Discrete Random Variables&Prob dist (4.0).ppt

statistics

1. © 2011 Pearson Education, Inc
Statistics for Business and
Economics

Random Variables &
Probability Distributions

2. © 2011 Pearson Education, Inc
Content
1.Two Types of Random Variables
2.Probability Distributions for Discrete
Random Variables
3.The Binomial Distribution
4.Poisson and Hypergeometric Distributions
5.Probability Distributions for Continuous
Random Variables
6.The Normal Distribution

3. © 2011 Pearson Education, Inc
Content (continued)
7.Descriptive Methods for Assessing
Normality
8.Approximating a Binomial Distribution
with a Normal Distribution
9.Uniform and Exponential Distributions
10.Sampling Distributions
11.The Sampling Distribution of a Sample
Mean and the Central Limit Theorem

4. © 2011 Pearson Education, Inc
Learning Objectives
1.Develop the notion of a random variable
2.Learn that numerical data are observed values of
either discrete or continuous random variables
3.Study two important types of random variables
and their probability models: the binomial and
normal model
4.Define a sampling distribution as the probability of
a sample statistic
5.Learn that the sampling distribution of follows a
normal model
x

5. © 2011 Pearson Education, Inc
Thinking Challenge
•You’re taking a 33 question
multiple choice test. Each
question has 4 choices. Clueless
on 1 question, you decide to
guess. What’s the chance you’ll
get it right?
•If you guessed on all 33
questions, what would be your
grade? Would you pass?

6. © 2011 Pearson Education, Inc
4.1
Two Types of Random Variables

7. © 2011 Pearson Education, Inc
Random Variable
A random variable is a variable that assumes
numerical values associated with the random
outcomes of an experiment, where one (and only
one) numerical value is assigned to each sample
point.

8. © 2011 Pearson Education, Inc
Discrete
Random Variable
Random variables that can assume a countable
number (finite or infinite) of values are called
discrete.

9. © 2011 Pearson Education, Inc
Discrete Random Variable
Examples
Experiment
Random
Variable
Possible
Values
Count Cars at Toll
Between 11:00 & 1:00
# Cars
Arriving
0, 1, 2, ..., ∞
Make 100 Sales Calls# Sales 0, 1, 2, ..., 100
Inspect 70 Radios # Defective0, 1, 2, ..., 70
Answer 33 Questions# Correct0, 1, 2, ..., 33

10. © 2011 Pearson Education, Inc
Continuous
Random Variable
Random variables that can assume values
corresponding to any of the points contained in
one or more intervals (i.e., values that are
infinite and uncountable) are called continuous.

11. © 2011 Pearson Education, Inc
Continuous Random Variable
Examples
Measure Time
Between Arrivals
Inter-Arrival
Time
0, 1.3, 2.78, ...
Experiment
Random
Variable
Possible
Values
Weigh 100 People Weight 45.1, 78, ...
Measure Part Life Hours 900, 875.9, ...
Amount spent on food$ amount54.12, 42, ...

12. © 2011 Pearson Education, Inc
4.2
Probability Distributions for
Discrete Random Variables

13. © 2011 Pearson Education, Inc
Discrete
Probability Distribution
The probability distribution of a discrete
random variable is a graph, table, or formula
that specifies the probability associated with each
possible value the random variable can assume.

14. © 2011 Pearson Education, Inc
Requirements for the
Probability Distribution of a
Discrete Random Variable x
1. p(x) ≥ 0 for all values of x
2. p(x) = 1
where the summation of p(x) is over all possible
values of x.

15. © 2011 Pearson Education, Inc
Discrete Probability
Distribution Example
Probability Distribution
Values, x Probabilities, p(x)
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
Experiment: Toss 2 coins. Count number of
tails.
© 1984-1994 T/Maker Co.

16. © 2011 Pearson Education, Inc
Visualizing Discrete
Probability Distributions
Listing Table
Formula
# Tails
f(x)
Count
p(x)
0 1 .25
1 2 .50
2 1 .25
px
n
x!(n – x)!
()
!
= p
x
(1 – p)
n – x
Graph
.00
.25
.50
0 1 2
x
p(x)
{ (0, .25), (1, .50), (2, .25) }

17. © 2011 Pearson Education, Inc
Summary Measures
1.Expected Value (Mean of probability distribution)
•Weighted average of all possible values
•  = E(x) = x

p(x)
2.Variance
• Weighted average of squared deviation about
mean
• 
2
= E[(x 

(x 

p(x)
3. Standard Deviation
2
 ●

18. © 2011 Pearson Education, Inc
Summary Measures
Calculation Table
xp(x)x p(x)x – 
Total (x 

p(x)
(x – 

(x – 

p(x)
xp(x)

19. © 2011 Pearson Education, Inc
Thinking Challenge
You toss 2 coins. You’re
interested in the number of
tails. What are the expected
value, variance, and
standard deviation of this
random variable(is it
Discrete or Continuous?),
number of tails?
© 1984-1994 T/Maker Co.

