Comment on Activity 4

The mean of the data set in part (a) is

`1/n sum_(k=1)^n (v_k-m) = 1/n sum_(k=1)^n v_k - (nm)/n = m - m = 0.`

Similarly, the mean of the data set in part (b) is

`1/n sum_(k=1)^n (v_k-m)/(sd) = 1/(n sd) sum_(k=1)^n v_k - (nm)/(n sd) = m/(sd) - m/(sd) = 0.`

The variance of this data set is

`1/n sum_(k=1)^n ((v_k-m)/(sd)-0)^2` `= 1/(n sd^2)sum_(k=1)^n (v_k - m)^2`
  `=` (variance of the original data set)/`sd^2`
  `= (sd^2)/(sd^2) = 1.`

Thus the standard deviation of the data set in part (b) is `sqrt(1) = 1`.