JEE Main 2021 · 22 July · evening · Numerical answer · +4 / −0
Consider the following frequency distribution :
If mean = 309/22 and median = 14, then the value (a - b)² is equal to _____________.
| Class : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
|---|---|---|---|---|---|
| Frequency : | a | b | 12 | 9 | 5 |
Answer and solution
Answer 4
Solution
Maths sir · 5 steps
Use the median first: it gives a + b at once, and then the mean gives you a second equation.
- Median class. The total frequency is N = a + b + 26. The median 14 lies in the class 12-18, so l = 12, h = 6, f = 12 and the cumulative frequency before it is cf = a + b.
14 = 12 + ((a+b+26)/2 - (a+b))/12 × 6
- Find a + b. This gives (a+b+26)/2 - (a+b) = 4, so 26 - (a+b) = 8 and a + b = 18. So N = 44.
- Use the mean. The class marks are 3, 9, 15, 21, 27, so Σ fx = 3a + 9b + 180 + 189 + 135 = 3a + 9b + 504. Setting (3a + 9b + 504)/44 = 309/22 gives 3a + 9b + 504 = 618, that is a + 3b = 38.
- Solve both. Subtract a + b = 18 from a + 3b = 38: 2b = 20, so b = 10 and a = 8.
- So. (a - b)² = (8 - 10)² = (-2)² = 4.
The common slip: -2. Stopping at a - b = 8 - 10 gives -2. The question asks for the square, which is 4.
The idea to keep. When a grouped table has two unknown frequencies, the median formula and the mean formula give you two equations. Start with the median, because it usually gives the total a + b directly.
Answer from the NTA final answer key. Solution by PrepWiser.