]> IB HL2 - Probability Distributions

Probability Distributions (Statistics Option: 8.2)

Recall that a distribution is something (a list or graph, for example) that shows possible values, and how probable those values are. Some of the following distributions were discussed in AP Statistics; others were not. What follows is part review and part new material!

On your calculator, the PDF functions give the height of the function at a certain point. That height is typically only useful for discrete distributions, where it also represents the probability of obtaining that value. IB calls PDF functions for discrete variables probability mass functions; and for continuous functions, it calls them probability density functions.

A. Discrete Types

A discrete distribution can only take certain (discrete!) values—typically integers. For all of the discrete types, to find the cumulative distribution function, simply sum up the results from the probability mass function (i.e., $P (X \leq x_{k}) = \sum_{i = 1}^{k} P (X = x_{i})$ ). Some of these are available on the calculator; some are not.

Recall that the mean of a discrete random variable is $μ_{x} = \sum_{i = 1}^{n} x_{i} \cdot P (X = x_{i})$ , and the variance is $σ_{x}^{2} = \sum_{i = 1}^{n} {(x_{i} - μ_{x})}^{2} \cdot P (X = x_{i})$ .

A1. Uniform (Discrete)

Description

The variable can only take on certain values, and each value occurs with equal frequency. The formulae list specifies that the possible values are integers from 1 to n. It is denoted $X ~ DU (n)$ .

Parameters

n, the final value.

Mean

$μ = \frac{n + 1}{2}$

Variance

$σ^{2} = \frac{n^{2} - 1}{12}$

Probability Mass Function

$P (X = x) = \frac{1}{n}$ . This function is too easy to be included on the calculator!

Example

An icosahedron (20 sided die) has the numbers 1 through 20 printed on its faces. The die is rolled, and the upper facing number is noted. What is the probability that the number is at least 15? What value should we expect?

If X = the facing number, then $X ~ DU (20)$ . Since there are 6 choices that are at least 15, $P (X \geq 15) = (6) (\frac{1}{20}) = \frac{6}{20} = \frac{3}{10} = 0.3$ , or 30%. We should expect $μ_{X} = \frac{20 + 1}{2} = \frac{21}{2} = 11.5$ on average (remember that the expected value is the mean).

A2. Bernoulli

Description

This is the result of a single experiment, that can result in either success or failure. It can be considered a special case of the binomial; hence its notation: $X ~ B (1, p)$ .

Parameters

p, the probability of success.

Mean

$μ = p$

Variance

$σ^{2} = p (1 - p)$

Probability Mass Function

$P (X = x) = p^{x} {(1 - p)}^{1 - x}$ . Since x is either 0 or 1, this will reduce to either p or $(1 - p)$ —thus, this function is not available on the calculator.

A3. Binomial

Description

A binomial occurs when counting the number of successes in a fixed number of independent trials of a Bernoulli experiment. Denoted $X ~ B (n, p)$ .

Parameters

n, the number of trials; and p, the probability of success on each trial.

Mean

$μ = n p$

Variance

$σ^{2} = n p (1 - p)$

Probability Mass Function

$P (X = x) = (\binom{n}{x}) p^{x} {(1 - p)}^{n - x}$ . This is available as binompdf( on your calculator. The cumulative distribution function is also available—binomcdf(.

Example

In January of 2006, approximately 38% of internet user in Finland used Firefox to browse the internet. In a random sample of 200 Finnish internet users, what is the probability that exactly 76 of them use Firefox? What is the probability that no more than 76 of them use Firefox?

Let X = number of Finnish Firefox users. Then $X ~ B (200, 0.38)$ . $P (X = 76) = 0.0580$ , and $P (X \leq 76) = 0.531$ (by GDC).

A4. Negative Binomial

Description

Also known as Pascal's Distribution…this occurs when we are counting the number trials required to achieve a certain number of successes in repeated independent trials of a Bernoulli experiment. Denoted $X ~ NB (r, p)$ .

Parameters

r, the number of successes that must be achieved; and p, the probability of success on each trial.

Mean

$μ = \frac{r}{p}$

Variance

$σ^{2} = \frac{r (1 - p)}{p^{2}}$

Probability Mass Function

$P (X = x) = (\binom{x - 1}{r - 1}) p^{r} {(1 - p)}^{x - r}$ .

Notice that this is very similar to the binomial formula. The difference comes from knowing that the final outcome must be success (that's why we stopped!), which reduces the arranged elements by one. Also note that $x \geq r$ (you must have at least as many trials as successes!).

This function is not available on the calculator.

