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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Statistics

The outcome of each of 30 items was observed; 10 items gave an...

The outcome of each of 30 items was observed; 10 items gave an outcome 12−d\frac{1}{2}-\mathrm{d} each, 10 items gave outcome 12\frac{1}{2} each and the remaining 10 items gave outcome 12+d\frac{1}{2}+\mathrm{d} each. If the variance of this outcome data is 43\frac{4}{3} then ∣d∣|\mathrm{d}| equals:

Options

  1. A23\frac{2}{3}
  2. B2
  3. C52\frac{\sqrt{5}}{2}
  4. D2\sqrt{2}

Correct answer

D. 2\sqrt{2}

Step-by-step solution

Outcomes are (12−d),(12−d),0…,10\left(\frac{1}{2}-d\right),\left(\frac{1}{2}-d\right), 0 \ldots, 10 times, 12,12\frac{1}{2}, \frac{1}{2}, …,10\ldots, 10 times, 12+d,12+d,…,10\frac{1}{2}+d, \frac{1}{2}+d, \ldots, 10 times \text{ Mean }=\frac{1}{30}\left(\frac{1}{2} \times30\right)=12\frac{1}{2}Variance of the outcomes is, σ2=130\Sigmaxi2−(xˉ)2=130[(12−d)2×10+(12)2×10+(12+d)2×10]−14⇒43=130[30×14+20d2]−14⇒43=14+23d2−14⇒d2=2⇒∣d∣=2\begin{aligned}\sigma^{2}&=\frac{1}{30} \Sigmax_i^{2}-(\bar{x})^{2} \\ &=\frac{1}{30}\left[\left(\frac{1}{2}-d\right)^{2} \times₁₀⁺\left(\frac{1}{2}\right)^{2} \times₁₀⁺\left(\frac{1}{2}+d\right)^{2} \times10\right]-\frac{1}{4}\\ \Rightarrow & \frac{4}{3}=\frac{1}{30}\left[30 \times \frac{1}{4}+20 d^{2}\right]-\frac{1}{4}\\ \Rightarrow & \frac{4}{3}=\frac{1}{4}+\frac{2}{3}d^{2}-\frac{1}{4}\\ \Rightarrow & d^{2}=2 \Rightarrow|d|=\sqrt{2} \end{aligned}
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