A set of n values X1, X2,...,Xn has a standard deviation of 5. The standard deviation of n values X1−7, X2−7,...,Xn−7 will be:
5
Standard deviation measures how widely the data spread about their own mean; it is unaffected by where that mean sits on the number line.
Let the new values be \(y_i=x_i-7\). Subtracting the same constant 7 from every value is a rigid shift of the whole data set.
The mean shifts by the same amount: \(\bar{y}=\bar{x}-7\).
So each deviation is \(y_i-\bar{y}=(x_i-7)-(\bar{x}-7)=x_i-\bar{x}\), exactly the original deviation, because the two 7's cancel.
Since standard deviation depends only on these deviations, \(\sqrt{\tfrac{1}{n}\sum(y_i-\bar{y})^2}=\sqrt{\tfrac{1}{n}\sum(x_i-\bar{x})^2}\), the value is unchanged.
The key concept: adding or subtracting a constant does not change the standard deviation (only multiplying by a constant would scale it).
Hence the standard deviation remains 5.
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