Question

The number of M&M's in a fun size pack has on average 10 candies with a variance of 4 candies. If you are given a single fun size pack of M&M's:

a. What can we say about the probability that your pack of M&M's does not have between 9 and 11 candies?

b. How useful do you find this information?

For a class activity, each of the 35 students gets a pack of M&M's to work with.

c. What is the distribution (including parameter values) of the total number of M&M's that will be used in class?

d. What is the probability that the average number of M&M's in a pack used in class is between 9.5 and 10.5 candies?

e. Would the variance in the average of candies in a pack increase or decrease if the same exercise was repeated in a class with 80 students?

Answer 1

mean = 10 variance = 4

a. What can we say about the probability that your pack of M&M's does not have between 9 and 11 candies?

1 - P[ 9 < X < 11 ]

We need to compute . The corresponding z-values needed to be computed are:

Therefore, we get:

1 - P[ 9 < X < 11 ] = 1 - 0.3829 = 0.6171

b. How useful do you find this information?

Since, most of the data (61.71% ) does not lies between 9 and 11 we can interpret a lot of things out of this.

c. What is the distribution (including parameter values) of the total number of M&M's that will be used in class?

For the sample, mean will remain same.

mean = 10

variance will be reduced to 4/35 = 0.114

variance = 0.114

d. What is the probability that the average number of M&M's in a pack used in class is between 9.5 and 10.5 candies?

Here we will use s = sqrt(0.114) = 0.338 to compute the probablity

We need to compute . The corresponding z-values needed to be computed are:

Therefore, we get:

e. Would the variance in the average of candies in a pack increase or decrease if the same exercise was repeated in a class with 80 students?

yes, as the sample size increases the variance decreases

variance will reduce to 4/80 = 0.05 from 0.114