Question

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
$5 - 14$	$6$
$15 - 24$	$11$
$25 - 34$	$21$
$35 - 44$	$23$
$45 - 54$	$14$
$55 - 64$	$5$
$65 - 74$	$3$

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

Answer 1

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 $208.5$

Age (in years)	No of cases ${\text{(}}{{\text{f}}_{\text{i}}}{\text{)}}$	Class mark ${\text{(}}{{\text{x}}_{\text{i}}}{\text{)}}$	${{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}$
$5 - 14$	$6$	$9.5$	$57$
$15 - 24$	$11$	$19.5$	$214.5$
$25 - 34$	$21$	$29.5$	$619.5$
$35 - 44$	$23$	$39.5$	$908.5$
$45 - 54$	$14$	$49.5$	$693$
$55 - 64$	$5$	$59.5$	$297.5$
$65 - 74$	$3$	$69.5$	$208.5$
{\sum {\text{f}} _{\text{i}}} = $$$$83		${\sum {\text{f}} _{\text{i}}}{{\text{x}}_{\text{i}}} = 2998.5$

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,
${\text{A = 39}}{\text{.5}}$
${\text{C = 10}}$
${{\text{d}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{x}}_{\text{i}}}{\text{ - A}}}}{{\text{c}}}$

Age (in years)	No of cases ${\text{(}}{{\text{f}}_{\text{i}}}{\text{)}}$	Class mark ${\text{(}}{{\text{x}}_{\text{i}}}{\text{)}}$	${{\text{d}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{x}}_{\text{i}}}{\text{ - A}}}}{{\text{c}}}$	$({{\text{d}}_{\text{i}}}{{\text{f}}_{\text{i}}}{\text{)}}$
$5 - 14$	$6$	$9.5$	$\dfrac{{{\text{9}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{ - 30}}{{10}} = - 3$	$- 3 \times 6 = - 18$
$15 - 24$	$11$	$19.5$	$\dfrac{{{\text{19}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{ - 20}}{{10}} = - 2$	$- 2 \times 11 = - 22$
$25 - 34$	$21$	$29.5$	$\dfrac{{{\text{29}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{ - 10}}{{10}} = - 1$	$- 1 \times 21 = - 21$
$35 - 44$	$23$	$39.5$	$\dfrac{{{\text{39}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{0}{{10}} = 0$	$0 \times 23 = 0$
$45 - 54$	$14$	$49.5$	$\dfrac{{{\text{49}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{10}}{{10}} = 1$	$1 \times 14 = 14$
$55 - 64$	$5$	$59.5$	$\dfrac{{{\text{59}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{20}}{{10}} = 2$	$2 \times 5 = 10$
$65 - 74$	$3$	$69.5$	$\dfrac{{{\text{69}}{\text{.5 - 39}}{\text{.5}}}}{{10}} = \dfrac{{30}}{{10}} = 3$	$3 \times 3 = 9$
${\sum {\text{f}} _{\text{i}}} = {\text{ 83}}$			${\sum {\text{d}} _{\text{i}}}{{\text{f}}_{\text{i}}} = - 28$

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.