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Probability and Statistics for Engineers
Chapter 4 Probability Distributions
ruochen Liu
Institute of Intelligent Information Processing, Xidian University
OutlineRandom variables The binomial distributionThe hypergeometric distributionThe mean and the variance of a probability distributionChebyshev’s Theorem
The Poisson approximation to the binomial distribution Poisson processesThe geometric distribution The multinomial distribution Simulation
Random variablesIn most statistical problems we are concerned with one number or a few numbers that are associated with the outcomes of experiments. they are values of random variables. In the study of random variables, we are usually interested in their probability distributions, namely, in the probabilities with which they take on the various values in their range.Section 4.1 introduce random variables and probability distribution. Some special probability distributions are followed.
OutlineRandom variablesThe binomial distributionThe hypergeometric distributionThe mean and the variance of a probability distributionChebyshev’s Theorem
The Poisson approximation to the binomial distribution Poisson processesThe geometric distribution The multinomial distribution Simulation
Random Variables
1P
2P
1C
2C
3C
1C
2C
3C
1C
2C
3C
1C
2C
3C
1C
2C
3C
1C
2C
3C
1E
2E
3E
1P
2P
1P
2P
Probability
Number ofpreferredratings
112223001112001112
0.070.130.060.050.070.020.070.140.070.080.110.030.020.030.010.010.020.01
Probability
x 0
0.22
21
0.26 0.50
3
0.02
The numbers 0,1,2 and 3 are values of a random variables: the number of preferred ratings.
The random variables are functions defined over the elements of a sample space
Let us refer to the lawn-mower-rating example on page 58. E1(easy to operate), P2 (inexpensive), and C3 (low average cost of repairs) as preferred ratings. (the number of preferred rating)
Fig.4.1
Random variables
Random variables are denoted by capital letters X, Y and so on, to distinguish them from their possible values given in lower case x, y and so on. Classification of random variables
Discrete random variables, which can take on only a finite number, or a countable infinity of valuesContinuous random variables, which take on an infinite number of values
A random variable is any function that assigns a numerical value to each possible outcome.
Probability distribution can be expressed by means of EquationTable Exhibit the correspondence between the value of the random variable and its probability （ rolling a balanced die. ）
Probability distributionThe table displays another function, called the probability distribution of the random variable.probability distribution is the function which assigns probability to each possible outcome x that is called the probability distribution：To denote the values of a probability distribution, use such symbols as f (x), g (x), h (z).
( ) ( )f x P X x= =
( ) 1/ 6 for 1,2,3,4,5,6f x x= =
Probability
x 0
0.22
21
0.26 0.50
3
0.02
Probability distribution
( ) [ ]
The probability distribution of a discrete random variable is a list of the possible values of together with their probabilities The probability distribution always sa
X X
f x P X x= =
( ) ( )all
tisfies the conditions
0 and 1x
f x f x≥ =∑
Of course, not every function defined for the values of a random variable can serve as a probability distribution, there has some demands for it.
Probability distributionExample: Checking for non-negativity and total probability equals one
Check whether the following can serve as probability distributions
2
( ) ( ) for 0,1,2,3,4,525xb h x x= =
2( ) ( ) for 1,2,3,42
xa f x x−= =
Solution. (a) this function cannot serve as a probability distribution because f(1) is negative.(b) This function also can not serve as a probability distribution because the sum of the five probabilities is 6/5 which is larger than 1.
Probability distribution
x0 21 3
( )f x
0.1
0.5
0.30.4
0.2
Probability histogram
Probability bar chart
x0 21 3
( )f x
0.1
0.5
0.30.4
0.2
It is often helpful to visualize probability distributions by means of graphs
Probability histogram: the areas of the rectangles are equal to the corresponding probabilities so their heights are proportional to the probabilities. The bases touch so that there are no gaps between rectangles representing the successive values of the random variable.
Probability bar chart: the heights of the rectangles are also proportional to the corresponding probabilities, but they are narrow and their width is of no significance.
Probability
x 0
0.22
21
0.26 0.50
3
0.02
Fig.4.2
Fig.4.2
Distribution function
Probability
x 0
0.22
21
0.26 0.50
3
0.02 F(x)
x 0
0.98
21
0.26 0.76
3
1
The probability F(x) that the value of a random variable is less than or equal to x. F(x) is called the cumulative distribution function or just the distribution function of the random variable.
