I. Suppose that 80% of the population likes cake. (a) Say we sample 3 people fro
ID: 3049856 • Letter: I
Question
I. Suppose that 80% of the population likes cake. (a) Say we sample 3 people from the population, with replacement, and let X be the number of those 3 people who like cake. Find the pmf of X. (Recall this means finding PX(k), or in other words P(X k), for all values of k.) Please calculate decimal approximations for your answers. (b) Suppose now that we sample 3 people without replacement from an overall population of size 10. (So the population consists of 8 people who like cake and 2 people who do not.) Again let X be the number of those 3 people who like cake, and find the pmf of X. As before, calculate decimal approximations. (Hint: You can solve this as a counting problem. Suppose you have a deck of 10 cards, numbered 1 through 10, and you draw 3 cards. What is the probability that k of the cards drawn have numbers between 1 and 8 inclusive?) (c) Repeat the previous part with an overall population of size 1000. To get a decimal approximation, please use Mathematica, Wolfram Alpha, or similar software that can handle very large numbers. (The command for () is binomial (n,k).) (d) Repeat the previous part with an overall population of size 1000000. (e) Compare your results, and comment on how population size affects the difference between sampling with and without replacement.Explanation / Answer
Since 80% people like the cake, so p = 0.8
For sample size of 10:
With replacement:
P(X=0) = 3C0 * 0.80 * (1-0.8)3 = 0.008
P(X=1) = 3C1 * 0.81 * (1-0.8)2 = 0.096
P(X=2) = 3C2 * 0.82 * (1-0.8)1 = 0.384
P(X=3) = 3C3 * 0.83 * (1-0.8)0 = 0.512
Without replacement:
P(X=0) = 0 (because when we choose 3 out 10 people and only 2 people don't like the cake, there will always be a person who will like it)
P(X=1) = 8C1 * 2C2 / 10C3 = 0.067
P(X=2) = 8C2 * 2C1 / 10C3 = 0.467
P(X=3) = 8C3 * 2C0 / 10C3 = 0.467
For sample size of 1000:
Without replacement:
P(X=0) = 800C0 * 200C3 / 1000C3 = 0.0079
P(X=1) = 800C1 * 200C2 / 1000C3 = 0.0958
P(X=2) = 800C2 * 200C1 / 1000C3 = 0.3846
P(X=3) = 800C3 * 200C0 / 1000C3 = 0.5116
In the same way compute for the population size of 10^6 as well.