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  • The probability distribution of a random variable x is given as under: P(X=x)=kx^2 where k is a constant. Calculate (i) E(X) (ii) E (3X2) (iii) P(X ≥ 4)

The probability distribution of a random variable x is given as under:

\mathrm{P}(\mathrm{X}=\mathrm{x})=\left\{\begin{array}{ll} \mathrm{kx}^{2} & \text { for } \mathrm{x}=1,2,3 \\ 2 \mathrm{kx} & \text { for } \mathrm{x}=4,5,6 \\ 0 & \text { otherwise } \end{array}\right.

where k is a constant. Calculate
(i) E(X) (ii) E (3X^2) (iii) P(X ≥ 4)

Answers (1)

Solution

Given-

\begin{array}{|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & \text { otherwise } \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{k} & 4 \mathrm{k} & 9 \mathrm{k} & 8 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 0 \\ \hline \end{array}

As we know, Sum of the probabilities  =1 
i, e . \sum_{i=1}^{n} p_{i}=1 \\\Rightarrow \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1 \\\Rightarrow 44 k=1 \\\Rightarrow k=\frac{1}{44}
(i) To find:

\mathrm{E}(\mathrm{X})$ As we know, $\mu=E(X) \because E(X)=\sum_{i=1}^{n} x_{i} p_{i}
or
\\E(X)=\Sigma X P(X) \\=1 \times k+2 \times 4 k+3 \times 9 k+4 \times 8 k+5 \times 10 k+6 \times 12 k \\=k+8 k+27 k+32 k+50 k+72 k \\=190 \mathrm{k}

\\=190 \times \frac{1}{44} \because\left[k=\frac{1}{44}\right] \\=\frac{95}{22} \\\\=4.32
(ii) To find: E\left(3 X^{2}\right)
We first find E\left(X^{2}\right)

As we know,
\\E\left(X^{2}\right)=\Sigma X^{2} P(X) \\=1^{2} \times \mathrm{k}+2^{2} \times 4 \mathrm{k}+3^{2} \times 9 \mathrm{k}+4^{2} \times 8 \mathrm{k}+5^{2} \times 10 \mathrm{k}+6^{2} \times 12 \mathrm{k} \\=\mathrm{k}+16 \mathrm{k}+81 \mathrm{k}+128 \mathrm{k}+250 \mathrm{k}+432 \mathrm{k} \\=908 \mathrm{k} \\=908 \times \frac{1}{44}

\begin{aligned} &=20.636\\ &\cong 20.64\\ &\therefore E\left(3 X^{2}\right)=3 \times 20.64=61.92\\ &\text { (iii) } P(X \geq 4)=P(X=4)+P(X=5)+P(X=6)\\ &=8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}\\ &=30 \mathrm{k}\\ &=30 \times \frac{1}{44}\\ &=\frac{15}{22} \end{aligned}

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