Transcript Chapter 5,Discrete Probability Distribution

Discrete Probability Distribution
• Probability Distribution of a Random Variable:
Is a table,
graph, or mathematical expression that specifies all possible values (outcomes) of a
random variable along with their respective probabilities.
Random
Variables
Ch.5
Probability
Distribution of a
Discrete
Random Variable
Probability
Distribution of a
Continuous
Random Variable
Ch.6
• A discrete probability distribution applied to countable values
(That is to a random variables resulting from counting, not
measuring)
• Example: Using the records for past 500 working days, a
manager of auto dealership summarized the number of cars
sold per day and the frequency of each number sold.
• Questions:
–
–
–
–
–
What is the average number of cars sold per day?
What is the dispersion of the number of cars sold per day?
What is the probability of selling less than 4 cars per day?
What is the probability of selling exactly 4 cares per day?
What is the probability of selling more than 4 cars per day?
• To answer these questions, we need the mean and the standard
deviation of the distribution.
• Expected Value (or mean) of a discrete distribution (Weighted Average)
N
  E(X)   X i *P( X i )
i 1
• Variance of a discrete random variable
N
2
σ   [X  E(X)] 2 * P(X )
i
i
i 1
• Standard Deviation of a discrete random variable
σ  σ2 
N
2
 [X i  E(X)] * P(X i )
i 1
where:
E(X) = Expected value of the discrete random variable X=Mean
Xi = the ith outcome of the variable X
P(Xi) = Probability of the Xi occurrence
Number of Cars Sold, X
Frequency
P(X)
X*P(X)
[X-E(X)]^2
[X-E(X)]^2*P(X)
0
40
0.08
0
9.339136
0.74713088
1
100
0.2
0.2
4.227136
0.8454272
2
142
0.284
0.568
1.115136
0.316698624
3
66
0.132
0.396
0.003136
0.000413952
4
36
0.072
0.288
0.891136
0.064161792
5
30
0.06
0.3
3.779136
0.22674816
6
26
0.052
0.312
8.667136
0.450691072
7
20
0.04
0.28
15.555136
0.62220544
8
16
0.032
0.256
24.443136
0.782180352
9
14
0.028
0.252
35.331136
0.989271808
10
8
0.016
0.16
48.219136
0.771506176
11
2
0.004
0.044
63.107136
0.252428544
Total 500
Mean =
3.056
Variance =
Std Dev =
6.068864
2.463506444
X is the number of cars sold per day; P(X) is the probability that that many are sold per
day.
• NOTE: The usefulness of the Table
• P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= (0.08 + 0.2 + 0.284 + 0.132) = 0.696
• P(X

4) = P(X < 4) + P(X = 4) = 0.696 + 0.072 = 0.768
• P(X  4) = 1 – P(X < 4) = 1 – 0.696 = 0.304
• P(X = 4) = 0.072
• P(X > 4) = 1 – P(X
 4)
• Go to handout example
= 1 – 0.768 = 0.232
Binomial Probability Distribution
(a special discrete distribution)
• Characteristics
• A fixed number of identical observations, n. Each observation is drawn
from:
– Infinite population without replacement or
– Finite population with replacement
• Two mutually exclusive (?) and collectively exhaustive (?) categories
– Generally called “success” and “failure”
– Probability of success is p, probability of failure is (1 – p)
• Constant probability for each outcome from one observation to observation
over all observations.
• Observations are independent from each other
– The outcome of one observation does not affect the outcome of the other
Binomial Distribution has many application in business
Examples:
–
•
A firm bidding for contracts will either get a contract or not
•
A manufacturing plant labels items as either defective or
acceptable
•
A marketing research firm receives survey responses of “yes I
will buy” or “no I will not”
•
New job applicants either accept the offer or reject it
•
An account is either delinquent or not
Example: Suppose 4 credit card accounts are examined for over the
limit charges. Overall probability of over the limit charges is known
to be 10 percent (one out of every 10 accounts).
Let,
p = probability of “success” in one trial or observation
n = sample size (number of trials or observations)
X = number of ‘successes’ in sample, (X = 0, 1, 2, ..., n)
P(X) = probability of X successes in n trials, with probability of success p on
each trial
Then,
1.
P(X success in a particular sequence (or order) =
2.
Number of possible sequences (or orders) 
Where
3.
n! =n(n - 1)(n - 2) . . . (2)(1)
X! = X(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)
p X ( 1  p )n X
n!
X ! ( n  X )!
P(X success regardless of the sequence or order)
P( X ) 
n!
p X ( 1  p )n X
X ! ( n  X )!
Back to our example:
•
What is the probability of 3 account being over the limit with the following
order? OL,OL,Not OL, and OL.
2.
How many sequences (order) of 3 over the limit are possible?
3.
What is the probability of 3 accounts being over the limit in all possible
orders? (all Possible sequences)
Solutions: Calculate 1, 2, and 3.
Characteristics of a Binomial Distribution
1.
2.
3.
4.
5.
6.
7.
8.
9.
1.
For each pair of n and p a particular probability distribution can be
generated.
The shape of the distribution depends on the values of p and n.
If p=0.5, the distribution is perfectly symmetrical
If p< 0.5, the distribution is right skewed
If p>0.5, the distribution is left skewed
The closer p to 0.5 and the larger the sample size, n, less skewed the
distribution
The mean of the distribution = μ  E(x)  np
The standard deviation = σ  np(1 - p)
Or you can download the binomial table on the text- book’s companion
Web site. (You need to learn how to use this table for the test).
http://wps.prenhall.com/wps/media/objects/9431/9657451/Ch_05/levine-smume6_topic_BINO.pdf