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.
x
P(x)
x
P(x)
2
1/36
8
5/36
3
2/36
9
4/36
4
3/36
10
3/36
5
4/36
11
2/36
6
5/36
12
1/36
7
6/36
Will do this together.
DISCRETE PROBABILITY
DISTRIBUTIONS:

Make up of Jury in area that is 65%
Hispanic.
x
P(x)
X
P(x)
0
0.000+
7
0.204
1
0.000+
8
0.237
2
0.001
9
0.195
3
0.005
10
0.109
4
0.020
11
0.037
5
0.059
12
0.006
6
0.128
DISCRETE PROBABILITY
DISTRIBUTIONS:



Mean of the
distribution:
Std. Dev. of the
Distribution:
    x * p ( x) 



x 2 * p ( x)   2
Calculator:
x  L1
p( x)  L 2
STATS  CALC  1  VARSTATS _ L1, L 2
EXPECTED VALUE:


Same as discrete probability distribution
but some outcomes may be negative.
Calculated same as mean of discrete
probability distribution:
E.V .  outcome * p(outcome)



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
p( x)  n Cx p q
x
(n x )
BINOMIAL PROBABILITY
DISTRIBUTION:

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 (2nd, 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:
  n* p
  n* p*q

Example
BINOMIAL PROBABILITY
DISTRIBUTION:



Cumulative Probability:
0
1
0.2
0.3
2
0.4
3
0.1
Find
p(x ≤ 2), p(x < 2), p(x ≥ 2), p(x > 2)
Use Calculator (always gives p(x ≤ c),
DISTR (2nd, VARS)A.binomcdf(n,p,x)
BINOMIAL PROBABILITY
DISTRIBUTION:


Cumulative Probability:
Use Calculator:
(always gives p(x ≤ c)
For p(x ≤ c)
 For p(x < c)
 For p(x ≥ c)
 For p(x > c)
USE NUMBER LINE

binomcdf(n,p,x)
binomcdf(n,p,x-1)
1 - binomcdf(n,p,x-1)
1 - binomcdf(n,p,x)
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
x

p ( x) 



 *e
x!
Mean of the Poisson Distribution = µ
Std. Dev. of the Poisson Distribution = 
Calculator: DISTR (2nd, 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)