There is a grouped data distribution for which mean is to be found by step deviation method.
Class IntervalNumber of Frequency(fi)Class mark(xi)di=xi−aui=dih0−1004050−200D......100−20039150B.....E.....200−3003425000300−400303501001400−50045450C.....F......TotalA=∑fi=......
Find the value of A, B, C, D, E and F respectively.
188 , -100, 200, -2 ,-1 and 2
∑fi= sum of frequencies of each class
= 40 + 39 + 34 + 30 + 45
= 188
∴A=188
Now, di=xi−a
where,
xi is ith row class mark and 'a' is assumed mean.
Using above formula for 1st class (0-100) we get:
−200=50−a⇒a=250 and class size h=100
Using the above information, lets complete the table,.
Class IntervalNumber of Frequency(fi)Class mark(xi)di=xi−aui=dih0−1004050−200−2100−20039150−100−1200−3003425000300−400303501001400−500454502002TotalA=∑fi=188
Thus,A, B, C, D, E and F are 188,−100,200,−2,−1 and 2 respectively.