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
What do your classnotes and textbook say? Probably similar to what is said here, for example: http://courses.wcupa.edu/rbove/Beren...section6_5.pdf
You want
where
We'll get an approximate answer by using another variable, Y
You have already worked out that
Do you understand why these values are used? We are making the mean and variance of the normal distribution equal to the mean and variance of the binomial distribution we want to approximate.
Now:
(continuity correction)
(standardise)