As we see later, there are many problems in which we are interested not only in the probability f(x) that the value of a random variable is x, but also in the probability F(x)
OutlineRandom variables The binomial distributionThe hypergeometric distributionThe mean and the variance of a probability distributionChebyshev’s Theorem
The Poisson approximation to the binomial distribution Poisson processesThe geometric distribution The multinomial distribution Simulation
The binomial distribution
Repeated trialsProbability that 1 of 5 rivets will rupture in a tensile testProbability that 9 of 10 DVD players will run at least 1000 hoursthe probability that 45 of 300 drivers stopped at a roadblock will be wearing seat belts. …we are interested in the Probability of getting x successes in n trials, or x successes and n-x failures in n attempts
Many statistical problems deal with the situations referred to as repeated trials
Bernoulli trialsThe assumptions of Bernoulli trials1. There are only two possible outcomes for each trial2. The probability of success is the same for each trial3. The outcomes from different trials are independentThere are a fixed number n of Bernoulli trials conducted
ExamplesBinomial probability distribution Over a period of a few months, an engineer found that her computer would often hang up while she was doing Internet searches. She postulates that the probability is 0.1 that any half-hour search session will require at least one reboot of the computer. Next week she will perform 3 half-hour searches, each on a different day.
(a) List all possible outcomes for the 3 searches in terms of success S, no hang up, and Failure F, at least one hang up, during each session.
(b) Find the probability distribution of the number of successes, X , among the 3 searches.
solution: (a) SSS SSF SFS SFF FSS FSF FFS FFF (SSS means all thee searches are successful. SSF means the first and second search are successful and the third is a failure).
(b)( ) ( ) ( ) ( )33
0.9 0.1 , for 0,1,2,3x xf x P X x xx
−⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
Binomial distributionLet X be the random variable that equals the number of successes in n trials. If p and 1-p are the probabilities of success and failure on any one trial, then probability of getting x successes and n-x failures, in some special order is
( )1 n xxp p −−
( ) ( ); , 1 0,1,2, ,n xxnb x n p p p x n
x−⎛ ⎞
= − =⎜ ⎟⎝ ⎠
Binomial distribution
binomial coefficients nx
⎛ ⎞⎜ ⎟⎝ ⎠
( )binomial expansion 1n
p p+ −⎡ ⎤⎣ ⎦
Binomial distributionExample: Evaluating binomial probabilities
It has been claimed that in 60% of all solar-heat installations the utility bill is reduced by at least one third. Accordingly, what are the probabilities that the utility bill will be reduced by at least one third in (a) four of five installations; (b) at least four of five installations.
( ) 4 5 454;5,0.6 0.6 (1 0.6) 0.259
4b −⎛ ⎞
= − =⎜ ⎟⎝ ⎠
( ) 5 5 555;5,0.6 0.6 (1 0.6) 0.078
5b −⎛ ⎞
= − =⎜ ⎟⎝ ⎠
And the answer is the sum of the two terms. We have 0.259+0.078=0.337
Example If the probability is 0.05 that a certain wide-flange column will fail under a given axial load, what are the probabilities that among 16 such columns (a) at most two will fail; (b) at least four will fail?
Binomial distributionTable 1 gives the values of (at this end of book)
( ) ( ) ( ); , ; , 1; , and ( 1) 0b x n p B x n p B x n p B= − − − =
( ) ( )0
; , ; , , for 0,1, 2, ,x
kB x n p b k n p x n
=
= =∑Relationship
Solution (a) Table 1 shows that B(2;16,0.05)=0.9571(b) Since 1- B(4;16,0.05)= 1-0.9930=0.007
Binomial distributionExample
If the probability is 0.2 that any one person will dislike the taste of a new toothpaste, what is the probability that 5 of 18 randomly selected persons will dislike it?
( ) ( ) ( )5;18,0.2 5;18,0.2 4;18,0.2b B B= −0.8671 0.7164 0.1507= − =
( ) 4 18 5185;18,0.2 0.2 (1 0.2)
5b −⎛ ⎞
= −⎜ ⎟⎝ ⎠
Example: A binomial probability to guide decision making
A manufacturer of fax machines claims that only 10% of his machines requires repairs within the warranty period of 12 months. If 5 of 20 of this machines required repairs within the first year, does this tend to support or refute the claim.Solution: let us find the probability that 5 of 20 of the fax machines will require repairs within a year when the probability that any one will require repairs within a year is 0.10.
b(5;20,0.1)=B(5;20,0.1)-B(4;20,0.1)=0.9887-0.9568=0.0324This probability is very small, The event that 5 of 20 of his machines required repairs occurs, this shows that this event should have a large probability. it would seem to reject the fax machine manufacturer’s claim.
