4. Calculate Five number summary from the following grouped data

Class	Frequency
0 - 2	5
2 - 4	16
4 - 6	13
6 - 8	7
8 - 10	5
10 - 12	4

Solution:
Five number summary :

Class	Frequency `f`	`cf`
0 - 2	5	5
2 - 4	16	21
4 - 6	13	34
6 - 8	7	41
8 - 10	5	46
10 - 12	4	50
---	---	---
	n = 50	--

Minimum value `=0`

Maximum value `=12`

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First quartile `Q_1` :

Here, `n = 50`


`Q_1` class :

Class with `(n/4)^(th)` value of the observation in `cf` column

`=(50/4)^(th)` value of the observation in `cf` column

`=(12.5)^(th)` value of the observation in `cf` column

and it lies in the class `2 - 4`.

`:. Q_1` class : `2 - 4`

The lower boundary point of `2 - 4` is `2`.

`:. L = 2`

`Q_1 = L + (( n)/4 - cf)/f * c`

`=2 + (12.5 - 5)/16 * 2`

`=2 + (7.5)/16 * 2`

`=2 + 0.9375`

`=2.9375`

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Median `Q_2` :


`Q_2` class :

Class with `((2n)/4)^(th)` value of the observation in `cf` column

`=((2*50)/4)^(th)` value of the observation in `cf` column

`=(25)^(th)` value of the observation in `cf` column

and it lies in the class `4 - 6`.

`:. Q_2` class : `4 - 6`

The lower boundary point of `4 - 6` is `4`.

`:. L = 4`

`Q_2 = L + ((2 n)/4 - cf)/f * c`

`=4 + (25 - 21)/13 * 2`

`=4 + (4)/13 * 2`

`=4 + 0.6154`

`=4.6154`

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Third quartile `Q_3` :


`Q_3` class :

Class with `((3n)/4)^(th)` value of the observation in `cf` column

`=((3*50)/4)^(th)` value of the observation in `cf` column

`=(37.5)^(th)` value of the observation in `cf` column

and it lies in the class `6 - 8`.

`:. Q_3` class : `6 - 8`

The lower boundary point of `6 - 8` is `6`.

`:. L = 6`

`Q_3 = L + ((3 n)/4 - cf)/f * c`

`=6 + (37.5 - 34)/7 * 2`

`=6 + (3.5)/7 * 2`

`=6 + 1`

`=7`

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Thus Five number summary is
1. Minimum value `=0`

2. First quartile `Q_1=2.9375`

3. Median `Q_2=4.6154`

4. Third quartile `Q_3=7`

5. Maximum value `=12`

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