MATHEMATICS-IIA. 1. Calculate the variance and standard deviation of the following continuous frequency distribution

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1 1. Calculate the variance and standard deviation of the following continuous frequency distribution d i = ( x i A ) c d i d i d i (A) d i = 15 d i = 105 Here =50, A=65, C=10, ( d i ) = 15, d i = 105 mean(x) = A + ( d i ) C = 65 + ( ) 10 = 65 3 = 6 Variance= C [( d i ) ( d i ) ] = [50(105) ( 15) ] = 100 [550 5] 500 = 1 [505] = 01 5 S.D= 01 = tutorial

2 . Calculate the variance and standard deviation of the following discrete frequency distribution x i x i x i (x i x) (x i x) (x i x) (x i x) = 1374 x i = 40 Here =30, x i = 40 = mean(x) = ( x i ) Variance= (x i x) = =45.8 (x i x) = 1374 = 14 S.D= 45.8 = 6.77 tutorial

3 3. The following table gives daily wages of workers in a factory. Calculate the standard deviation and coefficient of variation of the wages of the workers. wages o of workers d i = ( x i A ) c d i d i d i (A) d i = 31 d i = 39 Here =7, A=300, C=50, mean(x) = A + ( d i ) C = ( 31 7 ) 50 Variance= C [( d i ) ( d i ) ] S.D= 50 7 [7(39) ( 31) ] ( d i ) = 31, = = 88.5 d i = 39 = coefficient of variation = S.D 100 = x 3 tutorial

4 4. The scores of two cricketers A and B in 10 innings are given below, find who is better run getter and who is a more consistent player. Scores of A: x i Scores of B: y i (x i x) (x i x) (y i ) (y i y) (y i y) x i =540 (x i x) = y i =380 (y i y) = For cricketer A: x = x i =540 = 54; S. D = 10 (x i x) = = = For cricketer B: y = y i =380 = 38; S. D = 10 (y i y) = = 1068 = C.V of A = σ x = 100 = 60.8 and C.V of B = σ y x 54 y Since x > y, crickete A is a better run getter And C.V of A >C.V of B, A is also more consistent player = 100 = tutorial

5 5. Find the mean deviation from the mean of the following continuous frequency distribution d i = ( x i A ) c d i (A) =50 d i =10 = 47 Here =50, A=5, C=10, ( d i ) = 10, = 47 mean(x) = A + ( d i ) C x = 65 + ( ) 10 x = 65 3 x = 6 Mean D from mean = x i x = = tutorial

6 6. Find the mean deviation from the mean of the following continuous frequency distribution d i = ( x i A ) c d i (A) =100 d i =60 = 1040 Here =50, A=5, C=10, ( d i ) = 10, = 47 mean(x) = A + ( d i ) C x = 65 + ( ) 10 x = x = 71 Mean D from mean = x i x = = tutorial

7 7. Find the mean deviation from the median of the following continuous frequency distribution C.F x i M (A) , =1000 = 8175 Here =1000, =500, f=160, F=40 C=5, L=35 x i M = 8175 median(x) = L + ( P. C. F ). C f M = 35 + ( ) x = 35 + ( ) 5 Mean D from median = x i M = = x = = tutorial

8 8. Find the mean deviation from the median of the following continuous frequency distribution Arranging the observations in ascending order we get C.F x i M x i M (M) =30 = 149 Here =30, =15, Median=13 x i M = 149 Mean D from median = x i M = = tutorial

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