The following data shows that the age distribution of patients of malaria in a village during a particular month. Find the average age of the patients.

Age in years	No of cases
\[5 - 14\]	\[6\]
\[15 - 24\]	\[11\]
\[25 - 34\]	\[21\]
\[35 - 44\]	\[23\]
\[45 - 54\]	\[14\]
\[55 - 64\]	\[5\]
\[65 - 74\]	\[3\]

\[({\text{A) 36}}{\text{.12}}\]
\[(B{\text{) 36}}{\text{.13}}\]
\[(C{\text{) 13}}{\text{.36}}\]
\[({\text{D) 23}}{\text{.36}}\]

Hint: First, we have to find the class mark (mid-value of the intervals).
Multiply it with the f, frequencies to get \[\sum {{{\text{x}}_{\text{i}}}} {{\text{f}}_{\text{i}}}\] and using mean formula we will be able to find the answer.

Formula used: \[{\text{Mid value = }}\dfrac{{{\text{lower limit + upper limit}}}}{2}\]
To find mean,
\[\overline {\text{x}} {\text{ = }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}} {{\text{f}}_{\text{i}}}}}{{{{\sum {\text{f}} }_{\text{i}}}}}\]

Complete step-by-step solution:
This is a grouped data where class intervals are given. So, we need to find class marks.
Class mark is nothing but mid value of intervals (class mark is taken as \[{\text{(}}{{\text{x}}_{\text{i}}}{\text{)}}\])
Here the formula for \[{\text{Mid value = }}\dfrac{{{\text{lower limit + upper limit}}}}{2}\]
Then we get, \[\dfrac{{{\text{5}}\left( {{\text{lower limit}}} \right){\text{ + 14(upper limit)}}}}{{\text{2}}}\]
On adding the numerator terms and we get,
\[ = \dfrac{{19}}{2}\]
Let us divide,
\[ \Rightarrow 9.5\]
\[\dfrac{{15\left( {{\text{lower limit}}} \right){\text{ + 24(upper limit}})}}{2}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{39}}{2}\]
Let us divide,
\[ \Rightarrow 19.5\]
\[\dfrac{{{\text{25}}\left( {{\text{lower limit}}} \right){\text{ + 34(upper limit)}}}}{{\text{2}}}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{59}}{2}\]
Let us divide,
\[ \Rightarrow 29.5\]
\[\dfrac{{35\left( {{\text{lower limit}}} \right){\text{ + 44(upper limit}})}}{2}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{79}}{2}\]
Let us divide,
\[ \Rightarrow 39.5\]
\[\dfrac{{{\text{45}}\left( {{\text{lower limit}}} \right){\text{ + 54(upper limit)}}}}{{\text{2}}}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{99}}{2}\]
Let us divide,
\[ \Rightarrow 49.5\]
\[\dfrac{{{\text{55}}\left( {{\text{lower limit}}} \right){\text{ + 64(upper limit)}}}}{{\text{2}}}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{119}}{2}\]
Let us divide,
\[ \Rightarrow 59.5\]
\[\dfrac{{{\text{65}}\left( {{\text{lower limit}}} \right){\text{ + 74(upper limit)}}}}{{\text{2}}}\]
On adding the numerator terms and we get,
\[ \Rightarrow \dfrac{{139}}{2}\]
Let us divide,
\[ \Rightarrow 69.5\]
Now, we have to find \[{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}\] by multiplying \[{\text{(}}{{\text{f}}_{\text{i}}})\]and \[{\text{(}}{{\text{x}}_{\text{i}}}{\text{)}}\] to compute mean.
i.e., $6 \times 9.5 = 57$, like this we get \[214.5,619.5,908.5,693,297.5\] and \[208.5\]

Add all the \[{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}\] and \[{\text{(}}{{\text{f}}_{\text{i}}})\] to get \[{\sum {\text{f}} _{\text{i}}}{{\text{x}}_{\text{i}}}\] and \[{\sum {\text{f}} _{\text{i}}}\].
So here we have,
\[{\sum {\text{f}} _{\text{i}}}\] = 83
\[{\sum {\text{f}} _{\text{i}}}{{\text{x}}_{\text{i}}}\] = 2998.5
To find mean, the formula is \[\overline {\text{x}} {\text{ = }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}} {{\text{f}}_{\text{i}}}}}{{{{\sum {\text{f}} }_{\text{i}}}}}\]
Substituting the formula, we get
\[\overline {\text{x}} = \dfrac{{2998.5}}{{83}}\]
\[ = 36.126\]
\[ = 36.13\]

Therefore the correct answer is option \[({\text{B) }}36.13\]

Note: In this Alternative method:
We can avoid the tedious calculations of computing mean ${\text{(}}{{\text{x}}_{\text{i}}}{\text{)}}$ by using step-deviation method. In this method, we take an assumed mean which is in the middle or just close to it in the data.
${\text{A = Assumed mean}}$
${\text{C = Class length}}$ i.e., in the given class interval, there are \[10\] variables in between.
Formula used:
\[{{\text{d}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{x}}_{\text{i}}}{\text{ - A}}}}{{\text{c}}}\]
\[\overline {\text{x}} {\text{ = A + }}\dfrac{{\sum {{{\text{d}}_{\text{i}}}} {{\text{f}}_{\text{i}}}}}{{{{\sum {\text{f}} }_{\text{i}}}}}{\text{ }} \times {\text{ C}}\] So here,
\[{\text{A = 39}}{\text{.5}}\]
${\text{C = 10}}$
\[{{\text{d}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{x}}_{\text{i}}}{\text{ - A}}}}{{\text{c}}}\]

Add all the $({{\text{d}}_{\text{i}}}{{\text{f}}_{\text{i}}}{\text{)}}$ and \[{\text{(}}{{\text{f}}_{\text{i}}})\] to get \[{\sum {\text{d}} _{\text{i}}}{{\text{f}}_{\text{i}}}\] and \[{\sum {\text{f}} _{\text{i}}}\].
\[{\sum {\text{d}} _{\text{i}}}{{\text{f}}_{\text{i}}} = - 28\]
\[{\sum {\text{f}} _{\text{i}}} = {\text{ 83}}\]
Now,
\[\overline {\text{x}} {\text{ = A + }}\dfrac{{\sum {{{\text{d}}_{\text{i}}}} {{\text{f}}_{\text{i}}}}}{{{{\sum {\text{f}} }_{\text{i}}}}}{\text{ }} \times {\text{ C}}\]
Applying the formula,
\[\overline {\text{x}} {\text{ = 39}}{\text{.5 + }}\left[ {\dfrac{{( - 28)}}{{83}}{\text{ }} \times {\text{ 10}}} \right]\]
On dividing the bracket term and we get,
\[\overline {\text{x}} {\text{ = 39}}{\text{.5 + }}\left[ {{\text{ - 0}}{\text{.337 }} \times {\text{ 10}}} \right]\]
On multiply the terms and we get,
\[\overline {\text{x}} {\text{ = 39}}{\text{.5 + }}\left[ { - 3.37} \right]\]
Let us subtracting the terms and we get,
\[\overline {\text{x}} {\text{ = 36}}{\text{.13}}\]
We got the answer.