Question

Red

Orange

Yellow

Green

Purple

Total

1046.

1066

977

1029

969

5087

• 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) List the requirements of the binomial probability distribution and show that this meets them, including identifying the values for n and p. (6 points)
  • (b) What is the probability that exactly 4 of the 10 Skittles are yellow? (4 points)
  • (c) What is the probability that at most 2 of the 10 Skittles are yellow? (4 points)
  • (d) 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)

*Show your work

Answer 1

Answer 2

(a) The requirements of the binomial probability distribution are:

1. Fixed number of trials: The number of trials, denoted as n, is fixed. In this case, n = 10 as we are selecting 10 Skittles.

2. Two possible outcomes: Each trial has only two possible outcomes: yellow or not yellow.

3. Independent trials: The outcome of each trial does not affect the outcome of other trials.

4. Constant probability of success: The probability of success (p) remains constant for each trial.

Given the information, the values for n and p are: n = 10 (fixed number of trials) p = probability of selecting a yellow Skittle = (number of yellow Skittles) / (total number of Skittles) = 977 / 5087

(b) The probability that exactly 4 of the 10 Skittles are yellow can be calculated using the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

where X is the random variable representing the number of successes (yellow Skittles), k is the specific number of successes, n is the total number of trials, p is the probability of success, and C(n, k) is the combination function.

Plugging in the values: P(X = 4) = C(10, 4) * (977 / 5087)^4 * (1 - 977 / 5087)^(10 - 4)

(c) The probability that at most 2 of the 10 Skittles are yellow can be calculated by summing the probabilities of getting 0, 1, or 2 yellow Skittles:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

(d) For samples of size 10, the expected value (mean) and standard deviation for the number of yellow Skittles can be calculated as follows:

Expected value (mean): E(X) = n * p

Standard deviation: σ = √(n * p * (1 - p))

Plugging in the values: E(X) = 10 * (977 / 5087) σ = √(10 * (977 / 5087) * (1 - 977 / 5087))