Chapter SH, Problem 13.3.1EP

To determine

To find:the range, interquartile range, and any outliers for the given data

Answer to Problem 13.3.1EP

The range is $73$ , the inter quartile range is $42$ and here outliers are $\text{none }$

Explanation of Solution

Given:

  $44,37,23,35,61,95,49,96$

Calculation:

Range:

Order the data from least to greatest.

  $44,37,23,35,61,95,49,96$

Range is the difference between greatest and smallest value.

  $\begin{array}{l}\text{Range}=96-23\\ =73\end{array}$

Hence, the range is $73$

Inter quartile range:

Find the upper quartile and lower quartile.

Here median is $\frac{44+49}{2}=46.5$

Divide the data into two halves.

  $\stackrel{\text{Upper half}}{\overbrace{23,35,37,44,}}\stackrel{\text{median}}{\overbrace{46.5,}}\stackrel{\text{Upper half}}{\overbrace{49,61,95,96}}$

Median of lower half is

  $\begin{array}{l}LQ=\frac{35+37}{2}\\ =36\end{array}$

Median of upper half is

  $\begin{array}{l}\text{UQ}=\frac{61+95}{2}\\ =78\end{array}$

Inter quartile range is

  $\begin{array}{l}\text{UQ}-\text{LQ}=78-36\\ =42\end{array}$

Hence, the inter quartile range is $42$

OutlierData that are more than 1.5 times the value of the interquartile range beyond the quartiles is called outlier.

Here inter quartile range is 42

  $42\times 1.5=63$

Now subtract 63 from the lower quartile and add to upper quartile.

  $\begin{array}{l}LQ-63=-27\hfill \\ UQ+63=141\hfill \end{array}$

Thus the data which is not in the range of -27 to 141 is outlier.

Here outliers are $\text{none }$

Conclusion:

Therefore, the range is $73$ , the inter quartile range is $42$ and here outliers are $\text{none }$