For the probability distribution given by following $\begin{array}{|l|l|l|l|l|l|l|l|} \hline \mathrm{X} & 5…

For the probability distribution given by following $\begin{array}{|l|l|l|l|l|l|l|l|} \hline \mathrm{X} & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & 0.07 & 0.2 & 0.3 & \mathrm{k} & 0.07 & 0.04 & 0.02 \\ \hline \end{array}$ \operatorname{Var}(\mathrm{X})=$
  1. 2.56
  2. 2.85
  3. 1.65
  4. 3.85

Solution

We have $0.07+0.2+0.3+\mathrm{k}+0.07+0.04+0.02=1$ $\therefore \mathrm{k}=0.3$ $\begin{array}{|l|l|l|l|} \hline \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2 \\ \hline 5 & 0.07 & 0.35 & 1.75 \\ \hline 6 & 0.2 & 1.2 & 7.2 \\ \hline 7 & 0.3 & 2.1 & 14.7 \\ \hline 8 & 0.3 & 2.4 & 19.2 \\ \hline 9 & 0.07 & 0.63 & 5.67 \\ \hline 10 & 0.04 & 0.4 & 4 \\ \hline 11 & 0.02 & 0.22 & 2.42 \\ \hline & \text{Total} & 7.3 & 54.94 \\ \hline \end{array}$ Variance $(\mathrm{x})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=(54.94)-(7.3)^2=1.65$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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