Find the mean deviation about the mean for the following data: 1. income per day - 0-100,100-200, 300-400, 400-500, 500-600

Question

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

1. 

income per day	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Number of person	4	8	9	10	7	5	4	3

2.

Height	95-105	105-115	115-125	125-135	135-145	145-155
Number of boy	9	13	26	30	12	10

Answer 1

1.

income	xi	fi	xi - fi	|xi - 358|	|xi - 358| fi
0- 100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	2250	392	1176
		50	17900		7896

Mean = \(\bar x\) = \(\frac{\displaystyle\sum_{i=1}^{n} x_i f_i}{\displaystyle \sum_{i - 1}^{n}f_i}\) = \(\frac{17900}{50}\) = 358

M.D(\(\bar x\)) = \(\frac{\displaystyle\sum_{i=1}^{n}f_i|x_i - \bar x|}{\displaystyle \sum_{i - 1}^{n}f_i}\) = \(\frac{7896}{50}\) = 157.92

2.

Height	xi	fi	xi - fi	|xi - 125.3|	|xi - 125.3| fi
95-105	100	9	900	25.3	227.7
105-115	110	13	1430	15.3	198.9
115-125	120	26	3120	5.3	137.8
125-135	130	30	3900	4.7	141
135-145	140	12	1680	14.7	176.4
145-155	150	10	1500	24.7	247
		100	12530		1128.8

Mean = \(\bar x\) =  \(\frac{\displaystyle\sum_{i=1}^{n} x_i f_i}{\displaystyle \sum_{i - 1}^{n}f_i}\) = \(\frac{12530}{100} = 125.3\)

M.D (\(\bar x\)) = \(\frac{\displaystyle\sum_{i=1}^{n}f_i|x_i - \bar x|}{\displaystyle \sum_{i - 1}^{n}f_i}\) = \(\frac{1128.8}{100} = 11.28\)