Central Tendency & DispersionMTP June 22/ MTP Sep 24 IQuestion 2980 of 473
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The first Quartile is 142\displaystyle 142 and Semi-inter Quartile Range is 18\displaystyle 18, then the value of Median is:

Options

A151\displaystyle 151
B160\displaystyle 160
C178\displaystyle 178
DNone of these
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Correct Answer

✅ Option b — 160\displaystyle 160

All Options:

  • A151\displaystyle 151
  • B160\displaystyle 160
  • C178\displaystyle 178
  • DNone of these

Detailed Solution & Explanation

We are given: - First quartile: Q1=142\displaystyle Q_1 = 142 - Semi-interquartile range (Quartile Deviation, Q.D.): Q.D.=18\displaystyle \text{Q.D.} = 18 For a symmetric distribution, the quartiles are symmetric about the median, which means: Q.D.=Q3−Q12  ⟹  Q3−Q1=2(18)=36  ⟹  Q3=142+36=178\text{Q.D.} = \frac{Q_3 - Q_1}{2} \implies Q_3 - Q_1 = 2(18) = 36 \implies Q_3 = 142 + 36 = 178 Also, the median is the midpoint of Q1\displaystyle Q_1 and Q3\displaystyle Q_3: Median=Q1+Q32=142+1782=3202=160\text{Median} = \frac{Q_1 + Q_3}{2} = \frac{142 + 178}{2} = \frac{320}{2} = 160 Alternatively, we can directly compute: Median=Q1+Q.D.=142+18=160\text{Median} = Q_1 + \text{Q.D.} = 142 + 18 = 160 Hence, **Option B** is the correct answer.

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