Question

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

$$\begin{array}{ccccc}ClassInterval& NumberofFrequency\left(f_i\right)& Classmark\left(x_i\right)& d_i=x_i-a& u_i=\frac{d_i}{h}\\ 0-100& 40& 50& -200& D......\\ 100-200& 39& 150& B.....& E.....\\ 200-300& 34& 250& 0& 0\\ 300-400& 30& 350& 100& 1\\ 400-500& 45& 450& C.....& F......\\ Total& A=\sum f_i=......& & & \end{array}$$

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

• A. 188 , -100, 200, -2 ,-1 and 2
• B. 186 , 100, -200, -2 ,-1 and 2
• C. 188 , 100, -200, 2 ,-1 and -2
• D. 186 , -100, -200, -2 ,-1 and 2

Answer 1

The correct option is A

188 , -100, 200, -2 ,-1 and 2

$\sum f_i$= sum of frequencies of each class
= 40 + 39 + 34 + 30 + 45
= 188
$\therefore A=188$

Now, $d_i=x_i-a$

where,
$x_i$ is $i^{th}$ row class mark and 'a' is assumed mean.

Using above formula for $1^{st}$ class (0-100) we get:

$-200=50-a\Rightarrow a=250\text{and class size}h=100$

Using the above information, lets complete the table,.
$$\begin{array}{ccccc}ClassInterval& NumberofFrequency\left(f_i\right)& Classmark\left(x_i\right)& d_i=x_i-a& u_i=\frac{d_i}{h}\\ 0-100& 40& 50& -200& -2\\ 100-200& 39& 150& -100& -1\\ 200-300& 34& 250& 0& 0\\ 300-400& 30& 350& 100& 1\\ 400-500& 45& 450& 200& 2\\ Total& A=\sum f_i=188& & & \end{array}$$

$Thus,A,B,C,D,E$ and $F$ are $188,-100,200,-2,-1$ and $2$ respectively.