CHAPTER 6: DISCRETE PROBABILITY DISTRIBUTIONS

PROBIBILITY DISTRIBUTION DEFINITIONS (6.1):  Random Variable is a measurable or countable outcome of a probability experiment.  Discrete Random Variables have countable and finite outcomes.  Continuous Random Variables have infinitely many possible outcomes from measurements and are described in ranges of values.

DISCRETE PROBABILITY DISTRIBUTIONS:  Looks like Discrete Relative Frequency Distributions from chapter 2.  The relative frequency is the probability [p(x)] of the discrete value (x) occurring and therefore has a value between 0 and 1.  The sum of all probabilities is 1.0.

DISCRETE PROBABILITY DISTRIBUTIONS:  Example of rolling a pair of dice. Will do this together. xP(x)x 21/3685/36 32/3694/36 43/36103/36 54/36112/36 65/36121/36 76/36

DISCRETE PROBABILITY DISTRIBUTIONS:  Make up of Jury in area that is 65% Hispanic. xP(x)X

DISCRETE PROBABILITY DISTRIBUTIONS:  Mean of the distribution:  Std. Dev. of the Distribution:  Calculator:

EXPECTED VALUE:  Same as discrete probability distribution but some outcomes may be negative.  Calculated same as mean of discrete probability distribution:  Example of raffle.  Example of lottery.  Example of insurance.

BINOMIAL PROBABILITY DISTRIBUTION (6.2):  Calculate the probability of rolling a die five times and getting a “1” exactly 2 times.  p=1/6, n=5, 1-p=5/6.  How many ways can this happen.  Do the same for 0, 1, 3, 4 and 5 times.

BINOMIAL PROBABILITY DISTRIBUTION:  Let n = number of trials p = probability of success on any given trial q = 1 – p = probability of failure on any given trial x = the desired number of successes out of n trials  Then probability of x successes from n trials

 REQUIREMENTS FOR AN EXPERIMENT TO BE A BINOMIAL PROBABITY PROCEDURE: Fixed number of trials (n); Fixed probability (p) of success for every trial; Each trial independent of all other trials; Each trial has only TWO possible outcomes. BINOMIAL PROBABILITY DISTRIBUTION:

 Example: n=8, p=0.85, find p(5). Use formula Use table Use calculator:  DISTR (2 nd, VARS)  0.binompdf(n,p,x)  Do also for x = 0, 1, 2, 3, 4, 6, 7, 8 and build discrete probability distribution; graph result.

BINOMIAL PROBABILITY DISTRIBUTION:  Do again with n=8, p=0.22  Build discrete probability distribution and graph.  Compare graphs.  Skew is result of mean (n*p)  As n increases the graph becomes more bell shaped.

BINOMIAL PROBABILITY DISTRIBUTION:  Mean of Binomial Distribution:  Std. Dev. Of Binomial Distribution:  Example

BINOMIAL PROBABILITY DISTRIBUTION:  Cumulative Probability:  Find p(x ≤ 2), p(x 2)  Use Calculator (always gives p(x ≤ c), DISTR (2 nd, VARS)  A.binomcdf(n,p,x)

BINOMIAL PROBABILITY DISTRIBUTION:  Cumulative Probability:  Use Calculator: (always gives p(x ≤ c) For p(x ≤ c)binomcdf(n,p,x) For p(x < c)binomcdf(n,p,x-1) For p(x ≥ c)1 - binomcdf(n,p,x-1) For p(x > c)1 - binomcdf(n,p,x) USE NUMBER LINE TO FIND X

BINOMIAL PROBABILITY DISTRIBUTION:  65% of population of city is Hispanic. A jury of 12 is selected.  Is this Binomial? Why or why not?  What is n, and p?  Find p(x=7); p(X=2); p(x≤6); p(x>6).  Would it be unusual to find a jury with only 2 Hispanics.  Find the mean and std. dev. of this distribution  Is it unusual to have less than 7 Hispanic jurors; how about 4?

POISSON DISTRIBUTION: (6.3)  Defined as the countable number of events in a fixed interval.  The number of defects in a square yard of material.  The number of Okapi’s in a square mile of rain forest.  The number speeding tickets issued in a day in a city.  Look at 8 x 11 inch paper with dots.

POISSON DISTRIBUTION:  REQUIREMENTS FOR X TO BE A POISSON RANDOM VARIABLE: The probability for two or more events in a sufficiently small interval is 0. The probability is the same for any two equal intervals. The number of successes in any interval is independent of the number of successes in any other interval, providing the intervals are not overlapping

POISSON DISTRIBUTION:  Let λ be the mean number of events per unit interval and t be the number of intervals.  Then the mean number of events in t intervals is t* λ = µ.  Then  Mean of the Poisson Distribution = µ  Std. Dev. of the Poisson Distribution =  Calculator: DISTR (2 nd, VARS)  B.poissonpdf(µ,x)

POISSON DISTRIBUTION:  In the Binomial Distribution the minimum value for x is 0 and the maximum is n.  In the Poisson distribution the minimum value of x is 0, but the maximum value is infinity.

POISSON DISTRIBUTION:  Example: 3 defects per sq. yd.; let n = 10 sq. yds.; Find P(x = 20)  Example: 1 ticket per day per mile; n = 4 miles; Find P(x = 3); also P(x ≤ 3).  Example: 6 Okapis per Sq. mile; let n = 10 sq. miles; Find P(x = 70); also P(x ≥ 70)