1. Chapter 4 STAT 315
Nutan S. Mishra

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Random variable
A random variable is a mapping from set of sample space to the real line.
Example: consider the experiment of tossing a coin then S = { H, T}
Instead of using H and T we can give the numerical values as X(H) =1 and X(T) =0
Thus we define X as a random variable which takes value one if head shows up and 0 if tail shows up.
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Random Variable
Consider experiment of tossing a coin then S = {1,2,3,4,5,6}
Here we define x= 1,2,3,4,5,6
Random variables are discrete if they take values on a discrete set and are continuous if they take values on an interval.
Example: Recording GPA of a student. Here the random variable x takes values on (0,4).
Thus x is a continuous random variable.
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4. Probability distribution of a Random Variable
We have seen that corresponding to every sample space we can define a probability distribution. Similarly corresponding to every random variable there is a probability distribution.
Example: S = { H,T} x = 0 or 1
Then f denotes the probability distribution
f(x)=p(x=0) = ½ and f(x) =p(x=1)= ½ .
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5. Probability distribution
Example: S = {1,2,3,4,5,6} then
f(x) =P(x =i) = 1/6 for i = 1,2,3,4,5,6
Example: S = {lemon , good}
x(lemon) = 1 x(good) =0
Then f(x)= p(x=1)= .001
f(x) =p(x=0)= .999 this can be written as
f(x) = .001 if x=1
= .999 if x=0
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6. Probability distribution
Two properties of probability distribution
If f(x) is a probability distribution then f(x)0 for all x
and
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Exercise 4.5
Given
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Exercise 3.1 and 4.1
(x,y) represents the event that the technician finds x suitable solid crystal lasers and y that of carbon dioxide lasers.
There are 3 solid crystal lasers and 2 carbon dioxide lasers thus
S ={(0,0), (0,1), (0,2), (1,0),(1,1),(1,2), (2,0),(2,1),(2,2),(3,0),(3,1),(3,2)}
Z = x+y that is Z is total number of suitable lasers found. Then Z takes values
0,1,2,3,4,5
Value of Z
f(Z=z)=P(Z=z)
F(z) =P(Z<z)
1/12 1
2/12
3/12 2
6/12 3
9/12 4
11/12 5
12/12
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Exercise 4.2
Experiment: tossing four coins
|S| = 16,
S = { HHHH,
HHHT, HHTH,HTHH,THHH,
HHTT, HTHT, HTTH, THTH,TTHH, THHT
TTTH, TTHT, THTT, HTTT
TTTT }
Let X(H) =1 and X(T) =0
Let Y = total number of heads in a toss
Then Y takes values 0,1,2,3,4
Y =y P(Y =y)
/16
/16
/16
/16
/16
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10. Binomial distribution
An experiment is called Bernoulli experiment if it has only two outcomes.
Eg tossing a coin, declaring a car lemon or good.
Bernoulli experiment is also called Bernoulli trial.
P(success) = p and P(failure) = 1-p (or q)
Consider the following experiment
Take a bernoulli trial
Repeat it n times ( conduct n trials)
Probability of success remains same in each trial
The n trials are independent.
Define X = number of successes in n trials.
Then x is called Binomial random variable
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11. Binomial distribution
To find the probability distribution of X.
Note range of X is 0 to n.
We want to find P(X=1), …P(X=n).
We find a rule for P(X=x)
If there are x successes then there are (n-x) failures attached to the experiment.
P(such outcome) = p*p*…xtimes*(1-p)*(1-p)…(n-x) times
The x successes can occur in (n choose x ways)
Thus P(X=x) =
Useful link:
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12. Binomial distribution
Thus for given values of p and n we can find the probabilities of binomial distribution using following formula
b(x; n,p) = for x = 0,1,2,…,n
B(x; n,p) =
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13. Binomial distribution
Exercise 4.16:
Let success s = noise level exceeds 2 db
Given that p(s) = p = .05 then q = 1 - p = .95
Given 12 such amplifiers, let X = # s type amplifiers out of 12.
Then P(X=1) =(12 choose 1) p1 q11
P(X≤2) = P(X=0)+P(X=1)+P(X=2)
=(12choose0)p0q12+(12choose1)p1q11+(12choose2)p2q10
(c) P(X≥2) = 1- P(X<2) = 1- {P(X=0)+P(X=1)}
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14. Binomial distribution
Exercise 4.19
s = jar contains less than claimed amount of coffee.
P(s) =p = .10, n = 16 X = # jars found with less coffee
Claim is accepted if X<3 i.e X≤ 2
(a) P(acceptance ) = B(2; 16, .05)
(b) P(acceptance) = B(2; 16, .1)
(c) P(acceptance ) = B(2; 16,.15)
(d) P(acceptance) = B(2; 16, .20)
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Mean of a distribution
Mean of a probability distribution is given by
refer to Exercise 4.1
Value of Z f(Z=z)
0 1/12
1 2/12
2 3/12
3 3/12
4 2/12
5 1/12
µ = 0*1/2 + 1*2/12 + 2*3/12 + 3*3/12 + 4*2/12 + 5*1/12
= 28/12 = 2.33
Meaning: if the experiment of finding suitability of the lasers is repeated times and again then on the average 2.33 systems will be found suitable.
