Question

There is a grouped data distribution for which mean is to be found by step deviation method. For this purpose, the below table is constructed. Fill in the blanks

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

The A, B, C, D, E and F respectively are

• 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

$d_i$ = $x_i$ - a

where $x_i$ = $i^{th}$ class mark

a = assumed mean

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

-200 = 50 – a

a = 250

Using above formula for $2^{nd}$ class(100-200) we get-

B = 150 – 250

B = -100

Similarly, C = 200.

$u_i$ = $\frac{\left(d_i\right)}{h}$

Here h = class interval = 100. h is constant.

For $1^{s}t$ class(0-100),

$u_1$ = $\frac{d_1}{h}$= -200/100 = -2

Thus, D = -2

Similarly E = -1, F = 2

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