We have 5 ft 9 in = 69 in, so
P(X > 69) = P((X - 64.1)/2.5 > (69 - 64.1)/2.5) = P(Z > 1.96)
where X and Z both represent women's heights, but Z is normally distributed with mean 0 and s.d. 1.
Then consulting a table of z-scores, you'll find that P(Z > 1.96) ≈ 0.025. So approximately 2.5% of women are taker than 5'9''.