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Mean of a distribution
Experiment: tossing four coins |S| = 16,
S = { HHHH,
HHHT, HHTH,HTHH,THHH,
HHTT, HTHT, HTTH, THTH,TTHH, THHT
TTTH, TTHT, THTT, HTTT
TTTT }
µ = Σ y*P(y)
= 0*1/16 + 1*4/16 + 2*6/16 + 3*4/16 + 4*1/16
= 32/16 = 2
Interpretation: if the experiment of tossing
Four coins is repeated large number of times
Than on the average 2 heads will show up
Let X(H) =1 and X(T) =0
Let Y = total number of heads in a toss
Then Y takes values 0,1,2,3,4
Y =y P(Y =y)
/16
/16
/16
/16
/16
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17. Mean of binomial distribution
The mean value of the binomial distribution is µ = np where n is the number of trials and p is the probability of success for each trial.
Binomial distribution:Mean: np
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18. Mean of binomial distribution
From the definition of the mean using a distribution function, the binomial mean is
The goal is to reduce this expression to just np. Since the first term in the sum is zero, since x=0, we can replace the sum with a sum starting from 1.
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19. Variance of the Binomial Distribution
Variance of a distribution is given by
σ2 = Σ(X-µ)2*f(X)
For a binomial distribution with parameters n and p, the variance is given by npq.
σ2 = Σ(X-np)2*f(X) = np(1-p)
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20. Hypergeometric distribution
Suppose a lot of electrical switches consists of N units out of which D units are defective.
We draw a sample of size n from the lot of size N and interested in number of defectives occurred in the sample
X=number of defectives in a sample of size n which has been drawn from a lot of size N with D defectives.
Then the distribution of X is hypergeometric.
Total number of ways we can draw a sample of size n from N is (N choose n)
Out of these many possible samples we want to how many sample would contain x defectives. This is
(D choose x)*(N-D choose n-x) thus
P(X=x) = (D choose x)*(N-D choose n-x) / (N choose n)
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21. Properties of hypergeometric distribution
Hypergeometric distribution has three parameters N, n, D.
If we draw this sample with replacement then distribution of x is binomial with n and p=D/N
If we draw this sample without replacement then x has hypergeometric distribution.
Its mean is n*D/N and variance is
n*a*(N-a)*(N-n)/ N2*(N-1)
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22. Moments of a distribution
Kth moment about origin is also known as kth raw moment
µk’ = Σxk*f(x)
µ1’ = Σx*f(x) =µ
That is first raw moment is the mean of the distribution
And kth moment about mean is also known as kth central moment
µk = Σ(x-µ)k*f(x)
µ1 = Σ(x-µ)*f(x) = 0
That is first central moment is 0
µ2 = Σ(x-µ)2*f(x) =σ2
That is second central moment is variance of the distribution
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23. Moments of a distribution
Variance can be expressed as linear combination of raw and central moments
σ2 = µ2’ - µ1’2
= µ2’ - µ2
Moments are alternatively expressed as
µk’ = E(Xk)
µk = E(X-µ)k
Thus mean and variance can be expressed as expectations
µ = E(X) = expected value of x and
σ2= E(X-µ)2
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Poisson Distribution
Random variable x= # of defects occurred /unit
Example : an assembled car has thousands of parts. X = # defects /car.
Example: X = # of defects occurred /yard of cloth.
Such an X has Poisson distribution given by
f(x; λ) = for x = 0,1,….
λ is the parameter of Poisson distribution.
Mean = variance = λ for the Poisson distribution.
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Properties
Prove that binomial probabilities sum up to one.
Prove the Poisson probabilities sum up to one.
Prove the mean of binomial distribution is np
Prove the variance of binomial distribution is npq
Prove the mean and variance of Poisson distribution is λ.
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26. Binomial probabilities sum up to one
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27. Poisson Distribution (cont.)
Mean and Variance
Proof:
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28. Poisson Approximation to Binomial
Proof:
Binomial distribution with parameters (n, p)
As n→∞ and p→0, with np=l moderate, binomial distribution converges to Poisson with parameter l
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29. Geometric Distribution
Consider the following example: A quality inspector inspecting the electrical switches right off the manufacturing belt. He is interested in the question: How may items are inspected before the first failure occurred?
Suppose the first failure occurred at xth item.
Then X-1 = # successes before first failure is a random variable
Let probability of failure for each item is constant and is equal to p.
P(X=x) = pqx-1 for x = 1,2,….
Verify that mean of geometric distribution is 1/p.
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30. Multinomial distribution
A multinomial experiment is an extended binomial probability. The difference is that in a multinomial experiment, there are more than two possible outcomes. However, there are still a fixed number of independent trials, and the probability of each outcome must remain constant from trial to trial.
Instead of using a combination, as in the case of the binomial probability, the number of ways the outcomes can occur is done using distinguishable permutations.
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31. Multinomial distribution
The probability that a person will pass a College Algebra class is 0.55, the probability that a person will withdraw before the class is completed is 0.40, and the probability that a person will fail the class is Find the probability that in a class of 30 students, exactly 16 pass, 12 withdraw, and 2 fail.
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Exercise 4.79, 4.81 X
f(X)
X*f(X) X2
X2*f(X)
.89 1
.07 2
.03
.06 4
.12 3
.01 9
.09 Σ
1.00
.16
.28
P(X>=2) = = .04
P(X=0) = .89
P(X>=1) = .11
a comparison of the probabilities shows that its more likely to have 0 defects
Mean = Σx*f(X) = .16
Variance = ΣX2*f(X) – (ΣX*f(X))2 = .28 – (.16)2
= = .055
Standard Deviation = .2345
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