Solving Box and Whisker Plots Practice

Introduction To solving box and whisker plots practice:

Box-and-whisker diagram is also called as box plot or Whisker plot.
It is a suitable way of graphically give a picture of groups of numerical data with the help of their five-number summaries.
Box and Whisker plots are generally used in the display of statistical analyses of a group of numbers.
The five number summary is a different name for the visual illustration of the box and whisker plot.
The five number summary consist of:

1. The 2nd quartile.
2. The 1st quartile.
3. The 3rd quartile.
4. The Largest value in a data set.
5. The minimum value in a data set.

Diagram of Box and Wisker Plot :-



Now Lets see an example problem that helps you to understand the topic Solving box and whisker plots practice .

Solved Example Problem on Solving Box and Whisker Plots Practice

Solve and draw  the box plot for the following set of numbers

35, 24, 53, 57, 14, 78, 95

Solution:-

To draw the box plot for the given set of the numbers we have to arrange the given set of numbers in increasing order.

35, 24, 53, 57, 14, 78, 95

A box plot is entirely based on medians. The basic step is to find the median of the given set of the numbers.

Practice Help Step 1:-

We can find the median of set of numbers by using the formula` (n+1)/2`

14, 24, 35, 53, 57, 78, 95

Here the numbers in the set is odd (n=7) so we use this formula.

`(n+1)/2 = (7+ 1) /2 `

By Solving the above step we get  4

So the number in the fourth position is the median.

Here the number is 53

So the median is 53.

Practice Help Step 2:-

The median of lower numbers is called as lower quartile. (1st quartile).

14, 24, 35

Here the numbers in the set is odd (n=3) so we use this formula.

`(n+1)/2 = (3+ 1) /2 `
By solving the above we get as 2

Medial of lower set of numbers is 2

So the lower quartile is 24

Practice Help Step 3:-

The median of upper numbers is called as lower quartile. (3rd quartile).

57, 78, 95

Here the numbers in the set is odd (n=3) so we use this formula.

`(n+1)/2 = (3+ 1) /2 `   By solving the above we get as 2
Medial of upper set of numbers is 78

So the upper quartile is 78

Practice Help Step 4:-

The sample maximum is the number with the largest value in the data set.

Here it is 95

The sample maximum is 95

Practice Help Step 5:-

The sample minimum is the number with the smallest value in the data set.Having problem with 9th class cbse question papers keep reading my upcoming posts, i will try to help you.

Here it is 14

The sample minimum is 14

The 5 Number Summary is

The solved box plot gives the following information

Median – 53
Lower Quartile -24
Upper Quartile - 78
The sample maximum - 95
The sample minimum - 14