Question

Find the median of the following:
$33$, $31$, $48$, $45$, $41$, $92$, $78$, $51$, $61$

Answer 1

Hint: In this question, first we have to arrange the given data in the form of ascending order. Then count the numbers in given data. Then determine whether the numbers of observations are even or odd. Finally calculate the median using the median formula for even or odd respectively.

Formula used: If n is odd, median $ = {\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$observation (after size-ordering)
If n is even, median$ = \dfrac{{{{\left( {\dfrac{n}{2}} \right)}^{th}}{\text{ observation}} + {{\left( {\dfrac{n}{2} + 1} \right)}^{{\text{th}}}}{\text{ observation}}}}{2}$, where n is the total number of distribution.

Complete step-by-step solution:
It is given that: $33$, $31$, $48$, $45$, $41$, $92$, $78$, $51$, $61$.
As we arrange these numbers in ascending order.
Here we discuss the definition of ascending order means we have to arrange the given numbers in increasing order, that is, we form that from smallest to largest.
Smallest number is $31$.
Here we have to write down the smallest number first, and then compare to the remaining numbers with the same number of digits.
For which the numbers have the same number of digits, and it starts with comparing the numbers from the leftmost digit. Also, write the number with the smallest digit.
Numbers with the smallest leftmost digit are $33$, $31$.
If the leftmost digits are the same, move to the digits to the right and compare them. Write the number with a smaller digit.
Numbers with the smallest leftmost digit in ascending order are $31$, $33$
Continuing like this, we’ll get the data in ascending order.
$31$, $33$, $41$, $45$, $48$, $51$, $61$, $78$, $92$
Find the number of observations in given data, i.e., number of digits in given data.
Counting the numbers, we get
Number of observations $ = 9$
Check whether the number of observations is even or odd.
Since, $9$ is not divisible by $2$.
So, the number of observations is odd.
Now we have to find the median year using the median formula.
That is we can write it as, Median $ = {\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ observation
Putting $n = 9$ in the median formula we get,
Median $ = {\left( {\dfrac{{9 + 1}}{2}} \right)^{th}}$observation
On simplification we get
\[ \Rightarrow \] Median = ${5^{th}}$ observation
Find the ${5^{th}}$ observation in ascending order of the data we calculated.
Since, the data in ascending order is
$31$, $33$, $41$, $45$, $48$, $51$, $61$, $78$, $92$
So, ${5^{th}}$observation\[ = {\text{ }}48\]

Thus, the median of the given data is \[48\].

Note: The median of a set of data is the value of that observation which lies exactly halfway along the set (which must be arranged into ascending or descending order of magnitude).
The median of a set of data is the middlemost number or centre value in the set.
Here we have to find the median; first we arrange the data should be in order of least to greatest or greatest to the least value.
The median is different for different types of distributions.