Shape of binomial distributions
Symmetrical distribution (p=0.5)
( ); ,0.5 0.5nnb x n
x⎛ ⎞
= ⎜ ⎟⎝ ⎠
,n n
n x x⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
( ) ( ); ,0.5 ; ,0.5b x n b n x n= − x0 21 3
( );5, 0.5b x
1032
532
132
4 5
Important information about the shape of binomial distributions can be shown in probability histograms
since
so
Fig.4.3
Shape of binomial distributionsA long-tailed or Skewed distribution (p≠ 0.5)
x0 21 3 4 5
0.1
0.5
0.30.4
0.2
Negatively skewed
( );5,0.8b x( );5, 0.2b x
x0 21 3 4 5
0.1
0.5
0.30.4
0.2
Positively skewed
Fig.4.4
P is less than 0.5 P is greater than 0.5
A probability distribution that has a probability histogram like either of those in Figure 4.4 is said to be a long-tailed or skewed distribution. It is said to be a positively skewed distribution if the tail is on the right, and it is said to be negatively skewed if the tail is on the left.
OutlineRandom variables The binomial distributionThe hypergeometric distributionThe mean and the variance of a probability distributionChebyshev’s Theorem
The Poisson approximation to the binomial distribution Poisson processesThe geometric distribution The multinomial distribution Simulation
The hypergeometric distributionSuppose that we are interested in the number of defectives in a sample of n units drawn without replacement from a lost containing N units, of which a are defective. Let the sample be drawn in such a way that at each successive drawing, whatever units are left in the lot have the same chanceof being selected. the first drawing will yield a defective unite is a/N, but for the second drawing it is (a-1)/(N-1) or a/(N-1), depending on whether or not the first unit drawn was defective. Thus the trials are not independent, the binomial distribution does not apply. Note that the binomial distribution would apply if we do sampling with replacement
Hypergeometric distribution
N an x−⎛ ⎞
⎜ ⎟−⎝ ⎠
ax
⎛ ⎞⎜ ⎟⎝ ⎠
( ); , , for 0,1, ,
a N ax n x
h x n a N x aNn
−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠= =
⎛ ⎞⎜ ⎟⎝ ⎠
L
Consider all the possibilities as equally likely, it follows that for sampling without replacement the probability of getting x successes (defective) in n trials is (of which, a are defective)
n objects can be chosen from a set of N objects in :Nn⎛ ⎞⎜ ⎟⎝ ⎠
The x successes (defectives) can be chosen in : ways
the n-x failures (nondefectives) can be chosen in : waysx successes and n-x failures can be chosen in: ways.a N a
x n x−⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠
Example A shipment of 20 digital voice recorders contains 5 that are defective. If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective.
( )
5 20 52 10 2
2;10,5,20 0.3482010
h
−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠= =
⎛ ⎞⎜ ⎟⎝ ⎠
However, when n is small compared to N, less than N /10, the composition of the lot is not seriously affected by drawing the sample without replacement, and the binomial distribution with the parameters n and p=a / N will yield a good approximation.
Solution, substituting N=20, a=5, x=2, and n=10 into the formula for the hypergeometric distribution,
Example A numerical comparison of the hypergeometric and binomial distributions
Repeat the preceding example for a lot of 100 digital voice recorders, of which 25 are defective, by using
(a) The formula for the hypergeometric distribution(b) The formula for the binomial distribution as an approximation
( )
25 100 25 25 752 10 2 2 8
2;10, 25,100 = 0.292100 10010 10
h
−⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( ) ( )2 10 2102;10,0.25 0.25 1 0.25 0.282
2b −⎛ ⎞
= − =⎜ ⎟⎝ ⎠
what is the probability that 2 of the 10 will be defectiveSolution (a) here x=2, a=25, n=10, and N=100
(b) Here p=25/100, x=2 and n=10
Observe that the difference between the two values is only 0.01
hypergeometric and binomial distributions
Observe that the difference between the two values is only 0.01In general, it can be shown that h(x;n,a,N) approaches b(x;n,p) with p=a/N when N approaches infinite, the binomial distribution is used as an approximation to the hypergeometric distribution if n<=N/10.because we can use Table 1 to check the probability of
events.
Next lecture and HomeworkIn the next lecture, the chapter 4 (4.4-4.5) in the textbook will be discussed. Please read the them.Page112
4.3, 4.5, 4.6, 4.12, 4.20, 4.25,
