Chapter 5: Discrete Probability Distributions
5.1 Discrete Probability Distribution and Expected Value
Assume an insurance company sells a $10,000 one-year
life policy for $250. Also assume that the probability
of dying is .02. Then we have the discrete probability
distribution:
Event
Lives
Dies
X
250
-9750
P(X)
0.98
0.02
1.00
The expected value and variance
  E ( X )   ( XP( X ))
 2  E (( X   ) 2 )   (( X   ) 2 P( X ))
Learning Activity 5.1-1 Expected value
 Calculate the expected value and variance for the above
distribution 250*.98+(-9750)*.02=50($)
Open Prob2.xls!Discrete and calculate the expected value
and variance by using Excel.
If the insurance company sold 1000 policies and 2% of
the policyholders died, how much would the company
average per policy? Use Excel as a cacculator.
See Prob2.xls!solution2
5.2 Binomial Distribution
n x
P ( X | n, p )    p (1  p ) n  x
 x
where X  number of occurrences (0 to n)
n  number of trials
p  probabilit y of an occurrence on any trial
μ  np, σ 2  np(1  p)
Learning Activity 5.2-1 Binomial Distribution
Run MegaStat|Probability|
Discrete Probability Distribution|Binomial. Use n = 6, p = ½.
(What is the probability of observing 2 heads if a coin is
Tossed six time?)
 Calculate the expected value
What is the probability of a salesperson making in 4 or more
sales out of five sales calls if the probability of a sale on
any call is 0.25?
Use the MegaStat|Probability|Discrete Probability
Distributions|Binomial
Run MegaStat|Probability|Discrete Probability
Distributions|Binomial. Use n = 20, p = ¼.
What is the expected value for the distribution?
Use X and P(X) in your MegaStat output to calculate the
expected value.
DiscreteDist.xls shows the distribution for someone who is
guessing on a multiple-choice examination with 40 questions
and four choices for each question (n = 40, p = ¼).
Solution1: probability of exactly 14 answers correct.
Solution2: probability of getting 12 or fewer answers correct.
Solution3: probability of getting 8 or more answers correct.
Solution4: probability of getting 6 through 12 answers correct.
5.4 Hypergeometric Distribution
 S  N  S 
 

X  n  X 

P ( X | N , n, S ) 
N
 
n
N
n
S
X  number of occurrence
N  Populatio n size
S  possible number of occurrence in the population
n  sample size
S  N  n 
S 2
 S 
,


n
1


 


N
 N  N  N  1 
  n
The hypergeometric distribution is used to calculate the
probability of X occurrences in trials when sampling without
replacement from a finite population.
Learning Activity 5.4-1 Hypergeometric Distribution
 Open Hypergeometric.xls!sheet1 and run MegaStat|
Probability|Discrete Probability Distributions|Hypergeometric.
Use N = 12, S = 4, n = 3. Find the probability of X = 2.
 Use N = 1000, S = 50, n = 10. Find the probability of X = 1
or more by summing the probabilities of 1 through 10. Get the
same answer by 1 – P(0).
5.5 Poisson Distribution
The Poisson distribution calculates the probability of X
occurrences when there is a mean rate of occurrence.
e  X
P( X |  ) 
X!
where
X  number of occurrence
  mean rate of occurrence (mean)
 Open Poisson.xls!sheet1 and run MegaStat|Probability|
Poisson. Use  = 2.3. Find the probability of X = 2 or more
Comparing the binomial, hypergeometric and Poisson
distributions
Open comparison.xls!sheet1 and use
MegaStat|Probability|
Discrete Probability Distribution to run the following dists.
Binomial with n = 20, p = 0.05.
Hypergeometric with N = 960, S = 48, n = 20. Note that
S/N = 0.05 = p.
Poisson with  = 1. Note that  = 20*.05 = np.
Compare the 3 distributions. Look at the Solution and
Side_by _side worksheets.
Experiment to show that the binomial and hypergeometric
become more similar as N gets larger, e.g. with N=9600
and S=480.
Experiment to show that the binomial and Poisson become
more similar as n gets larger and p gets smaller, e.g. with
n = 100 and p = 0.01 and  = 1.
5.A Discrete distribution simulation
We can simulate the outcomes of the insurance company
policy example in Excel. That is we want a cell in a
worksheet to represent the outcome of an insurance policy
such that 98% of the time the cell contains 250 and 2% of
the time contains -9750.
Use =IF(RAND()<=0.02,-9750,250).
Rand() generates random numbers from 0 to 1.
The first argument of the IF() statement checks to see if the
random number value is less than or equal to .02. The
next two arguments determine what to put in the cell if the
logical condition is true or false, respectively.
See EV0.xls and EV1.xls.
Click a cell in EV1.xls and select 格式|設定格式化的條件