Probability question help

[lilyungn]

Hey, I'm stuck on a homework problem that I've been working on...

Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of observing 518 or fewer drivers wearing seat belts in a sample of 700 drivers?

Hint: Use normal distribution to approximate the binomial distribution

What I have so far is

xbar = 518
? = mean = (n * p) = (700 * .80) = 560
? = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583

Any help will be greatly appreciated. Thanks

 

[daon]

\(\displaystyle P(X \le 518 | N=700) = \sum_{k=0}^{518} {700 \choose k} (0.8)^k(0.2)^{700-k}\)

\(\displaystyle \approx P(X < 518.5) = \Phi (\frac{518.5-560}{10.583})\)