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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
A discrete random variable $X$ has the probability distribution given as below:
$X$
$0.5$
$1$
$1.5$
$2$
$P(X)$
$k$
$k^2$
$2k^2$
$k$
  1. Find the value of $k.$
  2. Determine the mean of the distribution.
✓

Answer

We have,
$X$
$0.5$
$1$
$1.5$
$2$
$P(X)$
$k$
$k^2$
$2k^2$
$k$
  1. We know that $\sum\limits_\text{i=1}^{\text{n}}\text{P}_{\text{i}}=1,$ where $\text{P}_{\text{i}}\geq0$
$\Rightarrow P_1+ P_2 + P_3 + P_4= 1$
$\Rightarrow k + k^2 + 2k^2 + k = 1$
$\Rightarrow 3k^2 + 2k - 1 = 0$
$\Rightarrow 3k^2 + 3k - k - 1 = 0$
$\Rightarrow 3k(k + 1) - 1(k + 1) = 0$
$\Rightarrow (3k - 1)(k + 1) = 0$
$\Rightarrow\text{k}=\frac{1}{3}$ and $k = -1$
Since, $\text{K}\geq0,$ we take $\text{k}=\frac{1}{3}$
  1. Mean of the distribution $(\mu)=\text{E}(\text{X})=\sum\limits_{\text{i=1}}^{\text{n}}\text{x}_{\text{i}}\text{P}_{\text{i}}$
$=0.5(\text{k})+1(\text{k}^2)+1.5(2\text{k}^2)+2(\text{k})$
$=4\text{k}^2+2.5\text{k}$
$=4\cdot\frac{1}{9}+2.5\cdot\frac{1}{3}\Big[\because\text{k}=\frac{1}{3}\Big]$
$=\frac{4+7.5}{9}=\frac{23}{18}$
 

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