The formula I'm given: $\displaystyle P(a\leq X\leq b)\approx\Phi(\frac{b-np}{\sigma})-\Phi(\frac{a-np}{\sigma})$.Factory makes 1600 products each day.

Probability of a flawed product is 10%.

What is the probability of more than 175 flawed products?

Plugging in: $\displaystyle P(175<X<\infty)\approx\Phi(\infty)-\Phi(0.104)\overbrace{=}^{from\text{ }\Phi\text{ }table}\frac{1}{2}-0.0398=0.4602$

Which isnot even close to a given solution: "Laplace's approximation for P(175 < X) = 0.10565". (Which is close to second $\displaystyle \Phi$.)

The correct solution is $\displaystyle \sum_{k=176}^{1600} {1600 \choose k} 0.1^k 0.9^{1600-k}=0.09944$, so I must be doing something wrong.

SOLVED: I was plugging in variance instead of standard deviation.