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Question

Find the mean using step deviation method
WeightNo. of Packets200−20113201−20227202−20318203−20410204−2051205−2061


A
205.19
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B
203.19
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C
202.19
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D
201.19
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Solution

The correct option is D 201.19

1.Choose an arbitrary constant 'a'. (also called assumed mean)

2. xi = Upper limit+Lower limit2

3. Subtract the value of 'a' from xi.

4. The reduced value of (xia) is called the deviation of xi from 'a'.

5. Divide the deviation by constant where 'h' is the width of the class interval.

6. ui=xiah

WeightNo. of PacketsMid valueui=xiahfiui (fi) (xi) 20020113200.511320120227201.5=a0020220318202.511820320410203.52102042051204.5332052061205.544 fi=70 fiui=22

fiui=22h=1,n=70,a=201.5

Mean=a+(fiuifi)×h
=201.5+(2270)×1=201.50.31=201.19


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