Download Chapter 5. Discrete Probability Distributions

Chapter 5. Discrete Probability Distributions
Chapter Problem: Did Mendel’s result from plant hybridization
experiments contradicts his theory?
1. Mendel’s theory says that when there are two inheritable traits,
one of them will be dominant and the other will be recessive.
2. Experiment using pea plants. Green dominant, yellow recessive.
3. P(green pod) = ¾ = 0.75.
4. 580 offspring, 428 with green pods, 428/580 = 0.738.
Gene from
Parent 1
green
green
yellow
yellow
+
+
+
+
Gene from
Parent 2
green
yellow
green
yellow




Offspring
Genes
green/green
green/yellow
yellow/green
yellow/yellow
Color of
Offspring Pod
 green
 green
 green
 yellow
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5.1 Review and Preview
Combine the descriptive statistics presented in chapter 2
and 3 and those of probability present in chapter 4.
x f
Chapters
2 and 3
Collect sample
data, then
get statistics
and graphs
1
2
3
4
5
6
8
10
11
12
13
14
x = 3.6
s = 1.7
P(1) = 1/6
Chapters 4
Find the
probability for
Each outcome
and graphs
P(2) = 1/6
P(3) = 1/6
P(4) = 1/6
P(5) = 1/6
P(6) = 1/6
chapter 5
Create a theoretical model
describing how the
experiment is expected to
behave, then get its
parameter.
x
1
2
3
4
5
6
P(x)
1/6
1/6
1/6
1/6
1/6
1/6
 = 3.5
 = 1.7
Figure 5 – 1 Combining Descriptive Methods
and Probability to Form a Theorem model of Behavior
2
5.2 Random Variables
1.
2.
3.
4.
5.
6.
7.
Related Concepts
Graphs
Mean, Variance, and Standard Deviation
Rationale for Formula 5-1 through 5-5
Identifying Unusual Results with the Range Rule of Thumb
Identifying Unusual Results with Probabilities
Expected Value
3
5.2 Random Variables
1. Related Concepts
Definition
A random variable is a variable (typically represented by x)
that has a single numerical value, determined by chance, for
each outcome of a procedure
A probability distribution is a description that gives the
probability for each value of the random variable. It is often
expressed in the format of a graph, table, or formula
4
5.2 Random Variables
1. Related Concepts
e.g.1 Genetics Consider the offspring of peas from parents both
having the green/yellow combination of pod genes. Under these
combinations, the probability that the offspring has a green pod
is ¾ or 0.75. If 5 such offspring is obtained, and let
x = number of peas with green pods among 5 offspring peas
then x is a random variable. Table 5-1 is a probability
distribution. (5-3 will tell how the values of P(x) is obtained)
x ( Number of Peas with Green Pods)
P(x)
0
1
0.001 0.015
2
3
4
0.088 0.264 0.396
5
0.237
Table 5-1. Probability Distribution: Probabilities of Numbers of Peas with
Green Pods Among 5 Offspring Peas
5
5.2 Random Variables
1. Related Concepts
Definition.
A discrete random variable has either a finite number of
values or a countable number of values
A continuous random variable has infinitely many values,
those values can be associated with measurements on a
continuous scale without gaps or interruptions (voltmeter).
6
5.2 Random Variables
1. Related Concepts
e.g.2 Determine each variable as discrete or continuous.
Discrete
1). x = the # of eggs that a hen lays in a day
2). x = the # of stat students present in class on a given day Discrete
3). x = the amount of milk a cow produces Continuous
4). x = the measure of voltage for a particular smoke detector
Continuous
7
5.2 Random Variables
2. Graphs
Probability Histogram.
Probability
0.4
0.3
Figure 5-3 Probabilities Histogram
0.2
0.1
0
1
2
3
4 5
Number of Peas with
Green Pods Among 5
8
5.2 Random Variables
2. Graphs
Requirement for a Probability Distribution
1.  P(x) = 1, where x assumes all possible values (the sum
of all probabilities must be 1)
2. 0 ≤ P(x) ≤ 1, for every individual value of x, (that is, each
probability value must be between 0 and 1 inclusive.)
Example 3. Is Table 5-2 a probability distribution?
x
0
1
2
3
No
P(x)
0.19
0.26 0.33
0.13
Table 5-2 Cell Phones per Household
9
5.2 Random Variables
2. Graphs
Example 4. P(x) = x/10, where x = 0, 1, 2, 3, 4. Is P a probability
distribution?
Yes
3. Mean, Variance, and Standard Deviation
Formula 5-1
   x  P(x)

