Question:
Consider the following frequency distribution :
Class : 0-6 6-12 12-18 18-24 24-30 Frequency : \(a\) \(b\) 12 9 5
If mean \(= \frac{309}{22}\) and median \(= 14\), then the value \((a - b)^2\) is equal to _____________.
Consider the following frequency distribution :
If mean \(= \frac{309}{22}\) 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 |
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The median lies in the class 12-18; use the median formula to get \(a + b\), then the mean to get \(a\) and \(b\).
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
Let \(N = a + b + 12 + 9 + 5 = a + b + 26\).
Median \(= 14\) lies in the class 12-18 (width 6, frequency 12, lower limit 12, cumulative frequency before it \(= a + b\)):
\(14 = 12 + \frac{\frac{N}{2} - (a + b)}{12} \times 6\), so \(\frac{N}{2} - (a + b) = 4\).
\(\frac{a + b + 26}{2} - (a + b) = 4\), so \(26 - (a + b) = 8\), giving \(a + b = 18\) and \(N = 44\).
Mean: class marks 3, 9, 15, 21, 27. \(\sum f x = 3a + 9b + 180 + 189 + 135 = 3a + 9b + 504\).
\(\frac{3a + 9b + 504}{44} = \frac{309}{22}\), so \(3a + 9b + 504 = 618\), i.e. \(a + 3b = 38\).
With \(a + b = 18\): \(2b = 20\), \(b = 10\), \(a = 8\).
\((a - b)^2 = (8 - 10)^2 = 4\).
Answer: 4
Median \(= 14\) lies in the class 12-18 (width 6, frequency 12, lower limit 12, cumulative frequency before it \(= a + b\)):
\(14 = 12 + \frac{\frac{N}{2} - (a + b)}{12} \times 6\), so \(\frac{N}{2} - (a + b) = 4\).
\(\frac{a + b + 26}{2} - (a + b) = 4\), so \(26 - (a + b) = 8\), giving \(a + b = 18\) and \(N = 44\).
Mean: class marks 3, 9, 15, 21, 27. \(\sum f x = 3a + 9b + 180 + 189 + 135 = 3a + 9b + 504\).
\(\frac{3a + 9b + 504}{44} = \frac{309}{22}\), so \(3a + 9b + 504 = 618\), i.e. \(a + 3b = 38\).
With \(a + b = 18\): \(2b = 20\), \(b = 10\), \(a = 8\).
\((a - b)^2 = (8 - 10)^2 = 4\).
Answer: 4
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