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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Statistics

If the mean of the following probability distribution of a random...

If the mean of the following probability distribution of a random variable X\mathrm{X} : X02468P(X)a2aa+b2b3b\begin{array}{|c|c|c|c|c|c|} \hline \mathrm{X} & 0 & 2 & 4 & 6 & 8 \\ \hline \mathrm{P}(\mathrm{X}) & a & 2 a & a+b & 2 b & 3 b \\ \hline \end{array} is 469\frac{46}{9}, then thevarianceofthedistributionisthe variance of the distribution is

Options

  1. A17327\frac{173}{27}
  2. B56681\frac{566}{81}
  3. C15127\frac{151}{27}
  4. D58181\frac{581}{81}

Correct answer

B. 56681\frac{566}{81}

Step-by-step solution

∑Pi=1a+2a+a+b+2b+3b=1\begin{aligned} & \sum P_i=1 \\ & a+2 a+a+b+2 b+3 b=1\end{aligned}4 a+6 b=1...(I)...(I)\mathrm{E}(\mathrm{x})=meanmean=\frac{46}{9} ∑PiXi=469⇒4a+4a+4b+12b+24b=4698a+40b=469\begin{aligned} & \sum P_i X_i=\frac{46}{9} \Rightarrow 4 a+4 a+4 b+12 b+24 b=\frac{46}{9} \\ & 8 a+40 b=\frac{46}{9}\end{aligned} 4 a+20 b=\frac{23}{9}...(II)Subtract(I)from(II)weget...(II) Subtract (I) from (II) we getb=19a=112 Variance =E(xi2)−E(xi)2E(xi)2=02×92+22×2a+42(a+b)+62(2b)+82(3b)=24a+280b\begin{aligned} & \mathrm{b}=\frac{1}{9} & \mathrm{a}=\frac{1}{12} \\ & \text{ Variance }=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}{ }^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 \\ & \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2=0^2 \times 9^2+2^2 \times 2 \mathrm{a}+4^2(\mathrm{a}+\mathrm{b})+6^2(2 \mathrm{ b})+8^2(3 \mathrm{ b}) \\ & =24 \mathrm{a}+280 \mathrm{ b} \end{aligned}PutPut\mathrm{a}=\frac{1}{12} \quad \mathrm{ b}=\frac{1}{9} \begin{aligned} & \mathrm{E}\left(\mathrm{x}_{\mathrm{i}^2\right)=2+\frac{280}{9}=\frac{298}{9} \\ & \therefore \sigma^2=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 \\ & =\frac{298}{9}-\left(\frac{46}{9}\right)^2 \\ & \sigma^2=\frac{298}{9}-\frac{2116}{81} \\ & =\frac{566}{81} \end{aligned}$
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