Normal approximation to binomial distribution.
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
μ = mean = (n * p) = (700 * .80) = 560
σ = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583
Any help will be greatly appreciated. Thanks