If the S.D. of a variate X is σ, then the S.D. of a a X+bis
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a
|a|σ
b
σ
c
a σ
d
a σ +b
answer is A.
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Detailed Solution
Let x1,x2,…,xn be n values of X. Then, σ2=1n∑i=1n xi−X¯2 ….(i) The variable aX+b takes values ax1+b,ax2+b,…,axn+b with mean aX¯+b.∴ Var (aX+b)=1n∑i=1n axi+b−(aX¯+b)2 =a21n∑i=1n xi−X¯2⇒ (S.D. of aX+b)=a21n∑i=1n xi−X¯2=|a|σ