13 MomentsThe kth moment about the origin of a rv X, denoted by µk’ , is the expected value of Xk, , symbolically,µk’ = E(Xk) = x xk · p(x).The kth moment about the mean of a rv X, denoted by µk, is the expected value of (X - µ)k, , symbolically,µk = E[(X - µ)k] = x (x - µ)k · p(x).
14 Special CasesThe expectation, or the mean, is the 1st moment about the origin.µ = µ1’ = E(X) = x x · p(x).The variance is the 2nd moment about the mean2 = µ2 = E[(X - µ)2] = x (x - µ)2 · p(x).
16 Binomial Distribution For X ~ Bin(n,p), the cdf will be denoted by
17 Mean & Variance If X ~ Bin(n, p), then E(X) = np, V(X) = npq (where q = 1-p.)
18 Example(Cont) n = 5, p = 11/32 . Then E(X) = n · p = 5 · 11/32 = 1.72. V(X) = n · p · q = 5 · 11/32 · 21/32 = 1.13. = (1.13)1/2 = 1.06.
19 Hypergeometric and Negative Binomial Distribution
20 IntroductionThe hypergeometric and negative binomial distribution are both closely related to the binomial distribution.
21 IntroductionThe negative binomial distribution arises from fixing the number of S’s and letting the number of trials to be random.The hypergeometric distribution is the exact probability model for sampling without replacement from a finite dichotomous (S,F) population.
22 Negative Binomial Dist’n The experiment consists of a sequence of independent trials.Each trial results in either S or F.The probability of success, p, is constant from trial to trial.Trials are performed until a total of s successes have been observed, where s is a prespecified positive integer.
23 Negative Binomial RVX = the number of F’s that precede the rth success, is called a negative binomial rv.Possible values of X are 0, 1, 2, …
24 pmf Denote by nb(x; r, p) the pmf of X. Then Why? Total # of trials = x; The last trial must be a success. Among the first (x-1) trials, there are (s - 1) successes & x-s failures.
25 Review of Chapter 3Hypergeometric distribution,Poisson distribution.
26 ExampleWhat’s the probability that < 3 requests are received during a particular hour?P( X < 3) = P(0) + P(1) + P(2)= e-5 + 5· e · e-5/2=
27 ExampleWhat’s the probability that exactly 10 requests are received during a particular 2-hour period?Rate = 2 × 5 = 10.P(X = 10) = e /10! =
28 Example How many calls do they expect to get during a 45-min period?
30 Continuous RVAn rv X is continuous if its set of possible values is an entire interval of numbers.Example:X = the pH of a random soil sampleX = the weight of a randomly selectedperson.
31 pdfLet X be a continuous rv. Then a probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a b,For f(x) to be a pdf, f(x) must satisfy:f(x) 0 for all x, and
32 ExampleWaiting time at a bus station. A bus arrives every 10 minutes. So the waiting time is from 0 to 10. One possible pdf for waiting time X isThe probability of waiting between 3 to 5 minutes is:
33 Uniform DistributionA continuous rv X is said to have a uniform distribution on the interval [A, B] if the pdf of X isGraphs of uniform distributions.
34 Probability at a PointWhen X is a discrete rv, each possible value is assigned positive probability. This is no longer true for continuous rv.If X is a continuous rv, then for any number c, P(X = c) = 0. Consequently, P(a X b) = P(a < X b) = P(a X < b)= P(a < X < b).
35 ExampleLet X = the “time headway” for two randomly chosen consecutive cars on a freeway during a period of heavy flow. Suppose the pdf of X is given by:f(x) = 0.15 e-0.15( x - 0.5), x 0.5.f(x) = 0 for x < .5 and f(x) decreases exponentially fast as x increase from .5.
36 Example First, it clear that f(x) 0. Now we verify The probability that headway time is at most 5 seconds is
38 cdfThe cumulative distribution function (cdf) F(x) for a continuous rv X is defined for every number x byFor each x, F(x) is the area under the density curve to the left of x. It is the probability of observing X a value smaller than or equal to x.
39 Example Let X have a uniform distribution on the interval [A, B]. Then So, for x < A, F(x) = 0 and for x B, F(x) = 1. For A x B,
40 ExampleThe entire cdf is:The graph of the cdf looks like:
41 PropositionsCompute probabilities using F(x): P(a x b) = F(b) - F(a).Obtaining pdf from cdf:If X is a continuous rv with cdf F(x) differentiable at every point x, then the pdf f(x) =F ’(x).
42 Example For uniform distribution on [A, B], the cdf is So, for example, if A < a < b < B, then P(a < X < b) = F(b)-F(a) = (b-a)/(B-A).The pdff(x) = F ’(x) = 1/(B-A) for A < x < B.
