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### STATISTICS

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• 20. Illustration – Discrete Frequency Distribution 4 68 6 66 10 64 18 62 12 60 No. of Students Height (in inches)
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• 26. Obtaining Cumulative Frequency Distribution 73 + 2 = 75 2 2 45-50 65 + 8 = 73 2 + 8 = 10 8 40-45 55 + 10 = 65 10 + 10 = 20 10 35-40 49 + 6 =55 20 + 6 = 26 6 30-35 15 +34 =49 26 + 34 = 60 34 25-30 15 60 + 15 = 75 15 20-25 Less than type More than type Cum.frequency cum.frequency Frequency Class -Intervals
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• 31. Averages Mathematical Averages Positional Averages A.M G.M H.M Median Mode
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• 34. μ = 3144 / 50 = 62.88 Ans. Illustration 4 272 50 = N 3144 = Σ fx 68 6 396 66 10 640 64 18 1116 62 12 60 x 12 = 720 60 No. of Students f fX Height (in inches) X
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• 42. Illustration Proceed as usual 2-0=2 90-100 2 90 0 100 9-2=7 80-90 9 80 25-9=16 70-80 25 70 45-25=20 60-70 45 60 75-45=30 50-60 75 50 100-75=25 40-50 100 40 118-100=18 30-40 118 30 133-118=15 20-30 133 20 140-133= 7 10-20 140 10 Freq. C.I Cum. Freq. Marks X or more
• 43. What if… ? N=40 Total 3 0-9 15 10-19 10 20-29 8 30-39 3 40-49 1 50-59 Frequency C.I
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• 45. Determining missing frequency when A.M is known – Illustration Mean = 16.82 Σ fd = -12 N = 70 + f 4 24 3 32.5 8 30-35 20 2 27.5 10 25-30 14 1 22.5 14 20-25 0 0 17.5 = A ? = f 4 15-20 -16 -1 12.5 16 10-15 -24 -2 7.5 12 5-10 -30 -3 2.5 10 0-5 fd d= (x –A)/i M.V (x) Freq. Marks
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• 55. Illustration – Weighted A.M 106000 = Σ wX 350 = Σ w 15000 150 100 Lower Staff 25000 100 250 Clerical Staff 35000 70 500 Subordinate Staff 16000 20 800 Class II officers 15000 10 1500 Class I Officers wX No. of employees (w) Monthly Salary (in Rs.) (X) Designation
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• 62. Calculation of Median-Illustration (Discrete Freq. Distribution) Here N = 50 (i) N/2 = 25 (ii) Cum. Frequency just greater than N/2 = 30 (iii)Corresponding value of item is 62. Median = 60 Ans. 12 12 60 30 18 62 40 10 64 46 6 66 50 4 68 N = 50 Cum. Freq. No. of students Height (in inches)
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• 64. Calculation of Median-Illustration (Grouped Freq. Distribution) N/2 = 3600/2 = 1800 Cum.freq. just greater than 1800 is 2600. Hence median class is 25-30. Hence L1 = 25 L2 = 30 C = 1800 f = 800 Md = 25 + 1800 - 1800 (30 – 25 ) 800 = 25 Ans. Σ f= 3600 3600 400 35-40 3200 600 30-35 2600 800 25-30 1800 900 20-25 900 700 15-20 200 200 10-15 Cum. Freq Freq.(f) C.I
• 65. Calculation of Missing Frequencies when median is known : Illustration : Median = 50 N = 100 15 56 + f1 + f2 80-100 ? = f 2 41+ f 1 +f 2 60-80 27 41 + f 1 40-60 ? = f 1 14 + f 1 20-40 14 14 0-20 No. of Families Cumulative Freq. Expenditure
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• 74. Mode: Formula for Continuous Frequency Distribution Mode = L1 + h(f1 – f0) 2f1-f0-f2
• 75. Empirical Relationship between Mean, Median & Mode Mode = 3 Median – 2 Mean
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