1
Question
Find the mean using step deviation method
WeightNo. of Packets200−20113201−20227202−20318203−20410204−2051205−2061
Find the mean using step deviation method
WeightNo. of Packets200−20113201−20227202−20318203−20410204−2051205−2061
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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 (xi−a) 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=xi−ah
WeightNo. of PacketsMid valueui=xi−ahfiui (fi) (xi) 200−20113200.5113201−20227201.5=a00202−20318202.5−1−18203−20410203.5−2−10204−2051204.5−3−3205−2061205.5−4−4 ∑fi=70 ∑fiui=−22
⇒∑fiui=−22h=1,n=70,a=201.5
Mean=a+(∑fiui∑fi)×h
=201.5+(−2270)×1=201.5−0.31=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 (xi−a) 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=xi−ah
WeightNo. of PacketsMid valueui=xi−ahfiui (fi) (xi) 200−20113200.5113201−20227201.5=a00202−20318202.5−1−18203−20410203.5−2−10204−2051204.5−3−3205−2061205.5−4−4 ∑fi=70 ∑fiui=−22
⇒∑fiui=−22h=1,n=70,a=201.5
Mean=a+(∑fiui∑fi)×h
=201.5+(−2270)×1=201.5−0.31=201.19
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