calculate Mean, median and mode from the following data:

$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
$17$	$6$	$37$	$0$	$25$	$13$	$12$

Hint: Here we will find the mean, median, and mode of the given grouped data by using certain formulae for each.
We will first find the mean of the distribution using the help of class interval and frequency.
Followed by cumulative frequency and mid-value.
Formula used:
The mean formula is given by,$Mean = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$ the sum of the values
$Mode = l + \dfrac{{h\left( {{f_1} - {f_0}} \right)}}{{2{f_1} - {f_0} - {f_2}}}$ and $Median = l + \dfrac{{\dfrac{n}{2} - cf}}{f}.\,h$

Complete step-by-step solution:
For finding the mean, median and mode tabulate the required data,


Now finding the mean of given data by using the formula
$Mean = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$
Here $\sum\limits_{i = 1}^n {{f_i}{x_i}} = 4820$ and $\sum\limits_{i = 1}^n {{f_i} = 110} $then applying in the formula we get,
$Mean = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{4820}}{{110}} = 43.81$
Now using mode formula that is
$Mode = l + \dfrac{{h\left( {{f_1} - {f_0}} \right)}}{{2{f_1} - {f_0} - {f_2}}}$
Now define the modal class, here modal class is $30 - 40$
Here $l$ is the lowest value of the modal class here it is $l = 30$
${f_1}$ is the frequency of the modal class ${f_1} = 37$
${f_0}$ is the frequency preceding ${f_0} = 6$
${f_2}$is the frequency succeeding ${f_2} = 0$
And $h$ is the width of the class interval $h = 10$ applying these values we get,
$Mode = 30 + \dfrac{{10\left( {37 - 6} \right)}}{{2\left( {37} \right) - 6 - 0}}$
On simplifying it we get,
$Mode = 30 + \dfrac{{310}}{{68}}$
$Mode = \dfrac{{2350}}{{68}} = 34.55$
Now calculating median by using the formula
$Median = l + \dfrac{{\dfrac{n}{2} - cf}}{f}.\,h$
$n$ be the total frequency $n = 110 \Rightarrow \dfrac{n}{2} = 55$ we can choose the median class just greater than this value,
So here the median class is $30 - 40$
Where Here $l$ is the lower boundary value of the median class here it is $l = 30$
Cumulative frequency preceding median class $cf = 60$
$f$ frequency of the median class $f = 37$
And $h$ is the width of the class interval $h = 10$ applying these values we get,
$Median = 30 + \dfrac{{55 - 60}}{{37}}.\,10$
$Median = 30 + \dfrac{{ - 50}}{{37}}$
$Median = \dfrac{{1110 - 50}}{{37}} = \dfrac{{1060}}{{37}} = 28.64$
So, the mean, median, and mode of the given data were found.

Note: In the case of positively skewed frequency distribution, the mean is always greater than the median and the median is always greater than the mode.
$Mean. > Median > Mode$
Arrange in ascending, then it n is odd, the median is the $\dfrac{{n + 1}}{2}$ . and if n is even, then the median will be the average of the $\dfrac{n}{2}$ th and the $\dfrac{n}{{2 + 1}}$ th observation (median).