Question

The average age of 6 students is 11 years. If two more students of age 14 and 16 years join, their average will become \[\]
A. 13 years\[\]
B. 12 years\[\]
C. $12\dfrac{1}{2}$ years\[\]
D. $11\dfrac{1}{2}$ years\[\]

Answer 1

Hint: We find the sum of the ages of students $S$ by multiplying the average age of students $\overline{x}$ and number of students $n$. We use the given data in the question and find a new number of students ${{n}_{new}}$ and a new sum of the ages of students ${{S}_{new}}$ after joining of extra students. We find the new average ages as ${{\left( \overline{x} \right)}_{new}}=\dfrac{{{S}_{new}}}{{{n}_{new}}}$\[\]

Complete step-by-step solution:
We know that mean or average is the expected value of the data sample set. It is the value we expect without going detailed into data and it is a measure of central tendency. It is denoted by $\overline{x}$. If there are $n$ number of data values with equal weights in the samples say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then the mean is calculated by first finding the sum of data values and then by dividing the sum by $n.$ So the sample mean is given by
\[\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}=\dfrac{S}{n}\]
If the weights of data values are not equal and are assigned weights say ${{w}_{1}},{{w}_{2}},...,{{w}_{n}}$ then the sample mean is given by
\[\overline{x}=\dfrac{{{w}_{1}}{{x}_{1}}+{{w}_{2}}{{x}_{2}}+...+{{w}_{n}}{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{w}_{i}}{{x}_{i}}}}{n}\]
We are given the question that the average age of 6 students is 11 years. Let us denote the average age of the students as $\overline{x}$ and the number of students as $n$. We have $\overline{x}=11,n=6$.The sum of the ages of students is
\[S=n\times \overline{x}=6\times 11=66\]
Let us denote the age of students as ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}},{{x}_{6}}$.So we have
\[S={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}=66\]
We are given in the question that two more students of age 14 and 16 years join the class. So we have a new number of students: ${{n}_{\text{new}}}=6+2=8$. Let us denote the age of new students as ${{x}_{7}}=14,{{x}_{8}}=16$. The new sum of ages is ${{S}_{\text{new}}}$ in years is
\[\begin{align}
&{{S}_{\text{new}}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}} \\
 & \Rightarrow {{S}_{\text{new}}}=66+{{x}_{7}}+{{x}_{ 8}} \\
 & \Rightarrow {{S}_{\text{new}}}=66+14+16=96 \\
\end{align}\]
So the new average of age of students in years is
\[{{\left(\overline{x}\right)}_{\text{new}}}=\dfrac{{{S}_{\text{new}}}}{{{n}_{\text{new}}}}=\dfrac{96}{8}=12\]
So the correct option is B.

Note: Here the mean is the arithmetic means AM. Other types of mean are geometric mean (GM) and harmonic mean (HM). The relation among mean is $AM\ge GM\ge HM$. We can directly find a new average using the formula $\dfrac{S+{{S}_{new}}}{n+h}$ where $h$ is the increase in the number of values. The relation between old and new average is given by ${{\left( \overline{x} \right)}_{new}}-{{\left( \overline{x} \right)}_{old}}=k$ where $k$ is the change in mean.