Question

Find the median of 63, 17, 50, 9, 25, 43, 21, 50, 14 and 34
A) 29.5
B) 29
C) 20
D) 30

Answer 1

Hint: The given data is ungrouped. So here we have to find the median of an ungrouped data set. The median refers to the middle data point of an ordered data set at the 50% percentile. It is important to know that if a data set has an odd number of observations, then the median is the middle value and if it has an even number of observations then the median will be the average of the two middle values.

Complete step-by-step solution:
In general the formula to find the median of an ungrouped data set is
\[\widetilde{x}=\dfrac{n+1}{2}\] Ranked value
Here, n= number of observations
Firstly, for an ungrouped data set, it is important to sort the data in an increasing order
So here let’s first sort 63, 17, 50, 9, 25, 43, 21, 50, 14, and 34 into an increasing order
Increasing order: 9, 14, 17, 21, 25, 34, 43, 50, 50, 63
Now, let us find the number of observations. As there are 10 observations
Number of observations= n = 10
Now here let us find \[\widetilde{x}\] by substituting it in the formula \[\widetilde{x}=\dfrac{n+1}{2}\]
\[\widetilde{x}=\dfrac{10+1}{2}=\dfrac{11}{2}=5.5\]
Here it can be observed that the value of n that is the number of observations is even so the median will be the average of the two middle values.
As calculated the median is at 5.5 which means the average of 5th and 6th data points.
$5^{th}$ data point = 25
$6^{th}$ data point= 34
Average = \[\dfrac{{{5}^{th}}data+{{6}^{th}}data}{2}\]
Average of 5th and 6th data point = \[\dfrac{25+34}{2}=\dfrac{59}{2}=29.5\]
Hence the median of the following ungrouped data is 29.5

Note: While solving for the median in an ungrouped distribution of data the number of observations should be counted first. The value of $\dfrac{n+1}{2}$ should not be mistaken with the value of the median. Not always we will have a value at the $\dfrac{n+1}{2}$position, in these cases the average of the data points near the $\dfrac{n+1}{2}$ should be calculated. The average calculated will be the median of the ungrouped data.