Question

If the median of the data ${x_1},{x_2},{x_3},{x_4},{x_5},{x_6},{x_7},{x_8}$ is $\alpha$ and ${x_1} < {x_2} < {x_3} < {x_4} < {x_5} < {x_6} < {x_7} < {x_8}$ , then the median of ${x_3},{x_4},{x_5},{x_6}$ is
A. $\alpha $
B. $\dfrac{\alpha }{2}$
C. $\dfrac{\alpha }{3}$
D. $\dfrac{\alpha }{4}$

Answer 1

Hint: As we know, median of ungrouped data is the middle term of the sorted data given in the question. Also, the median of the ungrouped data is different for even and odd numbers of terms. Firstly, we need to observe the data whether it has an even or odd number of terms then we will find the middle term of the sorted data.

Complete step-by-step answer:
The ungrouped data is given in the question.
So, we can write all the given values of $x$
The different values of $x$ are ${x_1},{x_2},{x_3},{x_4},{x_5},{x_6},{x_7},{x_8}$
In this, we are also given that ${x_1} < {x_2} < {x_3} < {x_4} < {x_5} < {x_6} < {x_7} < {x_8}$
To calculate the median of the ungrouped data, we have to observe the number of terms given.
There are 8 terms which are even.
So, median can be calculated by finding middle terms.
The middle terms of the given data are:
 $\dfrac{n}{2},\dfrac{n}{2} + 1$
Here, we have 8 numbers of terms which means $n = 8$.
So, middle terms will be 4 and 5.
4th and 5th term is ${x_4},{x_5}$
Therefore, median is $\dfrac{{{x_4} + {x_5}}}{2}$
The median is given in question to be $\alpha $
So, $\alpha = \dfrac{{{x_4} + {x_5}}}{2}$
Now, we have to find median of ${x_3},{x_4},{x_5},{x_6}$
The middle terms will be:
$
   = \dfrac{n}{2},\dfrac{n}{2} + 1 \\
   = \dfrac{4}{2},\dfrac{4}{2} + 1 \\
   = 2,3 \\
 $
2nd and 3rd terms are ${x_4},{x_5}$
So, the median is $\dfrac{{{x_4} + {x_5}}}{2}$ which is equal to $\alpha $ .
Hence, the median of ${x_3},{x_4},{x_5},{x_6}$ is $\alpha $ .
Hence, option A is correct.

Note: For an odd number of terms, the middle term can simply be calculated by observing the middle term or by adding one to the total number of terms and dividing by 2 and this term will be the median. For even number of terms, there will be two middle terms which is calculated by $\dfrac{n}{2},\dfrac{n}{2} + 1$ and then these two terms can be added and divided by 2 to get the median of data.