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If the mean and sd of x are a and b respectively, then the sd of (x-a)/b is?
Verified Answer
If the mean and sd of x are a and b respectively, then the sd of (x-a)...
 (x1-a)/b + (x2-a)/b + ... + (xn-a)/b = ( (x1+x2+x3+...+xn) - n*a ) / b =(n*a - n*a)/b = 0 
So the new SD is: 
sqrt( ((x1-a)/b - 0)^2 + ... + ((xn-a)/b - 0)^2) / n ) 
= sqrt( ((x1-a)^2 + (x2-a)^2 + ... + (xn-a)^2)/(n*b^2) ) 
= sqrt( ((x1-a)^2 + (x2-a)^2 + ... + (xn-a)^2)/n ) / b 
= b / b 
= 1 
So the result is 1 

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Most Upvoted Answer
If the mean and sd of x are a and b respectively, then the sd of (x-a)...
We know that std deviation is dependent of change in scale. and sd of x is b so put value of sd ie b in the given equation in place of x so, req. sd is x/b ie b/b and hnce ans will be 1
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If the mean and sd of x are a and b respectively, then the sd of (x-a)/b is?
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