AP Statistics Curriculum 2007 Distrib Dists

General Advance-Placement (AP) Statistics Curriculum - Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments

Geometric

• Definition: The Geometric Distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. The name geometric is a direct derivative from the mathematical notion of geometric series.

• Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is \(P(X = x) = (1 - p)^{x-1} \times p\), for x = 1, 2, 3, 4,....

• Expectation: The Expected Value of a geometrically distributed random variable X is \({1\over p}.\)

• Variance: The Variance is \({1-p\over p^2}.\)

• Example: See this SOCR Geometric distribution activity.

HyperGeometric

The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table:

Type	Drawn	Not-Drawn	Total
Defective	k	m-k	m
Non-Defective	n-k	N+k-n-m	N-m
Total	n	N-n	N

• Explanation: Suppose there is a shipment of N objects in which m are defective. The Hypergeometric Distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective.

• Mass function: The random variable X follows the Hypergeometric Distribution with parameters N, m and n, then the probability of getting exactly k successes is given by

\[ P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.\]

This formula for the Hypergeometric Mass Function may be interpreted as follows: There are \({{N}\choose{n}}\) possible samples (without replacement). There are \({{m}\choose{k}}\) ways to obtain k defective objects and there are \({{N-m}\choose{n-k}}\) ways to fill out the rest of the sample with non-defective objects.

The mean and variance of the hypergeometric distribution have the following closed forms:

Mean\[n \times m\over N\]

Variance\[{ {nm\over N} ( 1-{m\over N} ) (N-n)\over N-1}\]

Examples

• SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can be varied with scroll bars.

• A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to:
  • Compute the probability mass function of the number of tagged fish in the sample of 20.
  • Compute the expected value and the variance of the number of tagged fish in this sample.
  • Compute the probability that this random sample contains more than 3 tagged fish.

• Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use binomial approximation to hypergeometric. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be \(\hat{p}={5\over 100}=0.05\). Thus, we can estimate that \(0.05=\hat{p}={200\over N}\), and \(N\approx 4,000\), as shown on the figure below.

• You can also see a manual calculation example using the hypergeometric distribution here.

Negative Binomial

The family of Negative Binomial Distributions is a two-parameter family; p and r with 0 < p < 1 and r > 0. There are two (identical) combinatorial interpretations of Negative Binomial processes (X or Y).

X=Trial index (n) of the r^th success, or Total # of experiments (n) to get r successes

• Probability Mass Funcrtion\[ P(X=k) = {k-1 \choose r-1}\cdot p^r \cdot (1-p)^{k-r} \!\], for k = r,r+1,r+2,.... (k=trial number of the r^th success)
• Mean\[E(X)= {r \over p}\]
• Variance\[Var(X)= {r(1-p) \over p^2}\]

Y = Number of failures (k) to get r successes

• Probability Mass Function\[ P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!\], for k = 0,1,2,.... (k=number of failures before the r^th successes)
• \(Y \sim NegBin(r, p)\), the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial.
• Mean\[E(Y)= {r(1-p) \over p}\].
• Variance\[Var(Y)= {r(1-p) \over p^2}\].
• Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y).

Application

Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail.

• What's the probability mass function of the number of failures (k=n-r) to get r=6 successes?

In other words, What's the probability mass function that the last 6^th state she succeeds to secure all electoral votes happens to be the at the n^th state she campaigns in?

NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, k = n-6, and \(X\sim NegBin(r=6, p=0.3)\). Thus, for \(n \geq 6\), the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:

\[ P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k = {k + 6 - 1 \choose k} \cdot 0.3^6 \cdot 0.7^{k} \] \[= {k+5 \choose k} \cdot \frac{3^6 \cdot 7^{k}}{10^{k+6}} \]

• What's the probability that Jane finishes her campaign in the 10^th state?

Let \(Y\sim NegBin(r=6, p=0.3)\), then P(Y=10) = {{10+6-1 \choose 10}\cdot 0.3^6 \cdot 0.7^{10} = 0.022054.</math>

• What's the probability that Jane finishes campaigning on or before reaching the 8^th state?

\[ P(X\leq 8) = 0.011292\]

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• Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below)! What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?

\[ P(X\geq 50) = 0.632391\]

• SOCR Activity: If you want to see an interactive Negative-Binomial Graphical calculator you can go to this applet (select Negative Binomial) and see this activity.

References

• Negative-Binomial Activity

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• SOCR Home page: http://www.socr.ucla.edu

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