20. © 2011 Pearson Education, Inc
Expected Value & Variance
Solution*
0.25 –1.001.00
1.50 0 0
2.25 1.001.00
0
.50
.50
 = 1.0
xp(x)x p(x)x –  (x – 

(x – 

p(x)
.25
0
.25







21. © 2011 Pearson Education, Inc
Probability Rules for Discrete
Random Variables
Let x be a discrete random variable with probability
distribution p(x), mean µ, and standard deviation .
Then, depending on the shape of p(x), the following
probability statements can be made:
Chebyshev’s Rule Empirical Rule
Pxxµ  0 .68
Px2xµ2 
3
4
.95
Px3xµ3 
8
9
1.00

22. © 2011 Pearson Education, Inc
The Binomial Distribution

23. © 2011 Pearson Education, Inc
Binomial Distribution
Number of ‘successes’ in a sample of n
observations (trials)
•Number of reds in 15 spins of roulette wheel
•Number of defective items in a batch of 5 items
•Number correct on a 33 question exam
•Number of customers who purchase out of 100
customers who enter store (each customer is
equally likely to purchase)

24. © 2011 Pearson Education, Inc
Binomial Probability
Characteristics of a Binomial Experiment
1.The experiment consists of n identical trials.
2.There are only two possible outcomes on each trial. We
will denote one outcome by S (for success) and the other
by F (for failure).
3.The probability of S remains the same from trial to trial.
This probability is denoted by p, and the probability of
F is denoted by q. Note that q = 1 – p.
4.The trials are independent.
5.The binomial random variable x is the number of S’s in
n trials.

25. © 2011 Pearson Education, Inc
Binomial Probability
Distribution
!
( ) (1 )
!( )!
x n x x n x
n n
p x p q p p
x x n x
  
  
 
 
p(x) = Probability of x ‘Successes’
p=Probability of a ‘Success’ on a single trial
q=1 – p
n=Number of trials
x=Number of ‘Successes’ in n trials
(x = 0, 1, 2, ..., n)
n – x=Number of failures in n trials

26. Find the Combination (5,3)
•
5C3 = ?
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27. nC
x 
•Combination
5C
3 
= 
5!/[3!(5-3)!]=10
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28. © 2011 Pearson Education, Inc
Binomial Probability
Distribution Example
3 5 3
!
( ) (1 )
!( )!
5!
(3) .5 (1 .5)
3!(5 3)!
.3125
x n xn
p x p p
x n x
p


 

 


Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3 tails?
© 1984-1994 T/Maker Co.

29. © 2011 Pearson Education, Inc
Binomial Probability Table
(Portion)
n = 5 p
k .01 … 0.50… .99
0 .951… .031… .000
1 .999… .188… .000
21.000… .500… .000
31.000… .812… .001
41.000… .969… .049
Cumulative Probabilities
p(x ≤ 3) – p(x ≤ 2) = .812 – .500 = .312

30. © 2011 Pearson Education, Inc
Binomial Distribution
Characteristics(Analyze Graphs)
.0
.5
1.0
012345
X
P(X)
.0
.2
.4
.6
012345
X
P(X)
n = 5 p = 0.1
n = 5 p = 0.5

E(x)np
Mean
Standard Deviation

npq

31. © 2011 Pearson Education, Inc
Binomial Distribution
Thinking Challenge
You’re a telemarketer selling service
contracts for Macy’s. You’ve sold 20
in your last 100 calls (p = .20). If you
call 12 people tonight, what’s the
probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?

32. © 2011 Pearson Education, Inc
Binomial Distribution Solution*
n = 12, p = .20
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2)= p(0) + p(1) + p(2)
= .0687 + .2062 + .2835
= .5584
D. p(at least 2)= p(2) + p(3)...+ p(12)
= 1 – [p(0) + p(1)]
= 1 – .0687 – .2062
= .7251

33. •Exercise Time!
© 2011 Pearson Education, Inc

34. © 2011 Pearson Education, Inc
Properties of Variances:
If a random variable X 
is adjusted by multiplying by the value 
b 
and adding the value
 
a, then the variance is affected as follows:

Since the spread of the distribution is not affected
by adding or subtracting a constant,
the value
 
a 
is not considered.
And, since the variance is a sum of squared terms,
any multiplier value
 
b 
must also be squared
when adjusting the variance.

35. © 2011 Pearson Education, Inc
For independent random variables
 
X 
and 
Y,
the variance of their sum or difference is the sum of their variances:

Variances are added for both the sum
 
and 
difference of two independent
random variables because the variation in each variable contributes
to the variation in each case.

