Find the mean deviation about the median for the following data: Marks - 0-10, 10-20, 20-30, 30-40, 40-50, 50-60

Question

Find the mean deviation about the median for the following data:

1.

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of Girls	6	8	14	16	4	2

2.

Age	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
Number	5	6	12	14	26	12	16	9

Answer 1

1.

 

Marks	xi	fi	c.f	|xi - 27.86|	|xi-27.86| fi
0-10	5	6	6	22.86	137.16
10-20	15	8	14	12.86	102.88
20-30	25	14	28	2.86	40.04
30-40	35	16	44	7.14	114.24
40-50	45	4	48	17.14	68.56
50-60	55	2	50	27.14	54.28
			50		517.16

Median class is the class in which the \((\frac{N}{2})^{th}\) observation lies.

\(\frac{N}{2} = \frac{50}{2}= 25\)

Median class = 20 - 30

M = Median

= l + \(\cfrac{c(\frac{N}{2}-f_0)}{f_1}\)

= 20 + \(\frac{25-14}{14} \times 10 = 27.86\)

M.D(M) = \(\frac{\displaystyle \sum_{i=1}^n f_i|x_i - M|}{\displaystyle\sum_{i=1}^n f_i}\) = \(\frac{517.16}{50}\) = 10.34

2. 

Age	xi	fi	c.f	|xi - 38|	|xi-38| fi
16-20	18	5	5	20	100
21-25	23	6	11	15	90
26-30	28	12	23	10	120
31-35	33	14	37	5	70
36-40	38	26	63	0	0
41-45	43	12	75	5	60
46-50	48	16	91	10	160
51-55	53	9	100	15	135
		100			738

 Median class is the class in which the \((\frac{N}{2})^{th}\) observation lies.

\(\frac{N}{2} = \frac{100}{2} = 50\)

Median class = 35.5 - 40.5

M = Median

n = l + \(\cfrac{C(\frac{N}{2}-f_0)}{f_1}\)

= 35.5 + \(\frac{50-37}{2600} \times 5 = 38\)

M.D(M) = \(\frac{\displaystyle \sum_{i=1}^n f_i|x_i - M|}{\displaystyle\sum_{i=1}^n f_i}\) = \(\frac{735}{100} = 7.35\)