Question

The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population. Suppose that we purchase a mid-size bag of Skittles. Assume this size bag always has 200 candies. In this particular bag, 50 are green. What is the probability that a randomly selected bag of this size has 50 or more green candies? (Note: the answer may vary slightly due to rounding in the calculation of the Z-score.) Please show steps how to obtain standard error and z score Cheggs response Here, p = 0.20 n = 200 mean = np = 0.20 *200 = 40 Std.dev = sqrt(npq) = sqrt(200 *0.20 *0.80) = 5.6569 P(x> 50) = P(z> (x - mean)/std.dev) = P(z> (50-40)/5.6569) = P(z> 1.7678) S0, here, z = 1.7678 rounding to two decimal = 1.77 Std.error = 5.6569 Can you tell me where you got .80 And explain the SD=sqrt(npq) Why not use SD= sqrt p(1-p)/n. ???

Answer 1

Here, n = 200 , p = 0.20

We use the normal approximation o binomial so, formula in normal approximation for std.dev = sqrt(npq) = sqrt(200 * 0.20 *0.80) = 5.6569

Here, p =0.20
and q = 1- 0.20 = 0.80
So, q is 0.80 which is used in std.deviation formula