# Statistics – Exercise 1.5.3 – Class IX

1. Find the mean deviation about mean for the following date:

a) 14, 21, 28, 21, 18

Solution:

Mean = 14+21+28+21+18/5 = 102/5 = 20.4

 Score Deviation from mean |D| 14 14 – 10.4 = -6.4 6.4 18 18 – 20.4 = -2.4 2.4 21 21 – 20.4 = 0.6 0.6 21 21 – 20.4 = 0.6 0.6 28 28 – 20.4 = 7.6 7.6

(b)

 Score(x) 6 20 8 18 16 12 14 10 Frequency(f) 2 7 11 27 18 13 17 5

Solution:

 x f fxD= x – x |D| f|D| 6 2 12 – 8.58 8.58 17.16 20 7 140 + 5.42 5.42 37.94 8 11 88 – 6.58 6.58 72.38 18 27 486 – 3.42 3.42 92.34 16 18 288 – 1.42 1.42 25.56 12 13 156 – 2.58 2.58 33.56 14 17 238 – 0.58 0.58 9.86 10 5 50 – 4.58 4.58 22.9 N = 100 1458 311.68

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

a) 15, 18, 13, 16, 12, 24, 10, 20

Solution:

15+ 18+ 13+ 16+ 12+ 24+ 10+ 20 =128

N = 8

⅀x = 1258

Mean= x/N =128/8 = 16

 x D = x – x |D| 10 10-16=-6 6 12 12-16 = -4 4 13 13 – 16 = -3 3 15 15 – 16 = -1 1 16 16 – 16 = 0 0 18 18 – 16 = 2 2 20 20 – 16 = 4 4 24 24 – 16 = 8 8 ⅀28

MD =|D|/N = 28/8 = 3.5

(b)

 CI f 10-19 6 20-29 4 30-39 10 40-49 9 50-59 11 60-69 8 70-79 2

Solution:

 CI f x fx D = x – x |D| 10-19 6 14.5 87 -29.5 177 20-29 4 24.5 98 -19.5 78 30-39 10 34.5 345 -9.5 95 40-49 9 44.5 400.5 0.5 4.5 50-59 11 54.5 599.5 10.5 115.5 60-69 8 64.5 516 20.5 164.0 70-79 2 74.5 149 30.5 61 N  = 50 ⅀fx=2195 ⅀|D|=695

(c)

 Class interval Frequency 0-5 9 5-10 13 10-15 6 15-20 12 20-25 9 25-30 6 30-35 10 35-40 15 40-45 6 45-50 4

Solution:

 Class interval Frequency x fx D= x – x |D| 0-5 9 2.5 22.5 -20.5 184.5 5-10 13 7.5 97.5 -15.5 201.5 10-15 6 12.5 75.0 -10.5 63.0 15-20 12 17.5 210.0 -5.5 66.0 20-25 9 22.5 202.5 -0.5 4.5 25-30 6 27.5 165.0 -4.5 27.0 30-35 10 32.5 325.0 9.5 95.0 35-40 15 37.5 562.5 14.5 217.5 40-45 6 42.5 255.0 19.5 117.0 45-50 4 47.5 190.0 24.5 98.0 N = 90 ⅀fx=2105 ⅀|D|=1074.0

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

a) 18, 23, 9, 11, 26, 4, 14, 21

Solution:

4, 9, 1, 14, 18, 21, 23, 26

Median = N+1/2 = 8+1/2 = 9/2 =  4.5th

14+18/2 = 32/2= 16

median = 16

 x D =  x – median |D| 4 4 – 16 = -12 12 9 9 – 16 = -7 7 11 11 – 16 = -2 5 14 14  – 16 = -2 2 18 18 – 16 =  2 2 21 21 – 16 = 05 5 23 23 – 16 = 07 7 26 26 – 16 = 10 10 50

MD = 50/8 = 6.25

(b)

 Class interval Frequency 8 – 12 14 13   – 17 8 18 –  22 20 23 – 27 7 28 – 32 11 33 – 37 10 38 – 42 24 43  – 47 6

Solution:

 Class interval Frequency x fx D = x –  median f|D| 8 – 12 14 10 14 10 – 23 = -13 182 13   – 17 8 15 22 15 – 23 =  -8 64 18 –  22 20 20 42 20 – 23 = -3 60 23 – 27 7 25 29 25 – 23 = 2 14 28 – 32 11 30 60 30 – 23 = 7 77 33 – 37 10 35 70 35  –  23 = 12 120 38 – 42 24 40 94 40  – 23 = 17 408 43  – 47 6 45 100 45 – 23 = 22 132 N = 100 ⅀f|D|=1057

(c)

 Class interval frequency 20 – 30 9 30 – 40 18 40 – 50 7 50 – 60 21 60 – 70 11 70 – 80 4

Solution:

 Class interval frequency x fc D = x – median f|D| 20 – 30 9 25 19 25-42=-17 153 30 – 40 18 35 27 35-42=-7 126 40 – 50 7 45 34 45-42=3 21 50 – 60 21 55 55 55-42=13 273 60 – 70 11 65 66 65-42 = 23 253 70 – 80 4 75 70 75-42 =33 132 N = 70 ⅀f|D|=958

1. Find the mean deviation about mean and median for the following data:

a)

 Cl 1-5 6-10 11-15 16-20 21-25 f 2 9 5 4 10

Solution:

 Cl f fc x fx D = x – x f|D| 1-5 2 2 3 06 3 – 15 = -12 24 6-10 9 11 8 72 8 – 15 = -7 63 11-15 5 16 13 65 13 – 15 = -2 10 16-20 4 20 18 72 18-15 = 3 12 21-25 10 30 23 230 23 – 15 = 8 80 N=30 ⅀fx=445 189

Mean = 445/30 = 14.83 = 15

MD from Mean = 189/30 = 6.3

 CI f fc x D fx 1-5 2 2 3 -11.5 23 6-10 9 11 8 -6.5 58.5 11-15 5 16 13 -1.5 7.5 16-20 4 20 18 ±3.5 14.0 21-25 10 30 23 ±8.5 85.0 N=30 ⅀fx = 188

(b)

 CI 5 – 10 10-15 15-20 20-25 25-30 f 5 12 3 11 9

Solution:

 CI f fc x fx D=x-x f|D| x-median f|D| 5-10 5 5 7.5 37.5 7.5-18=10.5 52.5 -7.5 62.5 10-15 12 17 12.5 150 -5.5 66 -2.5 90 15-20 3 20 17.5 52.5 -0.5 1.5 2.5 7.5 20-25 11 31 22.5 247.5 4.5 49.5 7.5 29.5 25-30 9 40 27.5 247.5 9.5 85.5 12.5 0.5 N= 40 ⅀fx=735 235 255

Mean = 735/40 = 18.375 ≈18