Standard deviation

Standard deviation $\sigma$ is a measure of the spread of the data. By multiplying the data $\{a_n\}$ by -1 and add 2017, just shifts the data laterally without changing the spread and hence the standard deviation remains the same. $\implies \sigma_b = \sigma_a = \boxed{2017}$ .

Let us prove it as follows.

$\begin{aligned} \sum_{k=1}^n b_k & = \sum_{k=1}^n \left(N - a_{n+1-k}\right) = nN - \sum_{k=1}^n a_k \end{aligned}$

$\begin{aligned} \bar b & = \frac {\sum_{k=1}^n b_k} n = N - \frac {\sum_{k=1}^n a_k} n = N - \bar a \end{aligned}$

Now, we have:

$\begin{aligned} \sigma_b & = \sqrt{\frac {\sum_{k=1}^n \left(b_k-\bar b \right)^2}n} \\ & = \sqrt{\frac {\sum_{k=1}^n \left(N - a_{n+1-k}-\bar b \right)^2}n} \\ & = \sqrt{\frac {\sum_{k=1}^n \left(N - a_k-N + \bar a \right)^2}n} \\ & = \sqrt{\frac {\sum_{k=1}^n \left(- a_k + \bar a \right)^2}n} \\ & = \sqrt{\frac {\sum_{k=1}^n \left(a_k - \bar a \right)^2}n} \\ & = \sigma_a & \blacksquare \end{aligned}$