Chapter 10.4, Problem 2GP

To determine

To graph a box and whisker plot of the given data set

Explanation of Solution

Given information:

Data set : 40, 50, 35, 48, 43, 31 , 52, 42, 54, 38, 50, 46 , 49 , 4 3, 43, 40, 50, 32, 53, 51, 43, 47, 41, 49, 50, 34, 54, 51, 44, 54, 39, 47, 35, 51, 44, 48, 37

Formula used:

Lower quartile is,

  $Q1=(n+1)\frac{1}{4}term$

Upper quartile is,

  $Q3=(n+1)\frac{3}{4}term$

Median is,

  $M=\frac{(\frac{n}{2})+(\frac{n}{2}+1)}{2}term$

Graph:

  

Calculation:

Re-arranging the data in ascending order,

31, 32, 34, 35, 35, 37, 38, 39, 40, 40, 41, 42, 43, 43, 43, 44, 44, 46, 47, 47, 48, 48, 49, 49, 50, 50, 50, 50, 51,51, 51, 52, 53, 54, 54, 54.

For median, substituting the values,

  $M=\frac{(\frac{36}{2})+(\frac{36}{2}+1)}{2}term$

On solving,

  $M=\frac{{18}^{th}+{19}^{th}}{2}term$

Substituting the values,

  $M=\frac{46+47}{2}$

On solving,

  $M=46.5$

Hence, median and Q2 is 46.5

For lower quartile substituting the values,

  $Q1=(36+1)\frac{1}{4}$

So,

  $Q1=37\times \frac{1}{4}$

On solving,

  $Q1=9.25$

Rounding off the value,

  $Q1\approx 9$

Substituting the value of the lower quartile,

  $Q1=40$

Hence, lower quartile is 40

For upper quartil2 substituting the values,

  $Q3=(36+1)\frac{3}{4}$

So,

  $Q3=(37)\times \frac{3}{4}$

On solving,

  $Q3=27.75$

On rounding off the value,

  $Q3\approx 28$

Substituting the value of 28^th observation,

  $Q3=50$

Hence, upper quartile is 50

Minimum is the smallest value of the data set.

Hence, minimum=31

Maximum is the biggest value of the data set.

Hence, maximum= 54