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

ScoreDeviation from mean|D|
1414 – 10.4 = -6.46.4
1818 – 20.4 = -2.42.4
2121 – 20.4 = 0.60.6
2121 – 20.4 = 0.60.6
2828 – 20.4 = 7.67.6

Statistics - Exercise 1.5.3 – Class IX


(b)

Score(x)62081816121410
Frequency(f)2711271813175

Solution:

xffxD= x – x|D|f|D|
6212 – 8.588.5817.16
207140 + 5.425.4237.94
81188 – 6.586.5872.38
1827486 – 3.423.4292.34
1618288 – 1.421.4225.56
1213156 – 2.582.5833.56
1417238 – 0.580.589.86
10550 – 4.584.5822.9
 N = 1001458 311.68

Statistics - Exercise 1.5.3 – Class IX


  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

xD = x – x|D|
1010-16=-66
1212-16 = -44
1313 – 16 = -33
1515 – 16 = -11
1616 – 16 = 00
1818 – 16 = 22
2020 – 16 = 44
2424 – 16 = 88
  ⅀28

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


(b)

CIf
10-196
20-294
30-3910
40-499
50-5911
60-698
70-792

Solution:

CIfxfxD = x – x|D|
10-19614.587-29.5177
20-29424.598-19.578
30-391034.5345-9.595
40-49944.5400.50.54.5
50-591154.5599.510.5115.5
60-69864.551620.5164.0
70-79274.514930.561
 N  = 50 ⅀fx=2195 ⅀|D|=695

Statistics - Exercise 1.5.3 – Class IX


(c)

Class intervalFrequency
0-59
5-1013
10-156
15-2012
20-259
25-306
30-3510
35-4015
40-456
45-504

Solution:

Class intervalFrequencyxfxD= x – x|D|
0-592.522.5-20.5184.5
5-10137.597.5-15.5201.5
10-15612.575.0-10.563.0
15-201217.5210.0-5.566.0
20-25922.5202.5-0.54.5
25-30627.5165.0-4.527.0
30-351032.5325.09.595.0
35-401537.5562.514.5217.5
40-45642.5255.019.5117.0
45-50447.5190.024.598.0
 N = 90 ⅀fx=2105 ⅀|D|=1074.0

Statistics - Exercise 1.5.3 – Class IX


  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

xD =  x – median|D|
44 – 16 = -1212
99 – 16 = -77
1111 – 16 = -25
1414  – 16 = -22
1818 – 16 =  22
2121 – 16 = 055
2323 – 16 = 077
2626 – 16 = 1010
  50

MD = 50/8 = 6.25


(b)

Class intervalFrequency
8 – 1214
13   – 178
18 –  2220
23 – 277
28 – 3211
33 – 3710
38 – 4224
43  – 476

Solution:

Class intervalFrequencyxfxD = x –  medianf|D|
8 – 1214101410 – 23 = -13182
13   – 178152215 – 23 =  -864
18 –  2220204220 – 23 = -360
23 – 277252925 – 23 = 214
28 – 3211306030 – 23 = 777
33 – 3710357035  –  23 = 12120
38 – 4224409440  – 23 = 17408
43  – 4764510045 – 23 = 22132
 N = 100   ⅀f|D|=1057

Statistics - Exercise 1.5.3 – Class IX


(c)                                                       

Class intervalfrequency
20 – 309
30 – 4018
40 – 507
50 – 6021
60 – 7011
70 – 804

Solution:

Class intervalfrequencyxfcD = x – medianf|D|
20 – 309251925-42=-17153
30 – 4018352735-42=-7126
40 – 507453445-42=321
50 – 6021555555-42=13273
60 – 7011656665-42 = 23253
70 – 804757075-42 =33132
 N = 70   ⅀f|D|=958

Statistics - Exercise 1.5.3 – Class IX


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

a)

Cl1-56-1011-1516-2021-25
f295410

Solution:

ClffcxfxD = x – xf|D|
1-5223063 – 15 = -1224
6-109118728 – 15 = -763
11-15516136513 – 15 = -210
16-20420187218-15 = 312
21-2510302323023 – 15 = 880
 N=30  ⅀fx=445 189

Mean = 445/30 = 14.83 = 15

MD from Mean = 189/30 = 6.3

CIffcxDfx
1-5223-11.523
6-109118-6.558.5
11-1551613-1.57.5
16-2042018±3.514.0
21-25103023±8.585.0
 N=30   ⅀fx = 188

Statistics - Exercise 1.5.3 – Class IX


(b)

CI5 – 1010-1515-2020-2525-30
f5123119

Solution:

CIffcxfxD=x-xf|D|x-medianf|D|
5-10557.537.57.5-18=10.552.5-7.562.5
10-15121712.5150-5.566-2.590
15-2032017.552.5-0.51.52.57.5
20-25113122.5247.54.549.57.529.5
25-3094027.5247.59.585.512.50.5
 N= 40  ⅀fx=735 235 255

Mean = 735/40 = 18.375 ≈18

Statistics - Exercise 1.5.3 – Class IXStatistics - Exercise 1.5.3 – Class IX


 

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