Question

​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.

Answer 1

The probability p of an orangecandy is 0.2. The sample size = 100.
The mean is given by:
$n\times p=100\times0.2=20$
The standard deviation is given by:
$\sqrt{npq} = \sqrt{100\times0.2\times(1-0.2)} =4$
The answers are: Mean = 20. Standard deviation = 4.

Answer 2

Answer:

mean = 20; standard deviation = 4

Step-by-step explanation:

To find the mean, we multiply n, the number of trials, by p, the probability that the candy is orange:

n(p) = 100(0.2) = 20

To find the standard deviation, we first multiply n by p and by q, the probability that the candy is not orange.  Since p = 0..2, q = 1-0.2 = 0.8:

100(0.2)(0.8) = 16

Next we take the square root:

√16 = 4