Question

The number of peanut M&Ms in a 2 ounce package is normally distributed with a mean of 28 and standard deviation 2; The number of Skittles in a 2 ounce package is normally distributed with a mean of 60 and standard deviation 4.

Questions 1-3: Suppose that I purchase two 2-ounce packages of peanut M&Ms and one 2-ounce package of Skittles.

1. Let X= the total number of pieces of candy in all three bags combined. What is the distribution of X?

2. What is the probability that the total number of pieces of candy in all three bags combined is less than 110?

3. What is the probability that the total number of M&Ms (in both bags combined) is greater than the number of Skittles?

Answer 1

Let A be the 2-ounce packages of peanut M&Ms

A ~N(28, 2)

and

B be the 2-ounce package of Skittles.

B~N( 60, 4)

1) Let X= the total number of pieces of candy in all three bags combined.

X = 2A+B

E(X) = 2E(A)+E(B) = 2*28+60= 116

V(X) = 4V(A)+V(B) = 4*4+16 =32

SD(X) = 5.66

X ~N(116, 32)

2)

3)

Let U = 2A - B

E(U) = 2E(A)-E(B) = 2*28-60=-4

V(U) = 2V(A)+V(B) = 4*4+16 =32

SD(U) = 5.66