Question

The numbers $6,8,10,12,13$ and x are arranged in ascending order. If the mean of the observations is equal to the median, find the value of x.

Answer 1

Hint: In this question, we are given 6 numbers out of which one has to be found. We are also told that the mean of the numbers is equal to their median. So, first find the median- it will be the average of the middle two terms. Then we will find the mean in terms of x using the formula $\dfrac{{{\text{Sum of the observations}}}}{{{\text{Total observations}}}}$. Then we will equate the median and mean and then solve to find x.

Complete step-by-step solution:
We are given six numbers out of which one has to be found and we have been told that the mean of these observations is equal to their median.
Let us find the mean of the observations in terms of x using the formula-
$ \Rightarrow $ Mean = $\dfrac{{{\text{Sum of the observations}}}}{{{\text{Total observations}}}}$
Putting the values given in the question,
$ \Rightarrow $ Mean = $\dfrac{{6 + 8 + 10 + 12 + 13 + x}}{6}$
$ \Rightarrow $ Mean = $\dfrac{{49 + x}}{6}$ …………….... (1)
Now, let us find the median.
As we can see that there are six observations that are even numbers of terms. The average of the two middle most terms will be the median. The terms will be –
${\left( {\dfrac{n}{2}} \right)^{th}}$ and ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ term, where $n = 6$.
Therefore, the average of \[{3^{rd}}\] and ${4^{th}}$ term will be considered.
\[{3^{rd}}\] term = 10, ${4^{th}}$ term = 12
$ \Rightarrow $ Median = $\dfrac{{10 + 12}}{2} = \dfrac{{22}}{2} = 11$
Therefore, median = 11 …. (2) $ \Rightarrow 11 \times 6 - 49 = x$
Now, since it is given that mean is equal to the median, we will equate equation (1) and (2).
$ \Rightarrow 11 = \dfrac{{49 + x}}{6}$
Shifting to find the value,
$ \Rightarrow 11 \times 6 - 49 = x$
$ \Rightarrow 66 - 49 = x = 17$
$\Rightarrow $ The value of $x = 17$.

The value of x is equal to 17.

Note: While calculating median, we have used the average of ${\left( {\dfrac{n}{2}} \right)^{th}}$ and ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ term because there are even number of observations. If there had been an odd number of observations, the median would be ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term.