If the standard deviation of X is σ, then s.d. of the variable U=aX+bc where a,b,c are constants is
A
∣∣ca∣∣σ
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B
∣∣ac∣∣σ
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C
∣∣∣bc∣∣∣σ
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D
c2a2σ
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Solution
The correct option is B∣∣ac∣∣σ S.D of X=σ U=aX+bcX→x1,x2,...xnU→ax1+bc,ax2+bc,...xn+bc
We know that Var(pX+q)=p2Var(X)
So, Var(U)=Var(aX+bc)=Var(aXc+bc)=(ac)2Var(X)
So, S.D of U=√(ac)2Var(X)=∣∣ac∣∣σ