Question:

If the mean of the following data is \(12.5\), find the value of \(k\).

\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class interval} & 0\text{-}5 & 5\text{-}10 & 10\text{-}15 & 15\text{-}20 & 20\text{-}25 \\ \hline \text{Frequency} & 2 & k & 6 & 4 & 1 \\ \hline \end{array} \]

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For grouped data, \[ \boxed{ \bar{x} = \frac{\sum fx}{\sum f} } \] where \(x\) is the class mark of each interval.
Updated On: Jul 16, 2026
  • \(1\)
  • \(2\)
  • \(3\)
  • \(4\)
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The Correct Option is B

Solution and Explanation

Step 1: Find the class marks. \[ 2.5,\;7.5,\;12.5,\;17.5,\;22.5 \]

Step 2:
Compute \(\sum f\) and \(\sum fx\). \[ \sum f=2+k+6+4+1=13+k. \] Also, \[ \sum fx = 2(2.5)+k(7.5)+6(12.5)+4(17.5)+1(22.5) \] \[ =5+7.5k+75+70+22.5 \] \[ =172.5+7.5k. \]

Step 3:
Use the mean formula. Given, \[ \frac{172.5+7.5k}{13+k}=12.5. \] Therefore, \[ 172.5+7.5k = 162.5+12.5k. \] \[ 10=5k. \] \[ k=2. \]

Step 4:
Final conclusion. \[ \boxed{k=2} \] Hence, the correct option is \(\boxed{(B)}\).
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