There is a grouped data distribution for which mean is to be found by

1.	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.
Answer» There is a grouped data distribution for which mean is to be found by step deviation method. $\begin{matrix} C l a s s I n t e r v a l & N u m b e r o f F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & D . . . . . . 100 - 200 & 39 & 150 & B . . . . . & E . . . . . 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & C . . . . . & F . . . . . . T o t a l & A = \sum f_{i} = . . . . . . \end{matrix}$ Find the value of A, B, C, D, E and F respectively.

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.

Answer»

There is a grouped data distribution for which mean is to be found by step deviation method.

$\begin{matrix} C l a s s I n t e r v a l & N u m b e r o f F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & D . . . . . . 100 - 200 & 39 & 150 & B . . . . . & E . . . . . 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & C . . . . . & F . . . . . . T o t a l & A = \sum f_{i} = . . . . . . \end{matrix}$

Find the value of A, B, C, D, E and F respectively.

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