Q:

Let X be a r.v. has the probability function given by: Evaluate the S.D

0

Let X be a r.v. has the probability function given by:

5 4 3 2 1 x
0.1 0.3 c 0.3 0.1 (x)p

 Evaluate the S.D

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Since
∑〖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

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