Standard deviation is times of sqrt(MD x QD)

Option c: sqrt(15/8). Solution: For a normal distribution, the relationships between the standard deviation (S.D.), mean deviation (M.D.), and quartile deviation (Q.D.) are given by: text{Q.D.} approx 2/3text{S.D.} quad text{and} quad text{M.D.} approx 4/5text{S.D.} Taking the product of M.D. and Q.D.: text{M.D.} x text{Q.D.} approx 4/5text{S.D.} x 2/3text{S.D.} = 8/15text{S.D.}^2 sqrt(text{M.D.) x text{Q.D.}} approx sqrt(8/15)text{S.D.} implies text{S.D.} approx sqrt(15/8) sqrt(text{M.D.) x text{Q.D.}} Hence, Option C is the correct answer.

Central Tendency & DispersionPYQ June 19Question 3221 of 473

All Questions

Standard deviation is ________ times of $\displaystyle \sqrt{MD \times QD}$

Options

\displaystyle \frac{2}{3}

\displaystyle \frac{4}{5}

\displaystyle \sqrt{\frac{15}{8}}

\displaystyle \sqrt{\frac{8}{15}}

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Correct Answer

✅ Option c — $\displaystyle \sqrt{\frac{15}{8}}$

All Options:

A $\displaystyle \frac{2}{3}$
B $\displaystyle \frac{4}{5}$
C $\displaystyle \sqrt{\frac{15}{8}}$
D $\displaystyle \sqrt{\frac{8}{15}}$

Detailed Solution & Explanation

For a normal distribution, the relationships between the standard deviation (S.D.), mean deviation (M.D.), and quartile deviation (Q.D.) are given by:

\text{Q.D.} \approx \frac{2}{3}\text{S.D.} \quad \text{and} \quad \text{M.D.} \approx \frac{4}{5}\text{S.D.}

Taking the product of M.D. and Q.D.:

\text{M.D.} \times \text{Q.D.} \approx \frac{4}{5}\text{S.D.} \times \frac{2}{3}\text{S.D.} = \frac{8}{15}\text{S.D.}^2

\sqrt{\text{M.D.} \times \text{Q.D.}} \approx \sqrt{\frac{8}{15}}\text{S.D.} \implies \text{S.D.} \approx \sqrt{\frac{15}{8}} \sqrt{\text{M.D.} \times \text{Q.D.}}

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

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Standard deviation is ________ times of MD×QD\displaystyle \sqrt{MD \times QD}MD×QD​