Example

Bob's favorite cereal is currently promoting the wildly popular collectible card-trading game You-Go-Grrrl! by placing one of ten special cards in each box. Bob especially wants one of these ten cards (the gold foil hologram Dragon Lady card)—in fact, he wants three copies of this card (one to keep and two to trade). Assuming that the cards are randomly and equally placed in the cereal boxes, then what is the probability that Bob will have to buy nine boxes? How many boxes should he expect to open in order to get three of this card?

Let X = number of boxes opened to get three cards. Then $X ~ NB (3, 0.1)$ . $P (X = 9) = (\binom{8}{2}) {0.1}^{3} {0.9}^{6} = 0.0149$ (by GDC).

He should expect to open the mean— $μ_{X} = \frac{3}{0.1} = 30$ boxes.

A5. Poisson

Description

The Poisson occurs when we count the number of success that occur in an interval (of time, usually). The average number of successes must be the same for any interval, and occurrences in disjoint intervals must be independent of one another. Furthermore, the probability of two or more successes occurring at the same time (in a small subset of the given interval) must be approximately zero.

Another way to think about the Poisson is as the limit of the Binomial distribution—as the sample size n approaches infinity (or each trial is a measurement in a very small unit of time).

This distribution is denoted $X ~ Po (λ)$ .

Parameters

λ, the average number of successes that occur in an interval.

Mean

$μ = λ$

Variance

$σ^{2} = λ$

Probability Mass Function

$P (X = x) = \frac{λ^{x} e^{- λ}}{x!}$ . This function is available in the calculator by poissonpdf(. The cumulative distribution function is also available: poissoncdf(.

Example

A manufacturer of steel chain knows that there will be an average of one faulty link in every kilometer of chain produced. The company cuts the chain into 50 meter lengths before shipping. What is the probability that there are no faulty links in a 50 meter length of chain?

First, note that this is a Poisson scenario: we are counting the number of successes (faulty links) in an interval (50 meters of chain). We know the average—but be careful! The average was given for an interval of 1000 meters, and we are considering a length of 50 meters. There are 20 of these smaller intervals, and an average of one faulty link amongst them—thus, the mean $λ = \frac{1}{20} = 0.05$ . So if X = number of faulty links in a 50 meter length, then $X ~ Po (0.05)$ , and $P (X = 0) = \frac{{0.05}^{0} e^{- 0.05}}{0!} = \frac{1}{e^{0.05}} = 0.951$ (by GDC).

A6. Geometric

Description

This occurs when a Bernoulli experiment is repeated until a success occurs. We count the number of (independent) trials that were conducted. Denoted $X ~ Geo (p)$ . Note that this is a special case of the negative binomial ( $r = 1$ ).

Parameters

p, the probability of success on each trial.

Mean

$μ = \frac{1}{p}$

Variance

$σ^{2} = \frac{1 - p}{p^{2}}$

Probability Mass Function

$P (X = x) = p {(1 - p)}^{x - 1}$ . This is available on the calculator by geometpdf(. The cumulative distribution function is also available—geometcdf(.

Example

Dave is a pretty good shot with his vintage Red Rider BB Gun—he hits his target in 90% of his shots. What is the probability that he will have to shoot more than two shots to hit a target?

Let X = number of shots taken. Then $X ~ Geo (0.9)$ , and $P (X > 2) = 1 - P (X \leq 2) = 1 - 0.999 = 0.001$ (by GDC).

A7. Hypergeometric

Description

This occurs when we sample without replacement from a finite population, and each individual can be thought of as a success or failure. Denoted $X ~ Hyp (n, M, N)$ .

Parameters

n, the number of individuals sampled; M, the number of successes in the population; and N, the size of the population.

Mean

$μ = \frac{nM}{N}$

Variance

$σ^{2} = \frac{n M}{n} (1 - \frac{M}{N}) (\frac{N - n}{N - 1})$

Probability Mass Function

$P (X = x) = \frac{(\binom{M}{x}) (\binom{N - M}{n - x})}{(\binom{N}{n})}$ .

This formula makes perfect sense, if you think about it…in the numerator, you must get a certain number (x) of the potential successes (M), and the rest ( $n - x$ ) must be from the potential failures ( $N - M$ ). The denominator is the number of ways to select the sample from the population.

This function is not available in the calculator.

Example

A box contains 60 red balls and 40 blue balls. Ten balls are taken out and their colors noted. What is the probability that exactly five of the balls are blue?

Let X = number of blue balls. Then $X ~ Hyp (10, 40, 100)$ , and $P (X = 5) = \frac{(\binom{40}{5}) (\binom{60}{5})}{(\binom{100}{10})} = \frac{(658008) (5461512)}{17310309460000} = 0.208$ (by GDC).

