Mars, Inc. claims that $20 \%$ of its M\&M plain candies are orange. A sample of 100 such candies is randomly selected. Find the mean and standard deviation for the number of orange candies in such groups of 100 .
A. $\mu=0.20, \sigma=4.0$
B. $\mu=20, \sigma=0.20$
c. $\mu=20, \sigma=4.0$
D. $\mu=0.020, \sigma=0.20$

Step 1 ：Mars, Inc. claims that 20% of its M&M plain candies are orange. A sample of 100 such candies is randomly selected. We are asked to find the mean and standard deviation for the number of orange candies in such groups of 100.

Step 2 ：The mean and standard deviation of a binomial distribution can be calculated using the formulas: Mean = np and Standard Deviation = sqrt(np(1-p)) where n is the number of trials and p is the probability of success.

Step 3 ：In this case, n = 100 (the number of candies in the sample) and p = 0.20 (the probability that a candy is orange).

Step 4 ：Substituting these values into the formulas, we get: Mean = np = 100 * 0.20 = 20.0 and Standard Deviation = sqrt(np(1-p)) = sqrt(100 * 0.20 * (1 - 0.20)) = 4.0

Step 5 ：Final Answer: The mean and standard deviation for the number of orange candies in such groups of 100 are \(\boxed{20}\) and \(\boxed{4.0}\) respectively. So, the correct option is C. \(\mu=20, \sigma=4.0\).