The median of the following frequency distribution is 55. If the total frequency is 50, the values of \(x\) and \(y\) are:
| Class Interval | Class Frequency |
|---|---|
| 0-10 | 1 |
| 10-20 | 3 |
| 20-30 | \(x\) |
| 30-40 | 6 |
| 40-50 | 5 |
| 50-60 | 8 |
| 60-70 | \(y\) |
| 70-80 | 7 |
| 80-90 | 5 |
| 90-100 | 3 |
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x=6\), \(y=6\)
For grouped data the median lies in the median class, the class in which the cumulative frequency first passes \(\frac{N}{2}\), and
\(\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\)
where \(l\) is the lower limit of the median class, \(cf\) the cumulative frequency before it, \(f\) its frequency and \(h\) the class width.
From the total frequency:
\(1+3+x+6+5+8+y+7+5+3=50\)
\(38+x+y=50\), so \(x+y=12\) (1)
From the median (55 lies in the class 50-60):
\(\frac{N}{2}=25\); \(l=50\), \(f=8\), \(h=10\)
Cumulative frequency before 50-60: \(1+3+x+6+5=15+x\)
\(\displaystyle 55=50+\frac{25-(15+x)}{8}\times10\)
\(\displaystyle 5=\frac{(10-x)\times10}{8}\)
\(10-x=4\), so \(x=6\)
From (1): \(y=12-6=6\)
Note: NTA dropped this question. The original NTA paper printed the median as 52. With a median of 52 the working gives \(\displaystyle 52=50+\frac{(10-x)\times10}{8}\), so \(x=8.4\), which is not a whole number and cannot be a frequency, and none of the options fits. For the benefit of students and to keep a single answer, the median has been changed to 55, which gives \(x=6\) and \(y=6\), Option (1). Even NTA could not find a whole-number answer here, so it dropped the question.