Central Tendency & DispersionPYQ July 21Question 3094 of 473

The probable value of mean deviation when $\displaystyle Q_3 = 40$ and $\displaystyle Q_1 = 15$ is:

A15

B18.75

C17.50

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Correct Answer

✅ Option a — 15

We are given: - Third quartile:

\displaystyle Q_3 = 40

- First quartile:

\displaystyle Q_1 = 15

1. Calculate Quartile Deviation (Q.D.):

\text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{40 - 15}{2} = 12.5

2. For a normal distribution, the relationship between Quartile Deviation (Q.D.) and Mean Deviation (M.D.) is:

\text{Q.D.} \approx \frac{5}{6} \text{ M.D.} \implies \text{M.D.} \approx \frac{6}{5} \text{ Q.D.}

3. Calculate Mean Deviation:

\text{M.D.} \approx 1.2 \times 12.5 = 15

Hence, **Option A** is the correct answer.

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