Question

In: Statistics and Probability

Red Orange Yellow Green Purple Total 1046.    1066 977 1029 969 5087 Suppose all of...

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

Solutions

Expert Solution

(a)

Since, ten Skittles are selected with replacement, the probability of yellow Skittles in a selection = 977 / 5087 = 0.192

The number of yellow Skittles in 10 selected Skittles follow binomial probability distribution with n = 10, p = 0.192

For the binomial probability distribution, below conditions are met.

  • The experiment consists of 10 repeated trials.
  • Each trial can result in just two possible outcomes. Either the Skittles is yellow or not.
  • The probability of success, denoted by P, is the same on every trial because the Skittles are selected with replacement.
  • The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

Thus, all the requirements of the binomial probability distribution are met.

(b)

Probability that exactly 4 of the 10 Skittles are yellow = P(X = 4)

= 10C4  * 0.1924 * (1 - 0.192)10-4

= 210 * 0.1924 * 0.8086

= 0.0794

(c)

Probability that at most 2 of the 10 Skittles are yellow = P(X 2)

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

=  10C0  * 0.1920 * (1 - 0.192)10-0 + 10C1  * 0.1921 * (1 - 0.192)10-1 + 10C2  * 0.1922 * (1 - 0.192)10-2

= 0.1186 + 0.2818 + 0.3014

= 0.7018

Expected value = np = 10 * 0.192 = 1.92

standard deviation for the number of yellow skittles = = = 1.245536


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