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
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
44.
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)
63.
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