Mean for a probability distribution

Formula 5-2  2   ( x   ) 2  P( x) Variance for a probability distribution


Formula 5-3  2   x 2  P( x)   2 Variance for a probability distribution
Formula 5-4  
 x
2

 P( x)   2 Standard Deviation for a probability
distribution
10
5.2 Random Variables
4.
Rationale for Formula 5-1 through 5-4
•
Consider mean for example. From Formula 3-2, section 3-2, we
have
 f  x
 f  x




  x
N
•
•
•
 N 




f 
  x  P(x)

N
Similarly, one can derive 5-2 (using formula on page 103 for
Standard Deviation).
Formula 5-3 is derived from Formula 5-2
5-4 from 5-3 directly.
11
5.2 Random Variables
4.
Rationale for Formula 5-1 through 5-4
Round-off Rule for  ,  , and 
Round results by carrying one more decimal place than the
number of decimal places used for the random variable x.
If the value of x are integers, round  ,  , and  2 to one
decimal place
2
12
5.2 Random Variables
5.
Identifying Unusual Results with the Range Rule of Thumb
e.g.5 Continue use the Chapter example.
x
0
1
2
3
4
5
P(x)
0.001
0.015
0.088
0.264
0.396
0.237
x  P(x)
0.000
0.015
0.176
0.792
1.584
1.185
x2
0
4
9
16
25
0.352
2.376
6.336
5.925
x2  P(x)
1
0.000 0.015
sum = 3.752
sum = 15.004
Table 5-3 Calculating  ,  , and  2 for a Probability Distribution
 = 3.752  3.8
 2 = 15.004 – 3.7522 = 0.926496  0.9
  0.926496  0.9625  1.0
13
5.2 Random Variables
5.
Identifying Unusual Results with the Range Rule of Thumb
e.g.6 Identifying unusual results with the range rule of thumb.
By Range Rule of Thumb
Maximum usual value =  + 2 = 3.8 + 2(1.0) = 5.8
Minimum usual value =   2 = 3.8  2(1.0) = 1.8
Interpretation.
Based this result, we conclude that for groups of 5 offspring peas,
the number of offspring peas with green pods should usually
fall between 1.8 and 5.8. If 5 offspring peas are generated as
described, it would be unusual to get only 1 with a green pod.
14
5.2 Random Variables
6.
Identifying Unusual Results with Probabilities
Rare Event Rule
1
If, under a given assumption (such as the assumption that a
2
coin is fair), the probability of a particular observed event
(such as 992 heads in 1000 tosses of a coin) is extremely
3
small, we conclude that the assumption is probably not
correct.
15
5.2 Random Variables
6.
Identifying Unusual Results with Probabilities
Probabilities can be used to apply the rare event rule as follows:
Using Probabilities to Determine When Results Are Unusual
• Unusually high number of success: x successes among n trials is an unusually
high number of successes if P(x or more)  0.05.*
e.g.
9 success in 10 trials is high if P(9 or more successes)  0.05
• Unusually low number of success: x successes among n trials is an unusually
low number of successes if P(x or fewer)  0.05.*
e.g.
2 success in 10 trials is low if P(2 or fewer successes)  0.05
* The value of 0.05 is commonly used, but is not absolutely rigid.
16
5.2 Random Variables
6.
Identifying Unusual Results with Probabilities
• Let’s flip a coin to see it favors heads, 1000 tosses resulted in 501
heads.
• This is not an evidence that the coin favors heads
• P(501 heads in 1000 flips) = 0.0252, a very low probability.
• However, P(at least 501 heads in 1000 flips) = 0.487, a high
probability
17
5.2 Random Variables
6.
Identifying Unusual Results with Probabilities
e.g.7. Identifying Unusual Results with Probabilities Use
probability to determine whether 1 is an unusually low number
of peas with green pods when 5 offspring are generated from
parents both having the green/yellow pair of genes.
Solution. P( 1 or fewer ) = P(0 or 1)
= P(0) + P(1)
= 0.001 + 0.015 + = 0.016 < 0.05
Interpretation: the result of 1 pea with a green pod is unusually low.
There is a very small likelihood (0.016) of generating 1 or
fewer peas with green pods.
18
5.2 Random Variables
7.
Expected Value
Definition – the expected value of a discrete random variable is
denoted by E, and it represents the average value of a the
outcomes. It is obtained by finding the value of  x  P(x) ,
E =  x  P(x)
From formula 5-1, we see that E = . Thus the expected value of the
number of peas with green pods is also 3.8.
19
5.2 Random Variables
7. Expected Value
