Question

The median and mode of the following distribution are 33.5 and 34 rupees respectively. Find the missing frequencies.

Daily wages (in Rs.)	0-10	10-20	20-30	30-40	40-50	50-60	60-70	N
Frequencies	4	16	60	$x$	$y$	$z$	4	230

Answer 1

From the table we will first form an equation corresponding to the sum of all frequencies. Then we will use the formula of median given by ${\text{Median}} = l + \left( {\dfrac{{\dfrac{N}{2} - c.f.}}{f}} \right) \times h$ and formula of mode given by ${\text{Mode}} = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$ to form equations. Solve the equations to find the value of $x$ , $y$ and $z$.

Complete step-by-step answer:
We are given, that the sum of frequencies is 230.
Form as equation corresponding to the given condition.
Write the sum of frequencies in the form of an equation.
$4 + 16 + 60 + x + y + z + 4 = 230$
Add the numbers on the left side of the equation.
$84 + x + y + z = 230$
After subtracting 84 from both sides, we get,
$x + y + z = 146{\text{ }}\left( 1 \right)$
Add the previous frequencies to form the table with cumulative frequencies.

Daily wages (in Rs.)	0-10	10-20	20-30	30-40	40-50	50-60	60-70	N
Frequencies	4	16	60	$x$	$y$	$z$	4	230
Cumulative frequency (c.f.)	4	20	80	80+$x$	80+$x$+$y$	80+$x$+$y$+$z$	84+$x$+$y$+$z$

We are given that the median of the data is 33.5, this implies that the median class for the given data is 30-40.
Substitute the values in the formula of median, ${\text{Median}} = l + \left( {\dfrac{{\dfrac{N}{2} - c.f.}}{f}} \right) \times h$, where $l$ is the lower class limit of the median class, $N$ is the total number, c.f. refers to the cumulative frequency of the previous class, and f refers to the frequency of the class and $h$is the width of the class interval.
On substituting the values, we get,
${\text{33}}{\text{.5}} = 30 + \left( {\dfrac{{\dfrac{{230}}{2} - 80}}{x}} \right) \times 10$
Solve the above equation to find the value of $x$
${\text{33}}{\text{.5}} - 30 = \left( {\dfrac{{115 - 80}}{x}} \right) \times 10 \\\ \dfrac{{3.5}}{{10}} = \dfrac{{35}}{x} \\\ x = \dfrac{{35 \times 10}}{{3.5}} \\\ x = 100 \\\$
Also, it is given that the mode of the data is 34.
Substitute the values in the formula of mode, ${\text{Mode}} = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$, where $l$ is the lower class limit of the modal class which is 30-40, $N$ is the total number, ${f_1}$ refers to frequency of the modal class , and ${f_0}$ refers to the frequency of the previous modal class, ${f_2}$ refers to the frequency of the succeeding modal class and $h$is the width of the class interval.
On substituting the values, we get,
${\text{34}} = 30 + \left( {\dfrac{{x - 60}}{{2x - 60 - y}}} \right) \times 10$
We will substitute 100 for$x$and solve the equation to find the value of $y$
${\text{34}} = 30 + \left( {\dfrac{{100 - 60}}{{2\left( {100} \right) - 60 - y}}} \right) \times 10 \\\ 34 - 30 = \left( {\dfrac{{40}}{{200 - 60 - y}}} \right) \times 10 \\\ \dfrac{4}{{10}} = \dfrac{{40}}{{140 - y}} \\\ 140 - y = 100 \\\ y = 40 \\\$
Now, we can find the value of $z$by substituting the value of $x$ and $y$ in equation (1)
$100 + 40 + z = 146 \\\ 140 + z = 146 \\\ z = 6 \\\$
Hence, the missing entries are, $x = 100$, $y = 40$ and $z = 6$.

Note: In the formula of the median, we have to calculate cumulative frequencies and have to use cumulative frequency of the interval previous to the median class. Also, in the formula of mode, frequencies are used and not the cumulative frequencies.