Here we will learn how to solve the different types of problems on median of ungrouped data.
1. The heights (in cm) of 11 players of a team are as follows:
160, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
Solution:
Arranging the variates in the ascending order, we get
157, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
The number of variates = 11, which is odd.
Therefore, median = \(\frac{11 + 1}{2}\)^{th} variate = 6^{th} variate = 160.
2. Find the median of the first five odd integers. If the sixth odd integer is also included, find the difference of medians in the two cases.
Solution:
Writing the first five odd integers in ascending order, we get
1, 3, 5, 7, 9.
The number of variates = 5, which is odd.
Therefore, median = \(\frac{5 + 1}{2}\)^{th} variate = 3^{th} variate = 5.
When the sixth integer is included, we have (in ascending order)
1, 3, 5, 7, 9, 11.
Now, the number of variates = 6, which is even.
Therefore, median = mean of \(\frac{6}{2}\)^{th} and (\(\frac{6}{2}\) + 1)^{th} variates
= Mean of 3^{rd} and 4^{th} variates
= Mean of 5 and 7 = \(\frac{5 + 7}{2}\) = 6.
Therefore, the difference of medians in the two cases = 6 - 5 = 1.
3. If the median of 17, 13, 10, 15, x happens to be the integer x then find x.
Solution:
There are five (odd) variates. So, \(\frac{5 + 1}{2}\)^{th} variate, i.e., 3^{rd} variate when written in ascending order will the median x.
So, the variates in ascending order should be 10, 13, x, 15, 17.
Therefore, 13 < x < 15.
But x is an integer. So, x = 14.
4. The marks obtained by 20 students in a class test are given below.
Marks Obtained
6
7
8
9
10
Number of Students
5
8
4
2
1
Find the median of marks obtained by the students.
Solution:
Arranging the variates in ascending order, we get
6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10.
The number of variates = 20, which is even.
Therefore, median = mean of \(\frac{20}{2}\)^{th} and (\(\frac{20}{2}\) + 1)^{th} variate
= mean of 10^{th} and 11^{th} variate
= mean of 7 and 7
= \(\frac{7 + 7}{2}\)
= 7.
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