Download Problem 1: Suppose you are going to randomly select two Skittles

-REVISEDProblem 1: Suppose you are going to randomly select two Skittles from the bag YOU purchased.
(a) What is the probability that both Skittles are purple if you select them with replacement? Give your
answer correct to four decimal places. (4 points)
13/58 X 13/58= 0.0502
(b) What is the probability that both Skittles are purple if you select them without replacement? Give your
answer correct to four decimal places. (4 points)
13/58X12/57=.0472
(c) What is the probability that at least one Skittle is purple if you select them with replacement? (4
points)
1-(45/58)(45/58)= .3980
Problem 2: Suppose all of the Skittles in the class data set are combined into one large bowl and you are
going to randomly select one Skittle.
(a) What is the probability that you select a green Skittle? (4 points)
710/3551=0.1999
(b) What is the probability that you select a Skittle that is NOT green? (4 points)
1-0.1999=0.8001
(c) What is the probability that you select a Skittle that is red OR yellow? (4 points)
716+726=1442/3551=0.4061
(d) What is the probability that you select a Skittle that is orange GIVEN that it is a secondary color
(secondary colors are green, orange and purple)? (4 points)
698+710+701=2109 698/2109=0.3310
Problem 3: Suppose all of the Skittles in the class data set are combined into one large bowl and you are
going to randomly select ten Skittles with replacement and count how many are yellow.
(a) Show that this meets the requirements of the binomial probability distribution and identify n and p. (5
points)
Fixed number of trials n=number of skittles being selected=10
Independent trials since the skittles are being selected with replacement
Constant probability of success: p=726/351=.2044
And two outcomes are possible: yellow or not yellow
It does meet the requirements because there are two outcomes possible.
(b) What is the probability that exactly 4 of the 10 Skittles are yellow? (4 points)
Binompdf(10,726/3551,4) = .09302
(c) For samples of size 10, what is the expected value and standard deviation for the number of yellow
skittles that will be included? (4 points) Mean 2.0444945 SD= 1.2753
Problem 4: For this problem, treat a 2.17 ounce bag of Skittles as an individual. Suppose the values for
our class data are the parameter values for all 2.17 ounce bags of Skittles. In other words, assume μ =
mean number of candies per bag in our class data set and σ = standard deviation of number of candies per
bag in our class data set (you computed these values in Part 2).
Mean= 59.18 Standard Deviation=3.11
(a) Describe the sampling distribution for the mean number of candies per bag for samples of 32 bags.
Include center, spread and shape. Note: The shape of the SAMPLING DISTRIBUTION is different from
the shape of the population, which you determined in Part 2 of the project. (5 points)
Mean=59.18 3.11/sqrt(32)=0.549775522 = 0.5498
Shape of the sampling distribution: becomes approximately normal since n>30
(b) What is the probability that the mean number of candies per bag for a sample of 32 bags is greater
than 58.5? (4 points)
P(X>58.5)
z=(58.5-59.18)/(3.11/sqrt 32)=-1.24
Probability of 32 bags is greater than 58.5= 1-.1075 =0.8925