Chapter 6 Discrete Probability Distributions PDF

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Chapter 6 Discrete Probability Distributions

Random Variables (r.v’s) -

Numerical value assigned to the outcomes of an experiment. Capital letters X, Y, Z, with or without subscripts, are used to denote RV’s

Two types of r.v.’s: Discrete and Continuous a.

A discrete Random Variable. A random variable X is said to be discrete if it can assume only a finite or countable infinite number of distinct values. It has values that can be counted

Examples: • Toss a coin n time and count the number of heads. • Number of children in a family. • Number of car accidents per week. • Number of defective items in a given company. • Number of bacteria per two cubic centimeters of water. b.

A continuous random variable X is one for which the outcome can be any number in an interval or collection of intervals.

Examples:

The first two r.v’s above are ‘discrete’, and the second two r.v.’s are ‘continuous’.

PROBABILITY DISTRIBUTION – a statement, or table, that assigns probabilities to the possible values of a Random Variable.

a.

Discrete probability distribution is a table of values of the variable and the proportion of times (or probability) it occurs (which may be expressible in graphical form).

a.

Probability distribution of a continuous random variable: idealized curve (perhaps from a histogram) which represents probability that a value of the variable occurs as an area under the curve (which may be expressible in functional form).

Mean and Variance of Discrete Random Variables Let X be a discrete random variable. For convenience and brevity we will write the P(X = x) as P(x). Definition: The expected value of a discrete random variable or the mean of its distribution is defined as E(X) =  = x . P(x). Definition: The variance of a discrete probability distribution is defined by Var(X) = X 2 = x2 . P(x)   2. The standard deviation is SD(X) =

Var ( X )

= X =

X2

.

Binomial Experiment A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties:  

 

The experiment consists of n repeated trials. o E.g., 15 tosses of a coin; 10 light bulbs taken from a warehouse Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. o E.g., Heads or tails in each toss of a coin; defective or not defective light bulb. The probability of success, denoted by P, is the same on every trial. o E.g., Probability of getting a tail is the same each time we toss the coin The trials are independent. o The outcome of one trial does not affect the outcome of the other trials

Example: Consider the following statistical experiment. You flip a coin 2 times and count the number of times the coin lands on heads. This is a binomial experiment because:    

The experiment consists of repeated trials. We flip a coin 2 times. Each trial can result in just two possible outcomes - heads or tails. The probability of success is constant - 0.5 on every trial. The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials.

Notation The following notation is helpful, when we talk about binomial probability.      

x: The number of successes that result from the binomial experiment. n: The number of trials in the binomial experiment. P: The probability of success on an individual trial. q: The probability of failure on an individual trial. (This is equal to 1 - P.) b(x; n, P): Binomial probability - the probability that an n-trial binomial experiment results in exactly x successes, when the probability of success on an individual trial is P. nCx: The number of combinations of n things, taken x at a time.

Examples on Binomial distribution: Example 1:  E.g. flip a fair coin 10 times… – 1) Fixed number of trials  n=10 – 2) Each trial has two possible outcomes  {heads (success), tails (failure)} – 3) P(success)= 0.50; P(failure)=1–0.50 = 0.50  – 4) The trials are independent  (i.e. the outcome of heads on the first flip will have no impact on subsequent coin flips).  Hence flipping a coin ten times is a binomial experiment since all conditions were met. Example 2:

Example 3:

The mean and the variance of a binomial random variable

Example 4:

Cumulative Binomial Probability

Exercises: 1.

A company is considering taking on a project that has a 0.3 probability of producing a profit $2,000,000, a probability of 0.2 of producing a profit of $750,000, and a probability of 0.5 of creating a loss of $500,000. What is the company’s expected return on this project?

2.

A game is played in which a you can lose $3 with a probability of 0.3, lose $1 with a probability of 0.4, win $2 with a probability of 0.1 and win $6 with a probability of 0.2. Make a table for the distribution of the random variable X = the amount you win (a loss is a negative win, e.g. lose $1 = –1). Compute the expected value of X. Would you be willing to play this game? Explain why.

3.

Complete the following table and find the mean, variance, and standard deviation of the probability distribution. 2 x P(x) xP(x) x P(x) 2 0 1 3 4

0.10 0.25 0.30 0.20 0.15



X = ______ X2 = ______ X = ______

Poisson Distribution A Poisson experiment is a statistical experiment that has the following properties:    

The experiment results in outcomes that can be classified as successes or failures. The average number of successes (μ) that occurs in a specified region is known. The probability that a success will occur is proportional to the size of the region. The probability that a success will occur in an extremely small region is virtually zero.

Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc. Poisson Probability Distribution

μ x e− μ P( X )= x! , Where: x is the actual number of successes that result from the experiment,  = average (mean) number of successes and e is approximately equal to 2.71828.

Notation The following notation is helpful, when we talk about the Poisson distribution.    

e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.) μ: The mean number of successes that occur in a specified region, area, or period of time. x: The actual number of successes that occur in a specified region. P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is μ.

Mean and variance of Poisson The mean of the distribution is equal to μ x =  The variance is also equal to μ  2x = 

Examples on Binomial distribution: Example 1:

Example 2:

Example 3:

Example 4:

Cumulative Poisson Probability

A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit.

Example 5: Suppose that the number of messages that a person receives on his cellphone averages one per hour and that the number of messages received each hour is independent of all other hours. What’s the probability of his receiving two messages in the next hour? In this case, the value of lambda

is equal to 1, because the average number of messages

each hour equals 1. The probability of receiving two messages during the next hour is

Alternatively, you can get results from a Poisson table set up like this table.

The table shows the Poisson probabilities for different values of In the cellphone example, because x = 2 and the appropriate probability P(X = 2) is found in the ‘x = 2′ row and the

The probability is 0.1839.

Example 6 The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?

Solution: This is a Poisson experiment in which we know the following:   

μ = 2; since 2 homes are sold per day, on average. x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow. e = 2.71828; since e is a constant equal to approximately 2.71828.

We plug these values into the Poisson formula as follows: P(x; μ) = (e-μ) (μx) / x! P(3; 2) = (2.71828-2) (23) / 3! P(3; 2) = (0.13534) (8) / 6 P(3; 2) = 0.180 Thus, the probability of selling 3 homes tomorrow is 0.180 .

Example 7 Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari? Solution: This is a Poisson experiment in which we know the following:   

μ = 5; since 5 lions are seen per safari, on average. x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions. e = 2.71828; since e is a constant equal to approximately 2.71828.

To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula: P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5) P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ] P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738) (125) / 6 ] P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ] P(x < 3, 5) = 0.2650 Thus, the probability of seeing at no more than 3 lions is 0.2650....