36. © 2011 Pearson Education, Inc
4.4
Other Discrete Distributions:
Poisson and Hypergeometric

37. © 2011 Pearson Education, Inc
Poisson Distribution
1.Number of events that occur in an interval
• events per unit
— Time, Length, Area, Space
2.Examples
• Number of customers arriving in 20 minutes
• Number of strikes per year in the U.S.
• Number of defects per lot (group) of DVD’s

38. © 2011 Pearson Education, Inc
Characteristics of a Poisson
Random Variable
1.Consists of counting number of times an event
occurs during a given unit of time or in a given
area or volume (any unit of measurement).
2.The probability that an event occurs in a given unit
of time, area, or volume is the same for all units.
3.The number of events that occur in one unit of
time, area, or volume is independent of the number
that occur in any other mutually exclusive unit.
4.The mean number of events in each unit is denoted
by 

39. © 2011 Pearson Education, Inc
Poisson Probability
Distribution Function




p(x) = Probability of x given 
=Mean (expected) number of events in unit
e=2.71828 . . . (base of natural logarithm)
x=Number of events per unit
px
x
()
!

x


e
–
(x = 0, 1, 2, 3, . . .)

40. © 2011 Pearson Education, Inc
Poisson Probability
Distribution Function
.0
.2
.4
.6
.8
012345
X
P(X)
.0
.1
.2
.3
X
P(X)
= 0.5
= 6
Mean
Standard Deviation
 
E(x)

41. © 2011 Pearson Education, Inc
Poisson Distribution Example
Customers arrive at a
rate of 72 per hour.
What is the probability
of 4 customers arriving
in 3 minutes?
© 1995 Corel Corp.

42. © 2011 Pearson Education, Inc
Poisson Distribution Solution
72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

-
4
-3.6
( )
!
3.6
(4) .1912
4!
x
e
px
x
e
p



 

43. © 2011 Pearson Education, Inc
Poisson Probability Table
(Portion)
x
 0 … 3 4 … 9
.02 .980 …
: : : : : : :
3.4 .033 … .558 .744 … .997
3.6 .027 … .515 .706 … .996
3.8 .022 … .473 .668 … .994
: : : : : : :


Cumulative Probabilities
p(x ≤ 4) – p(x ≤ 3) = .706 – .515 = .191

44. © 2011 Pearson Education, Inc
Thinking Challenge
You work in Quality Assurance
for an investment firm. A clerk
enters 75 words per minute
with 6 errors per hour. What is
the probability of 0 errors in a
255-word bond transaction?
© 1984-1994 T/Maker Co.

45. © 2011 Pearson Education, Inc
Poisson Distribution Solution:
Finding *
•75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
• 6 errors/hr= 6 errors/4500 words
= .00133 errors/word
•In a 255-word transaction (interval):
 = (.00133 errors/word )(255 words)
= .34 errors/255-word transaction

46. © 2011 Pearson Education, Inc
Poisson Distribution Solution:
Finding p(0)*

-
0
-.34
( )
!
.34
(0) .7118
0!
x
e
p x
x
e
p



 

47. © 2011 Pearson Education, Inc
Characteristics of a
Hypergeometric
Random Variable
1.The experiment consists of randomly drawing n
elements without replacement from a set of N
elements, r of which are S’s (for success) and (N –
r) of which are F’s (for failure).
2.The hypergeometric random variable x is the
number of S’s in the draw of n elements.

48. © 2011 Pearson Education, Inc
Hypergeometric Probability
Distribution Function
where . . .
[x = Maximum [0, n – (N – r), …,
Minimum (r, n)]
px
r
x






Nr
nx






N
n






µ
nr
N

2

rNrnNn
N
2
N1

49. © 2011 Pearson Education, Inc

50. © 2011 Pearson Education, Inc
Hypergeometric Probability
Distribution Function
N = Total number of elements
r = Number of S’s in the N elements
n = Number of elements drawn
x = Number of S’s drawn in the n elements

51. © 2011 Pearson Education, Inc

52. © 2011 Pearson Education, Inc

53. © 2011 Pearson Education, Inc

54. © 2011 Pearson Education, Inc

55. Exercise 2:
•Assume we have an urne with 10 balls.
•4 of these balls are red
•6 of the balls are black
•We select 5 balls from a total of 10.
•Calculate the probability that we get 1 red ball
and 4 black balls.
© 2011 Pearson Education, Inc

56. •P (X= 0, 1, 2, 3, 4)  Maximum 4 red balls.
•N = 10 (Total number of balls in my sample)
•n = 5 ( number of balls selected)
•r = number of balls of the special kind in N (Red)
•X = is in the question (what we want to calculate).
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57. •P (X = 1) = 1 red ball and 4 black balls
•P(X = 1) = (4 choose 1) (6 choose 4) / (10
choose 5)
•Final answer = to be determined
© 2011 Pearson Education, Inc

58. © 2011 Pearson Education, Inc
4.5
Probability Distributions for
Continuous Random Variables

59. © 2011 Pearson Education, Inc
Continuous Probability
Density Function
The graphical form of the probability distribution for a
continuous random variable x is a smooth curve

60. © 2011 Pearson Education, Inc
Continuous Probability
Density Function
This curve, a function of x, is denoted by the symbol f(x)
and is variously called a probability density function
(pdf), a frequency function, or a probability
distribution.
The areas under a probability
distribution correspond to
probabilities for x. The area A
beneath the curve between two
points a and b is the probability
that x assumes a value betweena and b.