Question
Question: If the median of the distribution given below is \(28.5\), find the values of \(x\) and \(y\). C...
If the median of the distribution given below is 28.5, find the values of x and y.
Class interval | Frequency |
---|---|
0−10 | 5 |
10−20 | x |
20−30 | 20 |
30−40 | 15 |
40−50 | y |
50−60 | 5 |
Total | 60 |
Solution
In this question, we are given a table with class interval and frequency. The median of the data is also given. Using the median, find the median class. Draw the table calculating cumulative frequency. Use the median class and formula, Median =l+f2n−cfh to find x. After x has been found, use total frequency from the table to find y.
Formula used: Median =l+f2n−cfh where, l = lower limit of median class, n=∑fi, cf = cumulative frequency of the class before median class, h = class interval, f = frequency of median class.
Complete step-by-step solution:
We are given a table with class interval and its frequency and we are asked to find the value of x and y. First, we will make a table with a cumulative frequency (cf).
Class Interval | Frequency | Cumulative Frequency |
---|---|---|
0−10 | 5 | 5 |
10−20 | x | 5+x |
20−30 | 20 | 5+x+20=25+x |
30−40 | 15 | 25+x+15=40+x |
40−50 | y | 40+x+y |
50−60 | 5 | 40+x+y+5=45+x+y |
Total | 60 |
Now, we know that median = 28.5. Since median lies in the median class, its median class is 20−30. So, we will apply the formula, Median =l+f2n−cfh in this median class.
In this question, l=20, cf=5+x, h=10−0=10, f=20 and we know that n=∑fi=60. Therefore, 2n=260=30.
Putting all the values in the formula,
⇒28.5=20+(2030−(5+x))10
Solving for x,
⇒28.5=20+(230−5−x)
Shifting the terms,
⇒28.5−20=(225−x)
⇒8.5×2=25−x
⇒17=25−x
Shifting and finding the value of x,
⇒x=25−17=8
Now, we know from the table that 45+x+y=60. Putting x=8 to find the value of y.
⇒45+8+y=60
Shifting to find y,
⇒y=60−53=7
∴ The value of x=8,y=7.
Note: Students are often confused while finding cumulative frequency. Cumulative frequency is nothing but the sum of the frequency of that particular interval and the frequencies of predecessors. Basically, it is the sum of all the frequencies before that interval (including the frequency of that interval).