Now I have a little more progress:

Power(μ_a)

= P(X bar > 13 + [(1.645 x 4)/(sqrt n)] when μ = μ_a)

And X bar has the distribution N(μ, 16/n)

Here, we know the distribution of X bar, so I believe that we can express P(X bar > 13 + [(1.645 x 4)/(sqrt n)] when μ = μ_a) as an integral of the normal density and replace μ by μ_a, but how can I integrate it and get acompactanswer?

Thank you!