Question

If the median of the following frequency distribution is $46$ find the missing frequencies.

Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80	Total
Frequency	12	30	P	65	Q	25	18	230

Answer 1

Hint: Here we are given the median as $46$ so first of all we need to find the median class using the cumulative frequency and then the value of the median is given by the formula
${\text{median}} = l + \left( {\dfrac{{\dfrac{n}{2} - CF}}{f}} \right)h$
Here $l$ is the lower limit of the median class, $h$ is the class size and here $n$ is the sum of the frequencies and $CF$ is the cumulative frequency before the median class and $f$ is the frequency of the median class.

Complete step-by-step answer:
Here we are given the frequency distribution table and we are provided that the median of it is $46$

Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80	Total
Frequency	12	30	P	65	Q	25	18	230

And we are also given that the total frequency is $230$ and we know that the two frequencies are missing which are ${\text{P and Q}}$. So as we are given that the sum of the frequencies is $230$ so we can write that
$
  12 + 30 + P + 65 + Q + 25 + 15 + 18 = 230 \\
\Rightarrow P + Q = 230 - 150 \\
\Rightarrow P + Q = 80 - - - - - (1) \\
 $
Now we need to draw the cumulative frequency of the intervals to find the median and then we will get the second equation.
Cumulative frequency is the sum of the frequencies preceding to that interval and the frequency of that interval.

variable	frequency	Cumulative frequency
$10 - 20$	$12$	$12$
$20 - 30$	$30$	$42$
$30 - 40$	$P$	$42 + P$
$40 - 50$	$65$	$107 + P$
$50 - 60$	$Q$	$107 + P + Q$
$60 - 70$	$25$	$132 + P + Q$
$70 - 80$	$18$	$150 + P + Q$

And $\sum {f = 230} $
As we know that $46$ is our median so we can say that $40 - 50$ is our median class and we know the formula of the median which is
${\text{median}} = l + \left( {\dfrac{{\dfrac{n}{2} - CF}}{f}} \right)h$
Here $l$ is the lower limit of the median class which is$40$, $h$ is the class size which is $10$ and here $n$ is the sum of the frequencies which is$230$ and $CF$ is the cumulative frequency before the median class which is \[42 + P\] and $f$$ = 65$ is the frequency of the median class.
So we can write that
$46 = 40 + \left( {\dfrac{{\dfrac{{230}}{2} - 42 - P}}{{65}}} \right)10$
$
  6 = \left( {\dfrac{{73 - P}}{{65}}} \right)10 \\
\Rightarrow P = 73 - 39 = 34 \\
 $
And we know that from equation (1)
$
  P + Q = 80 \\
\Rightarrow 34 + Q = 80 \\
\Rightarrow Q = 46 \\
 $
Hence we get that
$P = 34,Q = 46$
Note: If we are given the value of mean and median then we can find the mode by using the given formula
${\text{mode}} = 3{\text{median}} - 2{\text{mean}}$