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Chapter 5

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.

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Chapter 5

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  1. Chapter 5 Discrete Probability Distributions

  2. 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).

  3. 2 Requirements for Prob. Dsn • The sum of the probabilities of all the events in the sample space must equal 1 OR ∑ P(X) = 1 • The probability of each event in the sample space must be between or equal to 0 and 1 OR 0 ≤ P(X) ≤ 1

  4. Is it a probability distribution? • Are either of the two following distributions probability distributions? • -

  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.

  6. 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:

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

  8. 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.

  9. 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 1.23.

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

  11. 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.

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

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

  14. 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

  15. 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

  16. 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

  17. 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%

  18. 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.

  19. 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!

  20. Example 2 (cont.) P(X = 3) = 0.132 P(X = 4) = 0.028 P(X = 5) = 0.002 Adding them up: 0.132 + 0.028 + 0.002 = 0.162 The probability that you select 5 teens and at least three of them have jobs is 16.2%

  21. 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 = σ2 = n ∙ p ∙ q (for standard deviation take the square root)

  22. 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

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

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