The median of the following data is 52.5. Find the values of x and y if the total frequency is 100.
|
Class Interval |
Frequency |
|
0-10 |
2 |
|
10-20 |
5 |
|
20-30 |
x |
|
30-40 |
12 |
|
40-50 |
17 |
|
50-60 |
20 |
|
60-70 |
y |
|
70-80 |
9 |
|
80-90 |
7 |
|
90-100 |
4 |
|
Total 100 |
|
C.l. |
fi |
c.f. |
|
0-10 |
2 |
2 |
|
10-20 |
5 |
7 |
|
20-30 |
x |
7 + x |
|
30-40 |
12 |
19 + |
|
40-50 |
17 |
36 + x |
|
50-60 |
20 |
56 + x |
|
60-70 |
y |
56 + x + y |
|
70-80 |
9 |
65 + x + y |
|
80-90 |
7 |
72 + x + y |
|
90-100 |
4 |
76 + x + y |
It is given that n = 100
∴ 76 + x + y 100 ⇒ x + y = 24 ...(i)
The median is 52.5 which lies in the class 50-60.
∴ l = 50,f = 20, c.f. = 36 + x, h = 10
Using the formula.
52.5 - 50 = (14 - x) x 0.5
2.5 = 7 -0.5x 
25 = 70 - 5x
5x = 70 - 25 = 45
x = 9
Putting this value of x in (i), we get
9 + y = 24
y = 24 - 9 = 15
Hence, x = 9 and y= 15
