Suppose the probability mass function of the discr – Class 12 Business Mathematics Statistics | ConceptFirst

Question 37 of 41

Q.

Suppose the probability mass function of the discrete random variable is :

$X=x$	0	1	2	3
$p(x)$	0.2	0.1	0.4	0.3

What is the value of $E(3X+2X^{2})$ ?

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2025Subjective· 3mImportance★★★★★

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Compute $E(X)=1.8$ and $E(X^2)=4.4$ , then $E(3X+2X^2)=3(1.8)+2(4.4)=14.2$ .

Step 1 — $E(X)=\sum x\,p(x)$ :

$E(X)=0(0.2)+1(0.1)+2(0.4)+3(0.3)=0+0.1+0.8+0.9=1.8.$

Step 2 — $E(X^{2})=\sum x^{2}p(x)$ :

$E(X^{2})=0^{2}(0.2)+1^{2}(0.1)+2^{2}(0.4)+3^{2}(0.3)=0+0.1+1.6+2.7=4.4.$

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 markMCQ
$E(6x)$
(a) $0$
(b) $6\,E(x)$
(c) $36\,E(x)$
(d) $E(x)$
›Reveal solutionSolution

Expectation is linear, so $E(6x)=6E(x)$ .

A key property of mathematical expectation is that a constant factor pulls out of the expectation. For any constant $a$ ,

$E(aX) = a\,E(X).$

With $a=6$ :

$E(6X) = 6\,E(X).$

…

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CBSE 2024Set MARCH1 markMCQ
$E(X)=5$
(a) $3$
(b) $7$
(c) $5$
(d) $-2$
›Reveal solutionSolution

$E(X-Y)=E(X)-E(Y)=5-(-2)=7$ .

By the linearity of mathematical expectation,

…

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CBSE 2023Set MARCH1 markMCQ
$E[X-E(X)]$
(a) $0$
(b) $E(X)$
(c) $E(X)-X$
(d) $V(X)$
›Reveal solutionSolution

Expectation is linear and $E(X)$ is a constant, so the mean deviation $E[X-E(X)]=0$ .

Let $\mu=E(X)$ , which is a constant. Using linearity of expectation:

$E[X-E(X)]=E(X)-E\big(E(X)\big)=E(X)-E(X)=\mu-\mu=0.$ …

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CBSE 2023Set MARCH1 markMCQ
$p(x)=\dfrac{1}{10}$
(a) $1$
(b) Zero
(c) $-1$
(d) $\dfrac{6}{8}$
›Reveal solutionSolution

Mathematical expectation is $\sum x\,p(x)$ ; here $10\times\tfrac{1}{10}=1$ .

The expected value of a discrete random variable is

$E(X)=\sum x\,p(x).$

…

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