e.g.8 How to be a Better Bettor. You are considering placing a bet
either on the number 7 in a roulette or on the “pass line” in the
dice game of craps at the Venetian casino in Las Vegas.
a. If you bet $5 on the number 7 in roulette, the probability of losing
$5 is 37/38 and the probability of making a net gain of $175 is
1/38. (the prize is $180, including your $5 bet, so the net gain is
$175.) Find your expected value if you bet $5 on the number 7
in roulette.
Table 5-4 Roulette
Event
Lose
Gain (net)
Total
x
P(x)
–$5
37/38
$175
1/38
xP(x)
–$4.87
$4.61
E = –$0.26 (or –26 ¢)
20
5.2 Random Variables
7. Expected Value
e.g.8 How to be a Better Bettor. Continue
b. If you bet $5 on the pass line in the dice game of craps, the
probability of losing $5 is 251/495 and the probability of
making a net gain of $5 is 244/495. (If you bet $5 on the pass
line and win, you are given $10 that includes your bet, so the
net gain is $5). Find your expected value if you bet on the pass
line.
which one is better: a $5 on the number 7 in roulette or a $5 on
the pass line in the dice game? Why?
Table 5-5 Dice
Event
Lose
Gain (net)
Total
x
P(x)
–$5
251/495
$5
244/495
xP(x)
–$2.54
$2.46
E = –$0.08 (or –8 ¢)
21
5.2 Random Variables
7.
Expected Value
Interpretation. The $5 bet in the roulette results in an expected value
of –26¢ and the $5 bet in the dice game results in an expected
value of –8¢. The bet in the dice game is better because it has
large expected value. That is, you are better off losing 8¢
instead of losing 26¢. Even though the roulette game provides
an opportunity for a larger payoff, the craps game is better in
the long run.
• So far, we introduced the discrete probability distribution
• Next, the binomial probability distribution, a special case of the
discrete distribution
• Other case, Poisson, geometric, hyper-geometric
22
5.3 Binomial Probability Distribution
1.
2.
Related Concepts
Rationale for Binomial Probability Formula
23
5.3 Binomial Probability Distribution
1.
Related Concepts
Definition – A Binomial probability Distribution results from a
procedure that meets all of the following requirements:
1)
2)
3)
4)
Fixed number of trials
The trials must be independent (the outcome of any individual trial
doesn’t affect the probabilities in the other trials)
Each trial must have all outcomes classified into two categories
(commonly referred as success and failure)
The probability of a success remains the same in all trials.
24
5.3 Binomial Probability Distribution
1.
Related Concepts
Notation for Binomial Probability Distributions
S and F (success and failure) denote the two possible categories of
all outcomes;
P(S) = p
( p = probability of a success)
P(F) = q
( q = probability of a failure, q = 1 – p )
n
denote the fixed number of trial
x
specific number of success in n trials, 0  x  n
p
denote the probability of success in one of n trials
q
denote the probability of failure in one of n trials
P(x)
the probability of getting exactly x success in n trials
25
5.3 Binomial Probability Distribution
1. Related Concepts
e.g.1. Genetics. Consider an experiment in which 5 offspring peas
are generated from 2 parents each having the green/yellow
combination of genes for pod color. P(green pod) = 0.75. We
want to find the probability that exact 3 of the 5 offspring peas
have a green pod.
a. Does this procedure result in a binomial distribution?
i.
ii.
iii.
iv.
The number of trails is fixed (5)
The 5 trails are independent, because the probability of any spring pea
having a green pod is not affected by the outcome of any other offspring
pea.
Each of the 5 trails has two categories of outcomes: the pea has a green
pod or it does not
For each offspring pea, the probability that it has a green pod is ¾ and
the probability remains the same for each of the 5 peas
26
5.3 Binomial Probability Distribution
1.
Related Concepts
b.
If yes, identify n, x, p and q.
a)
b)
c)
d)
n=5
x=3
p = 0.75
q = 0.25
Method 1: Using the Binomial Probability Formula.
Formula 5-5
n!
 p x  q n  x for x = 0, 1, …, n
P(x) =
x!(n  x)!
27
5.3 Binomial Probability Distribution
1. Related Concepts
e.g.2. Continued.
Method 1: Using the Formula 5-5
5!
 (0.75)3  (0.25)53
P(3) =
3!(5  3)!
5!

