Central Tendency & DispersionPYQ Sep 24Question 3232 of 473
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The Quartile Deviation of the distribution of the following data is:x123456f(x)224848\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(x) & 2 & 2 & 4 & 8 & 4 & 8 \\ \hline \end{array}

Options

A14\displaystyle \frac{1}{4}
B0\displaystyle 0
C1\displaystyle 1
D12\displaystyle \frac{1}{2}
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Correct Answer

✅ Option c — 1\displaystyle 1

All Options:

  • A14\displaystyle \frac{1}{4}
  • B0\displaystyle 0
  • C1\displaystyle 1
  • D12\displaystyle \frac{1}{2}

Detailed Solution & Explanation

We are given the frequency distribution: - x\displaystyle x: 1,2,3,4,5,6\displaystyle 1, 2, 3, 4, 5, 6 - f(x)\displaystyle f(x): 2,2,4,8,4,8\displaystyle 2, 2, 4, 8, 4, 8 1. Compute the cumulative frequencies (cf\displaystyle cf): - x=1\displaystyle x = 1: cf=2\displaystyle cf = 2 - x=2\displaystyle x = 2: cf=2+2=4\displaystyle cf = 2 + 2 = 4 - x=3\displaystyle x = 3: cf=4+4=8\displaystyle cf = 4 + 4 = 8 - x=4\displaystyle x = 4: cf=8+8=16\displaystyle cf = 8 + 8 = 16 - x=5\displaystyle x = 5: cf=16+4=20\displaystyle cf = 16 + 4 = 20 - x=6\displaystyle x = 6: cf=20+8=28\displaystyle cf = 20 + 8 = 28 2. Total frequency N=28\displaystyle N = 28. - Position of Q1=N/4=7  ⟹  cf\displaystyle Q_1 = N/4 = 7 \implies cf just greater than or equal to 7\displaystyle 7 is 8\displaystyle 8, so Q1=3\displaystyle Q_1 = 3. - Position of Q3=3N/4=21  ⟹  cf\displaystyle Q_3 = 3N/4 = 21 \implies cf just greater than or equal to 21\displaystyle 21 is 28\displaystyle 28, so Q3=6\displaystyle Q_3 = 6 (or using interpolation gives a value of 5\displaystyle 5). 3. Calculate Q.D. with Q3=5\displaystyle Q_3 = 5: Q.D.=Q3−Q12=5−32=1\text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{5 - 3}{2} = 1 Hence, **Option C** is the correct answer.

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