Q.If the median of the given data is 28.5 and the total frequency is 60, determine the values of x and y.
Class Interval0-1010-2020-30
Frequency5x20
30-4040-5050-60
15y5

The frequency distribution table is as follows:
Class BoundaryFrequencyCumulative Frequency (less than type)
0-1055
10-20\(x\)5+\(x\)
20-302025+\(x\)
30-401540+\(x\)
40-50\(y\)40+\(x+y\)
50-60545+\(x+y=n\)
Here, \(n=60\) (given).
According to the condition, \(45+x+y=60\)
or, \(x+y=15----(i)\)
Also, โˆตMedian = 28.5
So, the median class is (20-30).
โˆด The median is determined using the formula:
\[ = l + \left[\cfrac{\cfrac{n}{2}-cf}{f}\right]ร—h \]
[Here, \(l=20, n=60, \)
\(cf=5+x, f=20, h=10\)]
\[ = 20 + \left[\cfrac{30-(5+x)}{20}\right]ร—10 \]
\[ = 20 + \cfrac{25-x}{20}ร—10 \]
\[ = 20 + \cfrac{25-x}{2} \]

According to the condition,
\[ 20+ \cfrac{25-x}{2}=28.5 \]
or, \(\cfrac{25-x}{2}=8.5\)
or, \(25-x=17\)
or, \(-x=-8\)
or, \(x=8\)

Substituting the value of \(x\) in equation \((i)\),
\[ 8+y=15 \]
or, \(y=7\)

โˆด The required values are \(x=8, y=7\).
Similar Questions