}`
`\begin{gathered} |43-45|=|-2| \\ =2 \end{gathered}`
`\begin{gathered} |43-47|=|-4| \\ =4 \end{gathered}`
`\begin{gathered} |43-49|=|-6| \\ =6 \end{gathered}`
`\begin{gathered} |43-51|=|-8| \\ =8 \end{gathered}`

Determine the mean value for the absolute deviation.

`\begin{gathered} \text{MAD}=(7+7+5+1+2+4+6+8)/(8) \\ =(40)/(8) \\ =5 \end{gathered}`

So value of Mean Absolute Deviation (MAD) is 5.