Statistics please help

Question

A candy company claims that 20% of its plain candies are orange and a sample of 100 such candies is randomly selected

A. Find the mean and standard deviations for the number of orange candies in such groups of 100

B. A random sample of 100 candies contains 4 orange candies. Is the result unusual? Does it seem that the claimed rate of 20% is wrong?

Answer 1

a)

Each candy X_i∼Bernouili (p=.2) where p= chance of success which is .2 for this question. Since the 100 candies are independent B =∑X_i will have a binomial distribution and B∼ (n=100 , p = .2). A binomial distribution is used to model event that have n number of independent trials and p chance of success. For a Binomial Distribution the mean = np and the variance = np(1-p).

n=100

p=.2

(1-p) = .8

So the mean = 100*(.2) = 20

variance = 100*(.2)*(.8) = 16

We know the standard deviation is the square root of the variance so the standard deviation is the square root of 16 = 4

b) If a random sample of 100 candies contained 4 orange candies that would be unusual because this is very far from the mean of 20. Yes it would seem that the claim rate of .2 is wrong.