About 34 of Americans have type A blood A blood drive gets 2

About 34% of Americans have type A+ blood. A blood drive gets 200 donors. What is the probability that at most one third of the donors have type A+ blood donors?

Solution

n = 200 p = 0.34 For this case, applying normal approximation to binomial, we get: mean = n*p= 68 variance = n*p*(1-p) = 44.88 std dev = 6.6993 Find : P( X < 66.6667) We convert to standard normal form, Z ~ N(0,1) by z1 = (x1 - u )/s z1 =( x - u)/s = (66.6667 - 68 )/6.6993 = -0.2 P( X < x1) = P( Z < z1) = P( Z < -0.2) = 0.4207 _ _ _ _ _ _ _ _ _ (From Z-table)