Let X be a r.v. has the probability function given by:
Evaluate the S.D
Since∑〖xP(x)〗=E(X)Then(0.1)+2(0.3)+c+4(0.3)+5(0.1)=1But∑〖P(x)=1→∴0.1+0.3+c+0.3+0.1=1〗c=0.2∴E(X)=∑〖xP(x)〗=3E(X^2 )=∑〖X^2 P(x)〗=10.4∴V(X)=E(X^2 )-(E(X) ) ̅ ́^2=10.4-9=1.4∴σ_x=√(V(X))=√1.4=1.18
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need an explanation for this answer? contact us directly to get an explanation for this answer∑〖xP(x)〗=E(X)
Then
(0.1)+2(0.3)+c+4(0.3)+5(0.1)=1
But
∑〖P(x)=1→∴0.1+0.3+c+0.3+0.1=1〗
c=0.2
∴E(X)=∑〖xP(x)〗=3
E(X^2 )=∑〖X^2 P(x)〗=10.4
∴V(X)=E(X^2 )-(E(X) ) ̅ ́^2=10.4-9=1.4
∴σ_x=√(V(X))=√1.4=1.18