We have to calculate mean deviation from the median. So, first we calculate the median.
CI | x | F | cf | d = (x – med) | fd |
0 – 10 | 5 | 5 | 5 | 20 | 100 |
10 – 20 | 15 | 10 | 15 | 10 | 100 |
20 – 30 | 25 | 20 | 35 | 0 | 0 |
30 – 40 | 35 | 5 | 91 | 10 | 50 |
40 – 50 | 45 | 10 | 101 | 20 | 200 |
50 | 450 |
CI | x | f | xf | d = (x-mean) | fd |
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 |
Classes | fi | xi | di | fidi | ||
95 – 105 | 9 | 100 | -3 | -27 | 28.58 | 257.22 |
105 – 115 | 13 | 110 | -2 | -26 | 18.58 | 241.54 |
115 – 125 | 16 | 120 | -1 | -16 | 8.58 | 137.28 |
125 – 135 | 26 | 130 | 0 | 0 | 1.42 | 36.92 |
135 – 145 | 30 | 140 | 1 | 30 | 11.42 | 342.6 |
145 – 155 | 12 | 150 | 2 | 24 | 21.42 | 257.04 |
N = 106 | Total = -15 | Total = 1272.60 |
CI | x | f | xf | d = (x - mean) | fd |
0 – 10 | 5 | 6 | 30 | 22 | 132 |
10 – 20 | 15 | 8 | 120 | 12 | 96 |
20 – 30 | 25 | 14 | 350 | 2 | 28 |
30 – 40 | 35 | 16 | 560 | 8 | 128 |
40 – 50 | 45 | 4 | 180 | 18 | 72 |
50 – 60 | 55 | 2 | 110 | 28 | 56 |
50 | 1350 | 512 | |||
Mean | 27 | ||||
Deviation | 10.24 |
Find the mean deviation from the mean for the data:
Classes | 0 -10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |
We have to calculate mean deviation from the median. So, first we calculate the median.
CI | x | f | cf | D = (x – med) | fd |
17 – 19.5 | 18.25 | 5 | 5 | 20 | 100 |
20 – 25.5 | 22.75 | 16 | 21 | 15.5 | 248 |
26 – 35.5 | 30.75 | 12 | 33 | 7.5 | 90 |
36 – 40.5 | 38.25 | 26 | 59 | 0 | 0 |
41 – 50.5 | 45.75 | 14 | 73 | 7.5 | 105 |
51 – 55.5 | 53.25 | 12 | 85 | 15 | 180 |
56 – 60.5 | 58.25 | 6 | 91 | 20 | 120 |
61 – 70.5 | 65.75 | 5 | 96 | 27.5 | 137.5 |
96 | 980.5 |
We have N = 96 ⟹ N/2 = 48
The cumulative frequency just greater than N/2 is 59 and the corresponding value of x is 38.25.
Hence, median = 38.25
M.D from Median
Marks | Students | xi | Cum. Freq | fidi | |
0 – 10 | 5 | 5 | 5 | 55/3 | 275/3 |
10 – 20 | 8 | 15 | 13 | 25/3 | 200/3 |
20 – 30 | 15 | 25 | 28 | 3-May | 75/3 |
30 – 40 | 16 | 35 | 44 | 35/3 | 560/3 |
40 – 50 | 6 | 45 | 50 | 65/3 | 390/3 |
N = 50 | Total = 500 |
M.D from mean
Marks | Students | xi | fidi | |xi - 27| | fi|xi - 27| | |
0 – 10 | 5 | 5 | -3 | -15 | 22 | 110 |
10 – 20 | 8 | 15 | -2 | -16 | 12 | 96 |
20 – 30 | 15 | 25 | -1 | -15 | 2 | 30 |
30 – 40 | 16 | 35 | 0 | 0 | 8 | 128 |
40 – 50 | 6 | 45 | 1 | 6 | 18 | 108 |
N = 50 | Total = 40 | Total = 472 |
Converting the given data into continuous frequency distribution by sub tracing 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
Age | x: | f: | Cumulative frequency | |d:|=|x:-38| | f:|d:| |
15.5 – 20.5 | 18 | 5 | 5 | 20 | 100 |
20.5 – 25.5 | 23 | 6 | 11 | 15 | 90 |
25.5 – 30.5 | 28 | 12 | 23 | 10 | 120 |
30.5 – 35.5 | 33 | 14 | 37 | 5 | 70 |
35.5 – 40.5 | 38 | 26 | 63 | 0 | 0 |
40.5 – 45.5 | 43 | 12 | 75 | 5 | 60 |
45.5 – 50.5 | 48 | 16 | 91 | 10 | 160 |
50.5 – 55.5 | 53 | 9 | 100 | 15 | 135 |
N = Σf: = 100 | Σf:|d:| = 735 |
Clearly, N = 100 ⟹ N/2 = 50.
Cumulative frequency is just greater than N/2 is 63 and the corresponding class is 35.5 - 40.5.
I = 35.5, f = 26, h = 5, F = 37
Classes | fi | xi | fixi | |xi - 9.2| | fi|xi - 9.2| |
0 – 4 | 4 | 2 | 8 | 7.2 | 28.8 |
4 – 8 | 6 | 6 | 35 | 3.2 | 19.2 |
8 – 12 | 8 | 10 | 80 | 0.8 | 6.4 |
12 – 16 | 5 | 14 | 70 | 4.8 | 24 |
16 – 20 | 2 | 18 | 36 | 8.8 | 17.6 |
N = 25 | Total = 230 | Total = 96.0 |
Classes | fi | xi | fixi | |xi - 14.1| | fi|xi - 14.1| |
0 – 6 | 4 | 3 | 12 | 11.1 | 44.4 |
6 – 12 | 5 | 9 | 45 | 5.1 | 25.5 |
12 – 18 | 3 | 15 | 45 | 0.9 | 2.7 |
18 – 24 | 6 | 21 | 126 | 6.9 | 41.4 |
24 – 30 | 2 | 27 | 54 | 12.9 | 25.8 |
N = 20 | Total = 282 | Total = 139.8 |
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