**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 |

**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 |

**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} = 4.5^{th}

^{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 |

**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

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