Download Lecture 8 Discrete Random variables

Chapter 4.
Discrete Random Variables
A random variable is a way of recording a
quantitative variable of a random experiment.
A variable which can take on only finitely
many different values is called discrete.
Example: The number of girls in a family of 8
children
Example: The number of seeds which
successfully germinate when 50 seeds are
planted
Continuous random variables
• A random variable which can take on any
value (ie, all values) in a certain interval is
called a continuous random variable.
• EX. The height in centimeters of a 16 year
old Canadian male.
• Ex. The dosage in ml. of a certain pain
killer
Example
• Let 3 coins be tossed and let x denote the
number of heads
• Possible values for x are 0, 1, 2, and 3,
• As done earlier, it is easy to compute
• Pr(0) = 1/8, Pr(3) = 1/8, and Pr(1) = Pr(2)
= 3/8
• Notation: We will also use the notation
• P(x = 0) = 1/8, and so on.
Properties of Probability,
P( X = xi )
(1)
0  P ( X  xi )  1
n
(2)
 P( X  x )  1
i 1
i
Definition
Copyright © 2013
Pearson Education, Inc..
Example
Graph the probability distribution of the
random variable obtained by flipping an
unbiased coin two times and counting the
number of times heads comes up.
Solution
• Possible values of x are 0, 1, 2, and a
quick check shows P(0) = ¼, P(1) = 1/2,
and P(2) = ¼.
Probability distribution for a two-coin toss
Copyright © 2013
Pearson Education, Inc..
Definition
Copyright © 2013
Pearson Education, Inc..
Definition
Copyright © 2013
Pearson Education, Inc..
Definition
Copyright © 2013
Pearson Education, Inc..
Procedure
Copyright © 2013
Pearson Education, Inc..
Figure 4.6 Shapes of two probability
distributions for a discrete random variable x
Copyright © 2013
Pearson Education, Inc..
Example
• Medical research shows a certain type of
chemotherapy is successful 70% of the
time. Suppose 5 patients are treated and
let x denote the number of successes.
One can show
•
x
0
1
2
3
4
5
P(x)
.002
.029
.132
.309
.360
.168
Graph of p(x)
Copyright © 2013
Pearson Education, Inc..
Find the mean and interpret
• Applying the formulae we obtain
• Mean = 3.50
• Consider a large number of trials, each
consisting of treating 5 patients. On
average, the number of successes will be
3.5. Thus, if 200 trials each of 5 patients is
conducted, we would expect and average
of 3.5 successes per trial for a total of 700
successes for 1000 patients
Find the standard deviation and
interpret
• Using the formula you can check that
• Standard deviation = 1.02.
• Using Empirical Rule, would expect that
approximately 68% of times the trial is
repeated, the outcome will be between
3.5-1.02 and 3.5+1.02, i.e., will lie in the
interval [2.48, 4.52].
• What is the actual percentage of times the
outcome will be in that interval?
Binomial Experiment
A binomial experiment is one that:
1) Has a fixed number of trials (n)
2) These trials are independent
3) Each trial must have all outcomes
classified into two categories (Success
or Failure)
4) The probability of success remains
constant for all trials.
Notation:
•
•
•
•
S = success and P(S) = p
F = Failure and P(F) = q = 1- p
n = fixed number of trials
x = specific number of successes in n
trials
• P(x) = the probability of getting exactly x
successes among n trials
Factorials
0! = 1
1! = 1
2! = 2 * 1
3! = 3 * 2 * 1
4! = 4* 3 * 2 * 1
n! = n*(n-1)!
Factorials
0! = 1
1! = 1
2! = 2 * 1=2
3! = 3 * 2 * 1=6
4! = 4* 3 * 2 * 1=24
n! = n*(n-1)!
Binomial Probability Distribution
In a binomial experiment, with constant
probability p of success at each trial, the
probability of x successes in n trials is given
by
n!
x n x
P( x successes ) 
p q
(n  x)! x!
Example
Shaq is a basketball player who takes a lot
of free throws. The probability of Shaq
making a free throw is 0.60 on each throw.
With 3 free throws what is the probability
that he makes 2 shots?
Example
Shaq is a basketball player who takes a lot
of free throws. The probability of Shaq
making a free throw is 0.60 on each throw.
With 3 free throws what is the probability
that he makes 2 shots?
3!
2
3 2
P( x  2) 
(.6) (.4)
(3  2)!2!
 0.432
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
 1  P( x  1)  P( x  0)
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
 1  P( x  1)  P( x  0)
8! 1 7 8!
0
8
 1  (.4) (.6) 
(.4) (.6)
7!1!
8! 0!
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
 1  P( x  1)  P( x  0)
8! 1 7 8!
0
8
 1  (.4) (.6) 
(.4) (.6)  .894
7!1!
8! 0!
Table 4.4
Copyright © 2013
Pearson Education, Inc..
How to use the Binomial
Tables
• First find the appropriate table for the particular value
of n
• then find the value of p in the top row
• Find the row corresponding to k and find the
intersection with the column corresponding to the
value of p
• The value you obtain is the cumulative probability,
that is P(x ≤ k)
• N=10, p = 0.7: P(x ≤ 4) = 0.047
• N=10, p = 0.7: P(x = 4) = P(x ≤ 4) - P(x ≤ 3)
= 0.047-0.011=0.036
• N=10, p = 0.7: P(x > 4) = 1- P(x ≤ 4)
= 1 - 0.047 = 0.953
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
Example
Flipping a biased coin 8 times. The
probability of heads on each trial is 0.4.
What is the probability of obtaining at least 2
heads.
P( x  2)  P( x  2)  P( x  3)  ...  P( x  8)
 1  P( x  1)
 1  0.106
 .894
Mean and Standard deviation
  np   npq
q  1 p
Keys to success
Learn the binomial table.
Be able to recognize binomial distributions
and when you do apply the appropriate
formulas and tables.