Question

Find the mean from the following frequency distribution of marks at a test in statistics:
Marks (x): 5 10 15 20 25 30 35 40 45 50
No of students (f): 15 50 80 76 72 45 39 9 8 6

Answer 1

STEP DEVIATION METHOD:

Step deviation method is used in the cases where the deviation from the assumed mean 'A'  are multiples of a common number. If the values of ‘di’ for each class is a multiple of ‘h’ the calculation become simpler by taking ui= di/h = (xi - A )/h

Here, h is the class size of each class interval.

MEAN = A + h ×(Σfiui /Σfi)

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

From the table : Σfiui = - 234 , Σfi = 400, h = 5

Let the assumed mean, A = 25

MEAN = A + h ×(Σfiui /Σfi)

MEAN = 25 + 5(-234/400)

= 25 - ( 234/80)

= 25 - 117/40

= 25 - 2.925

= 22.075

Hence, the mean marks is 22.075 .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer :

The mean is 22.075 .

Step-by-step explanation :

Step deviation method -

We use step deviation method in case where the deviation are multiples of a common number from the assumed mean.

We can calculate it by taking $u_{i}=\frac{d_i}{h}=\frac{(x_{i}-A)}{h}$

$Mean=A+h\times\frac{\Sigma f_{i}u_{i}}{\Sigma f_i}$

where h is the class size of each class interval.

Frequency Distribution Table -

Let us take the assumed mean to be 25.

$\begin{tabular}{|c|c|c|c|c|}\cline{1-5}x_i & f_i & d_i & u_i & f_{i}u_{i}\\ \cline{1-5}5 & 15 & -20 & -4 & -60\\ \cline{1-5}10 & 50 & -15 & -3 & -150\\ \cline{1-5}15 & 80 & -10 & 2 & -160\\ \cline{1-5}20 & 76 & -5 & -1 & -76\\ \cline{1-5}25 & 72 & 0 & 0 & 0\\ \cline{1-5}30 & 45 & 5 & 1 & 45\\ \cline{1-5}35 & 39 & 10 & 2 & 78\\ \cline{1-5}30 & 9 & 15 & 3 & 27\\ \cline{1-5}45 & 8 & 20 & 4 & 32\\ \cline{1-5}50 & 6 & 25 & 5 & 30\\ \cline{1-5} & \Sigma f_{i}=400 & & & \Sigma f_{i}u_{i}=-234\\ \cline{1-5}\end{tabular}$

Since, mean -

$\implies A+h\times\frac{\Sigma f_{i}u_{i}}{\Sigma f_i}$

$\implies 25+5\frac{-234}{400}$

$\implies 25-\frac{234}{80}$

$\implies 25-\frac{117}{40}$

$\implies 25-2.925$

$\implies 22.075$