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Exercise 7.1 · Q16
Q.

Find expected value and variance of X for the following p.m.f.

X–2–1012
P (X)0.20.30.10.150.25
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E(X)=∑xipi=(−2)(0.2)+(−1)(0.3)+0(0.1)+1(0.15)+2(0.25)=−0.4−0.3+0+0.15+0.5=−0.05E(X)=\sum x_ip_i = (-2)(0.2)+(-1)(0.3)+0(0.1)+1(0.15)+2(0.25) = -0.4-0.3+0+0.15+0.5 = -0.05. E(X2)=∑xi2pi=4(0.2)+1(0.3)+0(0.1)+1(0.15)+4(0.25)=0.8+0.3+0+0.15+1.0=2.25E(X^2)=\sum x_i^2p_i = 4(0.2)+1(0.3)+0(0.1)+1(0.15)+4(0.25) = 0.8+0.3+0+0.15+1.0 = 2.25. $Var(X)= …

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