Question

The mean age of $5$ children is $12$ year. If four of them are respectively $6,11,13\,\&\, 16$ years. Then find the age of the fifth child?

Answer 1

Hint:Mean is the average of the given numbers. If $5$ children ages be $a,b,c,d\,\&\, e$ respectively.Then mean is given by$ = \dfrac{{a + b + c + d + e}}{5}$.You can find fifth child age by using this formula.

Formula used:
Complete step-by-step answer:
it is given that the mean age of $5$ children is year. This means when we take average of $5$ children age, we get $12$ years for example if we take average of $10\,\&\, 20$ then it will be$\dfrac{{10 + 20}}{2} = 15$
So mean is nothing but the average of the number. Therefore we can say that the mean of $10\,\&\, 20$will be $15$ .
In statement, mean is calculated by adding all the numbers, then dividing by total number there are. Or simply it is the sum divided by the count. Let's take an example: we need to find the mean of $a,b,c,d\,\&\, e$. So for mean we do
${\text{Mean = }}\dfrac{{{\text{Sum of number}}}}{{{\text{It's count}}}}$
So its count is $5$ because there are total $5$ number & sum of number is $a + b + c + d + e$ so mean can be given as
Mean$ = \dfrac{{a + b + c + d + e}}{5}$
So, in this question, we are given that mean is $12$ years & four ages out of five is given i.e. $6,11,13\,\&\, 16$
Let the fifth child age be $x$ years
So Mean of $5$ child${\text{ = }}\dfrac{{{\text{Sum of number}}}}{{{\text{It's count}}}}$
And given that Mean$ = 12$
So $12 = \dfrac{{6 + 11 + 13 + 16 + x}}{5}$
Now upon cross multiplication $\begin{gathered}
  12 \times 5 = 46 + x \\
  60 - 46 = x \\
\end{gathered} $
So we get $x = 14$
So age of fifth child will be $14$ years

Note:When frequency is given for e.g. In interval of $\left( {0 - 5} \right)$ years $5$ students are there and so on then $5$ is the frequency of that interval.
Then mean is given by$\dfrac{{\sum {x_i}{f_i}}}{{\sum {f_i}}} = \dfrac{{{x_1}{f_1} + {x_2}{f_2} + .....}}{{{f_1} + {f_2} + .....}}$
Here ${x_i} = $ it is the middle value of interval & ${f_i} = $ it is the frequency of interval.