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Question 73 of 88

Q.Let the p. m. f. of a random variable X be P(x)=3−x10P(x) = \dfrac{3-x}{10} for x=−1,0,1,2x = -1, 0, 1, 2, =0= 0 otherwise. Then E(X)E(X) is ______.

(a) 1
(b) 2
(c) 0
(d) -1
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018MCQ· 2mImportance★★★★★
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Compute each probability from the p.m.f., then E(X)=∑x P(x)E(X)=\sum x\,P(x).

P(−1)=3−(−1)10=410=0.4,P(0)=310=0.3,P(1)=210=0.2,P(2)=110=0.1P(-1)=\frac{3-(-1)}{10}=\frac{4}{10}=0.4,\quad P(0)=\frac{3}{10}=0.3,\quad P(1)=\frac{2}{10}=0.2,\quad P(2)=\frac{1}{10}=0.1

(Check: 0.4+0.3+0.2+0.1=10.4+0.3+0.2+0.1=1 ✓, a valid p.m.f.)

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