Question

Find the probability that in 200 tosses of a fair​ six-sided die, a five will be obtained at least 40 times.

Answer 1

n = number of trials = 200
p = prob. of success = 1/6 if we consider getting a five to be a success.

Then np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal. This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.

Now for the continuity correction. The rectangle over 40 goes from 39.5 to 40.5. Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.

z = (39.5-33.33)/5.27 = 1.17
Area to the right of 1.17 on the standard normal is 0.121