Question

# There is a grouped data distribution for which mean is to be found by step deviation method. For this purpose, the below table is constructed. Fill in the blanks Class IntervalNumber of Frequency(fi)Class mark(xi)di=xi−aui=dih0−1004050−200D......100−20039150B.....E.....200−3003425000300−400303501001400−50045450C.....F......TotalA.∑fi=...... The A, B, C, D, E and F respectively are 188 , -100, 200, -2 ,-1 and 2 186 , 100, -200, -2 ,-1 and 2 188 , 100, -200, 2 ,-1 and -2 186 , -100, -200, -2 ,-1 and 2

Solution

## The correct option is A 188 , -100, 200, -2 ,-1 and 2 ∑fi= sum of frequencies of each class = 40 + 39 + 34 + 30 + 45 = 188 di = xi - a where xi = ith class mark a = assumed mean Using above formula for 1st class(0-100) we get- -200 = 50 – a a = 250 Using above formula for 2nd class(100-200) we get-   B = 150 – 250  B = -100 Similarly, C = 200.  ui = (di)h Here h = class interval = 100. h is constant. For 1st class(0-100),  u1 = d1h= -200/100 = -2 Thus, D = -2 Similarly E = -1, F = 2  Answers are 188 , -100, 200, -2 ,-1 and 2

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