B. Continuous Types

A continuous random variable can take on any value in a continuum (like the real numbers). Because of this, the formulas that give the mean and variance (or standard deviation) for discrete random variables change into integrals. Thus, the mean of a continuous random variable is $μ = \int x \cdot f (x) d x$ , and the variance is $σ^{2} = \int {(x - μ)}^{2} f (x) d x$ . Also, since the CDF is found by summing results of the PDF, an integral is required (if the PDF is $f (x)$ , then the CDF is $\int_{x}^{a} f (t) d t$ ).

Recall that the PDF function only gives the height of the density curve—not a probability! Technically, for any continuous random variable X, $P (X = a) = 0$ . However, what is often meant by $P (X = a)$ is "what is the probability of obtaining a value that rounds to the integer a?" Now this can be done! In this case, $P (X = a) = P (a - 0.5 \leq x < a + 0.5)$ , which can be evaluated using the CDF function.

B1. Uniform (Continuous)

Description

This is just like the Discrete Uniform, except that the possible values are continuous over the interval $[a, b]$ —every value in that interval is equally likely. Denoted $X ~ U (a, b)$ .

Parameters

a, the lowest possible value; and b, the highest possible value.

Mean

$μ = \frac{a + b}{2}$

Variance

$σ^{2} = \frac{{(b - a)}^{2}}{12}$ .

Probability Density Function

$P (X = x) = \frac{1}{b - a}$ . This is too simple to include on the calculator!

Cumulative Distribution Function

Integrate! $P (X \leq x) = \int_{0}^{x} \frac{1}{b - a} d t = \frac{t}{b - a}]_{0}^{a} = \frac{x}{b - a}$ .

Example

Doc Holloman (the evil twin of Mr. Holloman) assigns grades randomly—he generates a random real number between zero and one, then converts that to a percentage. If some unlucky soul should chance to get Doc H for a class, then what is the probability that the person will pass the class?

Let X = the assigned grade. Then $X ~ U (0, 1)$ . Note that passing is a grade of at least 0.695 (because we round to two digits!). Thus, the probability of passing is $P (X \geq 0.695) = 1 - P (X \leq 0.695) = 1 - 0.695 = 0.305$ (by GDC).

B2. Exponential

Description

This occurs when the possible values are on the interval $[0, \infty)$ , and the probability of any value x is lower (exponentially lower) than any value below x. This is particularly used to model the amount of time (space) that occurs between successes in a Poisson distribution.

The exponential distribution is memoryless— $P (X > a + b | x > a) = P (X > b)$ ; or, the probability that you'll get a value a certain amount above a known value is the same as the probability of that certain value. Try an example: the probability that a fuse will last another 100 hours is the same as the probability that the fuse lasts a total of 100 hours—the fuse doesn't know how long it's been running; the probability of working into the next 100 hours is the same no matter how long it's already been running.

You will almost certainly have to be told that a variable has an exponential distribution.

Denoted $X ~ Exp (λ)$ .

Parameters

λ, the mean of the underlying Poisson distribution. Note that $λ > 0$ .

Mean

$μ = \frac{1}{λ}$

Variance

$σ^{2} = \frac{1}{λ^{2}}$

Probability Density Function

$P (X = x) = λ e^{- λ x}$

Cumulative Distribution Function

Integrate…watch out for the u-substitution! $P (X \leq x) = \int_{0}^{x} λ e^{- λ t} d t = - e^{- λ t}]_{0}^{x} = - e^{- λ x} + e^{0} = 1 - e^{- λ x}$

Example

The lifetime (in hours) of a certain electronic component follows an exponential distribution with $λ = \frac{1}{100}$ . What is the mean lifetime of this component? What is the probability that the component lasts for more than 200 hours?

We're told that $X ~ Exp (\frac{1}{100})$ is the lifetime (in hours) of this component. The mean (called, in this particular example, the Mean Time To Failure) is $μ = \frac{1}{λ} = 100$ hours.

The probability that the component lasts more than 200 hours is $P (X > 200) = 1 - P (X \leq 200) = 1 - (1 - e^{- (\frac{1}{100}) 200}) = e^{- 2} = 0.135$ .

B3. Normal

Description

The undisputed King of all distributions! It should need no description here. Denoted $X ~ N (μ, σ^{2})$ .

Parameters

μ, the mean; and σ², the variance.

Probability Density Function

$P (X = x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {(\frac{x - μ}{σ})}^{2}}$

Cumulative Distribution Function

Alas, a closed form of the integral of the PDF does not exist…your calculator uses approximate numerical integration to find the area under the curve.

Page last validated 2010-08-15