43 Expected ValuesThe expected value (or, mean) of a continuous rv X with pdf f(x) isIf X is a continuous rv with pdf f(x) and h(X) is any function of X, then
44 ExampleThe pdf of the waiting time (in minutes) at a checkout is given byf(x) = x/ for 0 x < 4.What’s the probability of waiting less than 3 min?What’s the expectation of the waiting time?
45 Example What’s the probability of waiting less than 3 min? What’s the expectation of the waiting time?
46 Variance & S.D.The variance of a continuous rv X with pdf f(x) and mean isThe standard deviation (S.D.) of X isV(X) = E(X2) - [E(X)]2.
47 Linear Transformation If h(X) = a X + b and V(X) = 2, thenV(h(X))=V(a X + b) = a 2 2andaX+b = |a| .
48 Example(Cont) The pdf of the waiting time at a checkout: f(x) = x/ for 0 x < 4.Find the variance of the waiting time. = E(X) =
50 IntroductionThe normal distribution is the most important distribution in all of probability and statistics.Many numerical populations have distributions that can be approximated very well by a normal curve.
51 Example Scores of standardized tests, Measurements of intelligence & aptitude,Returns of a stock (or a portfolio),Measurement errors …
52 DefinitionA continuous rv X is said to have a normal distribution with parameters and 2 if the pdf of X is
53 Remarks Notation: X ~ N(, 2). It’s clear that f(x; , 2) 0 and it can be shown thatE(X) = , and V(X) = 2.
54 Standard Normal Dist’n With = 0 and = 1, the normal distribution is called a standard normal distribution.The pdf of a standard normal rv Z isThe cdf of Z is denoted by (z).
55 Normal Probability Table Table A.3 on page 704 of the text tabulates the standard normal probabilities (cdf). This is one of the most useful statistical tables.Example: Using the table to compute:P(Z < 1.20),P(Z > 1.68), (= 1 - P( Z 1.68))P(-1.96 < Z < 1). (= P( Z < 1) - P( Z -1.96))
56 Inverse Reading of Table A.3 Z denotes the (100)th percentile of the standard normal distribution.The area under the standard normal curve to the right of Z (tail probability) is .Find: Z.30, Z.90.
57 Standardization If Z ~ N(0, 1), then X = + Z ~ N(, 2). Inversely, if X ~ N(, 2), thenZ = (X - )/ ~ N(0, 1).The transformationIs called standardization.P(X x) =P[Z (x - )/] = [(x - )/].
58 Standardization (100p)th percentile for N(, 2) So if X ~ N(, 2), thenX = + · Z .
59 Rule of Thumb If X is (approximately) normal, then about 68% of the x's are within 1 SD of the mean;about 95% of the x's are within 2 SDs of the mean;about 99.7% of the x's are within 3 SDs of the mean;
60 Example(Fish)The lengths of fish in a certain fish population follows a normal distribution with = 54 mm and = 4.5 mm.What percentage of the fish are between 50 and 60 mm long?Let Z = (X - )/. Thenz1=( )/4.5= -.89, z2=( )/4.5=1.33.Use Table A.3: P(50 X 60)=P(-.89Z1.33)= =
61 Example(Fish) What percentage of the fish are more than 48 mm long? z1 = ( )/4.5 =P( X > 48) = =What percentage of the fish are between 58 and 60 mm long?z1 = ( )/4.5 = 0.89, z2 = P(58 < X < 60) = =
62 Example(Fish)What is the 70th percentile of the fish length ? What is the 90th percentile?From Table A.3, Z.70 = So, X.70 = ·0.52 = 56.3Similarly, Z.90 = andX.90 = ·1.29 =
63 Example (Height)Among American women aged , 10% are less than 61.2 inches tall; 80% are between 61.2 and 67.4 inches and 10% are more than 67.4 inches. Assume the height can be well approximated by a normal distribution.Find the mean and the SD .
64 Example(Height) Z.10 = -1.29 and Z.90 = 1.29 , so Solving for and , we have = ( )/( ) = 2.4, and = 64.3.
65 Normal ApproximationThe normal distribution is often used to approximate the distribution of discrete populations.In particular, under certain conditions, the normal distribution can be used as an approximation to the binomial distribution.
66 Normal Approximation to Binomial Distribution For a binomial rv X , we haveWhen both np and nq are relatively large, the normal distribution with the same mean and SD is a very good approximation to Bin(n, p).
67 Normal Approximation to Binomial Distribution Let X ~ Bin(n, p), Then if np 5 and nq 5, X has approximately a normal distribution withThis is the area under the normal curve to the left of x+.5. “+.5” is the correction for discreteness. This is called continuity correction.