Binomial to Normal Approximation

Assume an airline manages to sell 160 seats for a plane that is only capable of seating 155. Assume that 0.05 of ticket holders cancel their reservation. What is the probability that the place will be overbooked?

I found these http://www.mathhelpforum.com/math-he...obability.html and http://www.mathhelpforum.com/math-he...rs-flight.html that deal with the exact same question, although they focus on solving the problem through the binomial method, which I have no trouble doing, nevertheless I would like to know how to solve it using the Normal Approximation. I am not sure what variables I have to use to get the result...

I would think that my variance would be npq = 160*0.05*0.95=7.6

then I would have:

$\displaystyle P\left(\frac{np-nr}{\sqrt{npq}}= \frac{8-7.75}{\sqrt{7.6}} = 0.90685\right) $

so I would have:

$\displaystyle P(Z\leq 0.90685) = 0.8186$

but the solution in the back of the book gives 0.8980 which would be $\displaystyle P(Z\leq 1.27) = 0.8980$