Chapter 4 STAT 315 Nutan S. Mishra

UNiversity of South Alabama 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. 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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. 12/31/2018 UNiversity of South Alabama

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 . 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

Probability distribution Two properties of probability distribution If f(x) is a probability distribution then f(x)0 for all x and 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama Exercise 4.5 Given 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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) 0 1/16 1 4/16 2 6/16 3 4/16 4 1/16 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

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: http://www-stat.stanford.edu/~naras/jsm/example5.html 12/31/2018 UNiversity of South Alabama

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) = 12/31/2018 UNiversity of South Alabama

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)} 12/31/2018 UNiversity of South Alabama

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) 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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. 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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) 0 1/16 1 4/16 2 6/16 3 4/16 4 1/16 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

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. 12/31/2018 UNiversity of South Alabama

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) 12/31/2018 UNiversity of South Alabama

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) 12/31/2018 UNiversity of South Alabama

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) 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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. 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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. 12/31/2018 UNiversity of South Alabama

Binomial probabilities sum up to one 12/31/2018 UNiversity of South Alabama

Poisson Distribution (cont.) Mean and Variance Proof: 12/31/2018 UNiversity of South Alabama

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 12/31/2018 UNiversity of South Alabama

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. Useful Link: http://www.stat.vt.edu/~sundar/java/applets/Distributions.html#BINOMIAL 12/31/2018 UNiversity of South Alabama

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. 12/31/2018 UNiversity of South Alabama

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 0.05. Find the probability that in a class of 30 students, exactly 16 pass, 12 withdraw, and 2 fail. 12/31/2018 UNiversity of South Alabama

UNiversity of South Alabama 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) = .03+.01 = .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 = .28 - .225 = .055 Standard Deviation = .2345 12/31/2018 UNiversity of South Alabama