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Question of 165
Q.
X=xX = x−2-2−1-100112233
P(X=x)P(X = x)0.10.1kk0.20.22k2k0.30.3kk

is the probability distribution of a random variable XX. Find the value of kk and the variance of XX.

Telangana TsbieTelangana Board of Intermediate Education 2026Subjective· 7mImportance★★★★★
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∑P=1⇒k=0.1\sum P=1\Rightarrow k=0.1; then E(X)=0.8E(X)=0.8, E(X2)=2.8E(X^2)=2.8, so Var(X)=2.16(X)=2.16.

The probabilities must sum to 1:

0.1+k+0.2+2k+0.3+k=1⇒0.6+4k=1⇒4k=0.4⇒k=0.10.1 + k + 0.2 + 2k + 0.3 + k = 1 \Rightarrow 0.6 + 4k = 1 \Rightarrow 4k = 0.4 \Rightarrow k = 0.1.

So the distribution is:

xx−2-2−1-100112233
P(x)P(x)0.10.10.10.10.20.20.20.20.30.30.10.1

Mean: μ=E(X)=∑x P(x)=(−2)(0.1)+(−1)(0.1)+(0)(0.2)+(1)(0.2)+(2)(0.3)+(3)(0.1)\mu = E(X) = \sum x\,P(x) = (-2)(0.1) + (-1)(0.1) + (0)(0.2) + (1)(0.2) + (2)(0.3) + (3)(0.1)

=−0.2−0.1+0+0.2+0.6+0.3=0.8= -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8.

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