Question

Find the mean, median and mode of the following data:

Classes	0-50	50-100	100-150	150-200	200-250	250-300	300-350
Frequency	2	3	5	6	5	3	1

Answer 1

Hint: We start solving the problem by checking the given frequency distribution continuous or not. We then find the midpoint of class interval, sum of frequencies and sum of multiplication of midpoint and frequencies. We then use the formula $m={\displaystyle \frac{\sum _{}^{}{f}_i{x}_i}{\sum _{}^{}{f}_i}}$ to find the mean of the given data.
We then find the cumulative frequency of each of the class intervals and then find the class interval in which the median lies. We then use $m_d=L_i+{\displaystyle \frac{{\displaystyle \frac{N}{2}}-c{f}_{i-1}}{f_i}}\times I$ to get the value of median. We then find the class interval with highest frequency which is the modal interval. We then use the formula $m_e=L+{\displaystyle \frac{f_m-f_{m-1}}{(f_m-f_{m-1})+(f_m-f_{m+1})}}\times I$ to get the value of mode.

Complete step by step answer:
According to the problem, we need to find the mean, median and mode of the given data in the distribution table.

Classes	0-50	50-100	100-150	150-200	200-250	250-300	300-350
Frequency	2	3	5	6	5	3	1

We can see that the lower boundary of the previous class interval is equal to the upper boundary of the next class interval, which makes the given distribution table continuous.
Let us first find the mean of the given frequency distribution.
Let us find the midpoint of the given classes

Classes	Mid-point	Frequency
0-50	25	2
50-100	75	3
100-150	125	5
150-200	175	6
200-250	225	5
250-300	275	3
300-350	325	1

Now, let us multiply frequency with the midpoint of the classes and find the sum of all the frequencies and the multiplied results.

Classes	Mid-point $\left(x_i\right)$	Frequency $\left(f_i\right)$	$f_i\times x_i$
0-50	25	2	50
50-100	75	3	225
100-150	125	5	625
150-200	175	6	1050
200-250	225	5	1125
250-300	275	3	825
300-350	325	1	325
	Total	$\sum _{}^{}{f}_i=25$	$\sum _{}^{}{f}_i{x}_i=4225$

We know that mean is defined as $m={\displaystyle \frac{\sum _{}^{}{f}_i{x}_i}{\sum _{}^{}{f}_i}}$.
$\Rightarrow m={\displaystyle \frac{4225}{25}}$.
$\therefore m=169$.
So, we have found the mean of the frequency distribution as 169.
Now, let us find the median of the given frequency distribution.
Let us find the cumulative frequency for each of the class intervals.

Classes	Frequency	Cumulative frequency
0-50	2	2
50-100	3	5
100-150	5	10
150-200	6	16
200-250	5	21
250-300	3	24
300-350	1	25

We know that the cumulative frequency of the last class interval is equal to the sum of all the frequencies given.
So, we get $\sum _{}^{}{f}_i=25$. We know that the median lies at the cumulative frequency ${\displaystyle \frac{25+1}{2}}=13$. We can see that this occurs at the class interval $150-200$.
We know that the median of the grouped data is defined as $m_d=L_i+{\displaystyle \frac{{\displaystyle \frac{N}{2}}-c{f}_{i-1}}{f_i}}\times I$.
Where $L_i$ = lower limit of the class interval in which we get the median = 150.
N = sum of all the frequencies given = 25.
$c{f}_{i-1}$ = cumulative frequency of the class interval which is just before the interval we get the median = 10.
$f_i$ = frequency of the interval where we get the median = 6.
I = width of the class interval = 50.
So, we get $m_d=150+{\displaystyle \frac{{\displaystyle \frac{25}{2}}-10}{6}}\times 50$.
$\Rightarrow m_d=150+{\displaystyle \frac{12.5-10}{6}}\times 50$.
$\Rightarrow m_d=150+{\displaystyle \frac{2.5}{6}}\times 50$.
$\Rightarrow m_d=150+20.83$.
$\therefore m_d=170.83$.
∴ We have found the median as 170.83.
Now, let us find the mode of the given frequency distribution.

Classes	Frequency
0-50	2
50-100	3
100-150	5
150-200	6
200-250	5
250-300	3
300-350	1

Now, let us find the class interval in which we get the higher frequency in which we get the mode.
We can see that the frequency is higher in the class interval $150-200$, which is the modal interval.
We know that the mode of a grouped data is defined as $m_e=L+{\displaystyle \frac{f_m-f_{m-1}}{(f_m-f_{m-1})+(f_m-f_{m+1})}}\times I$.
Where, L = Lower limit of the modal interval = 150.
$f_m$ = frequency of the modal interval = 6.
$f_{m-1}$ = frequency of the class interval which is before modal interval = 5.
$f_{m+1}$ = frequency of the class interval which is after modal interval = 5.
I = Width of the class interval = 50.
So, we get $m_e=150+{\displaystyle \frac{6-5}{(6-5)+(6-5)}}\times 50$.
$\Rightarrow m_e=150+{\displaystyle \frac{1}{1+1}}\times 50$.
$\Rightarrow m_e=150+25$.
$\therefore m_e=175$.
So, we have found the mode of the given frequency distribution as 175.

Note:
We should know that the obtained values of mean, median and mode are estimates that need not be the exact correct values. We should not confuse the formula of one with the other while solving this problem. We should not make calculation mistakes while solving this problem. Similarly, we can check the condition mean = 3median – 2mean after getting their values.