Q79 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

If a random variable X has the following probability distribution, then its variance is nearly $\begin{array}{|c|c|c|c|c|c|c|c|} \hline \mathbf{X}=x: & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline \mathbf{P}(\mathbf{X}=x): & 0.05 & 0.1 & 2 \mathrm{~K} & 0 & 0.3 & \mathrm{~K} & 0.1 \\ \hline \end{array}$

  • Correct answer
    A
    $2.8875$
  • B
    $2.9875$
  • C
    $2.7865$
  • D
    $2.785$

Solution

\begin{array}{|c|c|c|c|c|c|c|c|}\hline \mathbf{X}=\mathbf{x} & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\\hline \mathbf{P}(\mathbf{X}=\mathbf{x}) & 0.05 & 0.1 & 2 k & 0 & 0.3 & k & 0.1 \\\hline\end{array} $\begin{aligned} & \Sigma \mathrm{P}(\mathrm{X}=x)=1 \\ & \Rightarrow 0.05+0.1+2 k+0+0.3+k+0.1=1 \Rightarrow k=0.15\end{aligned}$ \begin{array}{|c|c|c|c|c|c|c|c|}\hline \mathbf{X = x} & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\\hline \mathbf{P}(\mathbf{X}=\mathbf{x}) & 0.05 & 0.1 & 0.3 & 0 & 0.3 & 0.15 & 0.1 \\\hline\end{array} $\begin{aligned} & \text { Mean }(\mu)=(-3)(0.05)+(-2)(0.1)+(-1)(0.3) \\ & +(1)(0.3)+2(0.15)+3(0.1)=0.25 \\ & \text { Variance }=\sum_{i=1}^7\left(x_i-\mu\right)^2 \times \mathrm{P}\left(x_i\right) \\ & =0.528125+0.50625+0.46875+0+0.16875 \\ & +0.459375+0.75625=2.8875\end{aligned}$

Hence, the correct answer is option: 'A'.

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