 (0.75) 3  (0.25) 2
3!2!
= (10)(0.421875)(0.0625)
= 0.263671875
The probability of getting exactly 3 peas with green pods among
among 5 offspring peas is 0.264 (round to 3 significant digits)
28
5.3 Binomial Probability Distribution
1.
Related Concepts
Method 2: Using Technology
• Use Excel (function name: binom(x, n, p))
• Use TI-84 (function name: binompdf(n, p, x))
29
5.3 Binomial Probability Distribution
1. Related Concepts
Method 3: Using Table A-1 in Appendix A
Table A-1 cannot be used for e.g.2 because p = 0.75 is not included.
e.g.3 McDonald’s Brand Recognition. The fast food chain has a
brand name recognition rate of 95% around the world.
Randomly select 5 people, find
a. The probability that exactly 3 of the 5 recognize McDonald’s.
b. The probability that the number of people who recognize
McDonald’s is 3 or fewer.
Solution.
a. P(3) = 0.021
b. P(3 or fewer) = P(0) + P(1) + P(2) + P(3) = 0.021 + 0.001 + 0 + 0 = 0.022
30
5.3 Binomial Probability Distribution
2.
Rationale for the Binomial Probability Formula
The number of outcomes
with exactly x successes
among n trials
The probability of x
successes among n trials
for any one particular order
n!
x
n x
P( x) 
 p q
x!(n  x)!
31
5.4 Mean, Variance, and SD for Binomial Distribution
For Any Discrete Prob. Distribution
Formula 5-1  =  [x  P(x)]
Formula 5-3
For Binomial distribution
Formula 5-6  = np
 2 =  [x2  P(x)] – 2
Formula 5-7
 2 = npq
2
2
[
x

P
(
x
)]



Formula 5-8
  npq
Formula 5-4  
Using the range rule of thumb, we can consider values to be
unusual if they fall outside of the limits obtained from the following:
maximum usual value:  + 2
minimum usual value:   2
32
5.4 Mean, Variance, and SD for Binomial Distribution
Example 1. Genetics. Use the formula 5-6 and 5-8 to find the mean
and standard deviation for the number of peas with green pods
when groups of 5 offspring peas are generated. Assume that
there is a 0.75 probability that an offspring pea has a green pod
(as described in the chapter problem).
.
Solution. Using n = 5, p = 0.75, and q = 0.25. Now
 = np = 5(0.75) = 3.75
(rounded)
  npq  (5)(0.75)(0.25)  0.968246  1.0
This results are consistent with the results on page example 5,
section 5-2. But computation is much easy.
33
5.4 Mean, Variance, and SD for Binomial Distribution
e.g.2. Genetics. In an actual experiment, Mendel generated 580
offspring peas. He claimed that 75%, or 435, of them would
have green pods. The actual experiment resulted in 428 peas
with green pods.
a. Assuming that groups of 580 offspring peas are generated, find
the mean and standard deviation for the numbers of peas with
green pods.
b. Use the range rule of thumb to find the minimum usual number
and the maximum usual number of peas with green pods. Based
on those numbers, can we conclude that Mendel’s actual result
of 428 peas with green pods is unusal? Does this suggest that
Mendel’s value of 75% is wrong?
34
5.4 Mean, Variance, and SD for Binomial Distribution
Solution.
a. Using n = 580, p = 0.75, and q = 0.25. Now
 = np = (580)(0.75) = 435.0
  npq  (580)(0.75)(0.25)  10.4
b.
Maximum usual value:  + 2 = 435.0 + 2(10.4) = 455.8
Minimum usual value:   2 = 435.0  2(10.4) = 414.2
Interpretation....
If Mendel generated 580 offspring peas and if his 75% rate is correct, the number
of peas with green pods should usually fall between 414.2 and 455.8. Mendel
actually got 428 peas with green pods, and that value does fall within the range of
usual values. So the experimental result is consistent with the 75% rate.
35