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Question 18 of 46

Q.For a normal variable, mean deviation is 4848 and its third quartile is 120120. Estimate its first quartile.

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2020Subjective· 2mImportance★★★★★
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σ=54(48)=60\sigma = \frac{5}{4}(48) = 60; Q.D.=23σ=40Q.D. = \frac{2}{3}\sigma = 40; Q1=Q3−2(Q.D.)=120−80=40Q_1 = Q_3 - 2(Q.D.) = 120 - 80 = 40.

For a normal distribution the standard relations of the measures of dispersion with σ\sigma are

Mean deviation (M.D.)=45 σ,Quartile deviation (Q.D.)=23 σ.\text{Mean deviation (M.D.)} = \frac{4}{5}\,\sigma, \qquad \text{Quartile deviation (Q.D.)} = \frac{2}{3}\,\sigma.

Step 1 — find σ\sigma from M.D.:

45σ=48  ⟹  σ=48×54=60\frac{4}{5}\sigma = 48 \implies \sigma = 48 \times \frac{5}{4} = 60

Step 2 — find Q.D.:

Q.D.=23σ=23(60)=40Q.D. = \frac{2}{3}\sigma = \frac{2}{3}(60) = 40

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