Question

Find the missing frequency..​

Answer 1

Answer:

Please Refer to the attachment.

Answer 2

$\large\underline{\sf{Solution-}}$

The frequency distribution table for calculations of mean using Step Deviation Method are given below :-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c|c|c}\sf Class\: interval&\sf Frequency\: (f_i)&\sf \: midvalue \: (x_i)&\sf \: u_i&\sf \: f_iu_i\\\frac{\qquad  \qquad}{}&\frac{\qquad  \qquad}{}\\\sf 0 - 30&\sf 12&\sf15&\sf - 2&\sf - 24\\\\\sf 30 - 60 &\sf 21&\sf45&\sf - 1&\sf - 21\\\\\sf 60-90 &\sf f_1 &\sf75&\sf0&\sf0\\\\\sf 90 - 120&\sf 52&\sf105&\sf1&\sf52\\\\\sf 120-150&\sf f_2&\sf135&\sf2&\sf2f_2\\\\\sf 150-180&\sf 11&\sf165&\sf3&\sf33\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & 150\sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Now, given that

$\rm \: \displaystyle\sum\rm f_i = 150$

$\rm \: 96 + f_1 + f_2 = 150$

$\rm\implies \:\boxed{ \rm{ \:f_1 + f_2 = 54 \: }} - - - (1)$

Now,

$\rm \: A = 75$

$\rm \: \overline{x} \: = \: 91$

$\displaystyle\sum\rm f_iu_i = 40 + 2f_2$

$\displaystyle\sum\rm f_i = 150$

$\rm \: h = 30$

Now, using Step Deviation Method, mean is evaluated as

$\rm \: \overline{x} \: = \: A + \frac{\displaystyle\sum\rm f_iu_i}{\displaystyle\sum\rm f_i} \times h$

So, on substituting the values, we get

$\rm \: 91 = 75 + \dfrac{40 + 2f_2}{150} \times 30$

$\rm \: 91 - 75 = \dfrac{40 + 2f_2}{5}$

$\rm \: 16 = \dfrac{40 + 2f_2}{5}$

$\rm \: 40 + 2f_2 = 80$

$\rm \: 2f_2 = 80 - 40$

$\rm \: 2f_2 = 40$

$\rm\implies \:\boxed{ \rm{ \:f_2 \: = \: 20 \: \: }}$

On substituting this value in equation (1), we get

$\rm\implies \:\boxed{ \rm{ \:f_1 \: = \: 34 \: \: }}$

$\rule{190pt}{2pt}$

Additional information :-

1. Mean using Direct Method :-

$\rm \: \overline{x} \: = \: \frac{\displaystyle\sum\rm f_ix_i}{\displaystyle\sum\rm f_i}$

2. Mean using Short Cut Method

$\rm \: \overline{x} \: = \: A + \frac{\displaystyle\sum\rm f_id_i}{\displaystyle\sum\rm f_i}$