Chapter 5 Discrete Probability Distributions

DISCRETE Discrete variables – have a finite number of possible values or an infinite number of values that can be counted. Discrete probability distribution – consists of the values a random variable can assume and the corresponding probabilities of the values. The graphical displays of these distributions are BAR GRAPHS (see your text).

2 Requirements for Prob. Dsn 1.The sum of the probabilities of all the events in the sample space must equal 1 OR ∑ P(X) = 1 2.The probability of each event in the sample space must be between or equal to 0 and 1 OR 0 ≤ P(X) ≤ 1

Is it a probability distribution? Are either of the two following distributions probability distributions? a) - b) X P(X)1/7 3/72/71/7 X51015 P(X).2.3.5

Example (Discrete Probability Dsn) During the summer months, a rental agency keeps track of the number of chain saws it rents each day during a period of 90 days. The number of saws rented per day is represented by the variable X. The results are shown below. Construct a probability dsn and graph for the data. X (saws per day)Number of Days Total = 90

Example (cont.) Find probability for each X – P(0) = 45/90 = 0.50, P(1) = 30/90 = 0.33, P(2) = 15/90 = 0.17 – Probability distribution: X (saws per day)Number of Days Total = 90 Number of saws rented (X)012 Probability (P(X))

Mean of Probability Distribution E(X) – called expected value E(X) = ∑ [X ∙ P(X)] 1.Take each X and multiply it by its probability 2.Add them all up

Example (Mean of Probability Dsn) The probability distribution shown represents the number of trips of five nights or more hat American adults take per year. (That is, 6% do not take any trips lasting five nights or more, etc.) Find the mean. Number of trips (X) Probability P(X)

Example (cont.) E(X) = ∑ [X ∙ P(X)] = (0)(0.06)+(1)(0.70)+(2)(0.20)+(3)(0.03)+(4)(0.01) = 1.23 The mean number of trips lasting five nights or more per year taken by American adults is Number of trips (X) Probability P(X)

Example (cont.) You can do this on your calculator! 1.Put the X’s into L1 2.Put the probabilities into L2 3.STAT → CALC → 1-VAR STAT L1, L2 4.This will give you mean and standard deviation (see next)

Example (Variance and S.D.) A talk radio station has four telephone lines. If the host is unable to talk (i.e., during a commercial) or is talking to a person, the other callers are placed on hold. When all lines are in use, others who are trying to call in get a busy signal. The probability that 0, 1, 2, 3, or 4 people will get through is shown below. Find the variance and standard deviation for the distribution. X01234 P(X)

Example (cont.) Putting the probability distribution in your calculator and running one-variable statistics on it yields: – Sigma (s.d.) = 1.12 X01234 P(X)

A Binomial Experiment Binomial experiment – probability experiment that satisfies the following four requirements: 1.There must be a fixed number of trials. 2.Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. (We call these success and failure) 3.The outcomes of each trial must be independent of each other. 4.The probability of a success must remain the same for each trial.

The Binomial Distribution Binomial distribution – the outcomes of a binomial experiment and the corresponding probabilities of these outcomes. Notation: – p = probability of a success – q = probability of a failure (note: 1-p = q) – n = number of trials – X = number of successes in n trials

The Binomial Formula Notation: – p = probability of a success – q = probability of a failure (note: 1-p = q) – n = number of trials – X = number of successes in n trials

Example 1 (Finding Binomial Prob.) A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month. – The probability that we are looking for is P(X=3) – We know we need p, q, n and x p = 1/5 = 0.2 q = (1 – 1/5 ) = 4/5 = (1 – 0.2) = 0.8 n = 10 x = 3

Example 1 (cont.) – p = 1/5 = 0.2 – q = (1 – 1/5 ) = 4/5 = (1 – 0.2) = 0.8 – n = 10 – x = 3 So, if you select 10 people at random the probability that 3 of them have been to the doctor (in the last month) is 20.1%

Example 2 (Finding Binomial Prob.) A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part- time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part- time jobs. This time the probability that we are looking for is P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X=5) That is, to do this we need for 3, 4, and 5 and add them up.

Example 2 (cont.) This time the probability that we are looking for is P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X=5) That is, to do this we need for 3, 4, and 5 and add them up. – p = 0.3 – q = (1 – 0.3) = 0.7 – n = 5 – X = 3, 4, and 5 Let’s use the table in the back to get these!

Example 2 (cont.) P(X = 3) = P(X = 4) = P(X = 5) = Adding them up: = The probability that you select 5 teens and at least three of them have jobs is 16.2%

Binomial Mean and Variance The formula for the probability distribution is complex but the formulas for mean and variance are much more user friendly. Mean = μ = n ∙ p Variance = σ 2 = n ∙ p ∙ q (for standard deviation take the square root)

Example (Binomial mean and variance) The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins. – n = 8000 – p =.02 – q =.98

Example (cont.) n = 8000 p =.02 q =.98 – Mean = n ∙ p = (8000)(0.02) = 160 – Variance = n ∙ p ∙ q = (8000)(0.02)(0.98) = – Standard deviation = √ (156.8) = 12.5