Explain and describe the following topic in statistic

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First: in your own word, explain and describe the following topics. and support your answer by real example including numbers (small set) (check the attachment)

Second: write at least four questions with your answer regarding the probability in chapter 4 (file is attached)

Explain and describe the following topic in statistic
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Chapter 4-1 Discrete Random Variables (cont.) The Negative Binomial PD ▪ Same characteristics as the binomial, ▪ The variable of interest (y) is the number of trials to the rth success. The probability of observing the rth success in y trials is:  y − 1 r y − r  p q P( y ) =   y − r With, r r 2  = , and  = q 2 p p Example A large lot containing 10% defective items. If items are drawn sequentially (item-by-item), what is the probability that the second defective is observed on the 5th item ? Let; p= probability of success then, for y = 5, and r = 2 5-1 2 5-2 P (y=5) = ( )p q 5-2 = 0.03 • For the special case where r =1, the probability of observing the first success in y trials is: P (y) = p qy-1 A.K.A. the Geometric Prob. Dist., with =1/p and 2= q /p2 Example: In the previous example, what is the probability that the first defective is observed on the 5th item drawn? P(y=5) = 0.1 (0.9)5-1= 0.066 The Poisson PD ▪ A model for the relative frequency distribution of the number of occurrences of a "rare event" over a given area of opportunity ( unit of time, area, volume, etc.) Examples: ▪ the number of 6.0 earthquakes in California per month. ▪ the number of accidents in a given intersection, etc. Characteristics of the Poisson Probability Distribution: 1. The experiment consists of counting the number of times y, a particular event occurs during a given area of opportunity. 2. The probability that an event occurs in a given unit of time, area, etc. is constant. 3. The number of events per unit of time, area, etc., is independent of the number that occurs in other units. The Poisson PD: P( y ) =  y y! e − Where, = mean number of occurrences, and  = =  2 Example: If the mean number of accidents occurring on a given highway each day is 3, what is the probability that on any day: a) No accident will occur? b) Less than 2 accidents will occur? Given;  = 3, P(0) = And, 30 0! e-3 = 0.0498 P(y<2)= P(0)+ P(1) = 0.199 See cumulative Poisson probabilities in Table 4 in Appendix B, p. 906-907 ▪ Also, the Poisson probability distribution can be used to approximate the binomial distribution when n is large and = np is small, say np < 7. Example: A large lot contains 5% defective items. In a random sample of 30, what is the probability of 2 or fewer defectives? Given:  = np = 30(0.05) P(y<2)= P(0)+P(1)+P(2) = 0.8088 Note: using the binomial, P(y<2)= 0.8122 Hypergeometric Distribution When sampling without replacement from a finite population of size N and the sample size n is comparatively large, i.e., n/N > 0.10 Properties : The experiment consists of drawing n elements without replacement from a set of N elements; r of which are S's and (N-r) of which are F’s. The Hypergeometric PD: The probability of y successes in n trials is given by:  r  N − r     y  n − y   P( y ) = N   n max [0, n − ( N − r )] for y =   min (r , n) Where, N = total number of elements r = # of S's in the N elements n = # of elements drawn y = # of S's in the n elements The mean of a Hypergeometric random variable is: nr = N The variance is:  = 2 r ( N − r ) n( N − n) 2 N ( N − 1) Example Suppose that a fuse box containing 20 fuses of which 5 are defective. If 2 fuses are selected at random without replacement, what is the probability that both fuses are defective? Given: N=20 r=5 n=2 y=2 Then:  5  20 − 5     2  2 − 2  1  P( y = 2) = = 19  20    2
Chapters 4 Discrete Random Variables Random Variables: Numerical Outcomes Of Experiments: • Discrete • Continuous Discrete Random Variables: • Result from a counting process, e.g. - No. of heads in n trials, - No. of defective items in a sample, - No. of machines operating / shift, - No. of accidents / given intersection. Discrete Random Variables: y 0 1 2 3 …. A Countable Number of Values Probability Distribution Functions: Theoretical models of the relative frequency distribution of discrete random variables when n N: Continuous Random Variables: • Result from a measuring process e.g. - Diameter of shafts, - Processing time, - Car speed, - % of DDT. Continuous Random Variables: y 0 1 2 3 …. An infinite Number of Values Probability Density Functions: Theoretical models of the relative frequency distribution of continuous random variables when n N: Chapter 4 Discrete Random Variables Probability Distribution: a table, formula, or graph which displays the probabilities for each possible value of a random variable y, which would be observed if an experiment were repeated an infinitely large number of times. Properties: 1.The probability of y taking on any given value: 0  p(y)  1 2. The values of y represent sets of mutually exclusive events, therefore  p(y) = 1 For example: A balanced coin is flipped twice and the number of heads, y, is observed. Find the probability distribution of y. Probability Distribution of y Tabular Format: y 0 1 2 p(y) 1/4 1/2 1/4 S p(y)=1 Probability Distribution of y Graphical Format: p(y) 0.5 0.4 0.3 0.2 0.1 0 0 1 2 y Probability Distribution of y Formula: 1  2 P( y ) =   4  y Where, (.) is the number of combinations of 2 taken y at a time. Expected Value of y: the mean or expected value of a discrete random variable, y, with probability distribution p(y) is:  = E( y) =  y.P( y) The variance of y, is the expected value of the squared deviation of y from its mean  that is:  = Var ( y) = E[( y −  ) ] 2 2 = E( y ) −  2 2 Example: What is the expected value and variance of the number of heads y, when you flip a balanced coin twice? y 0 1 2 p(y) 1/4 1/2 1/4 S p(y)=1 The mean, or expected value of y is as follows: S y p(y)= 0(1/4) + 1(1/2) + 2(1/4) = 1 Thus, The expected value  = 1 The variance of y is as follows: 2 = (0 - 1)2 (1/4) + (1-1)2 (1/2) + (2-1)2 (1/4) OR, 2 + (1)2 (1/2) + (2)2 (1/4)} - (1)2 = {(0)2 (1/4) 2 = 1/2 , and  = 0.707 The Binomial Experiment: Any experiment where there are 2 possible outcomes, the population N, is large, and the sample taken from the population n, is relatively small, (say n < 0.10 N) Characteristics of a Binomial Random Variable: 1. The experiment consists of n identical, independent trials 2. There are only 2 possible outcomes on each trial, say S (success) and F (failure) 3. The probability of success p, is the same from trial to trial and the probability of failure is denoted by q, Note: p+q=1 4. The binomial random variable y, is the number of Successes in n trials The Binomial Probability Distribution:  n y n− y P( y ) =   p q  y for y = 0,1,2,..n where: p = probability of a success on a single trial, and q = 1-p y = # of successes in n trials  n n!   =  y  y!(n − y )! The mean  = np, and The variance 2 = npq Example: Suppose that a large lot of fuses contains 10% defectives. Find the probability that a random sample of 5 fuses contains: A) Exactly one defective, B) At most one defective, C) At least one defective. The Cumulative Binomial Probability Table: Appendix B, Table 2, p 901-904. For n and p; find: P(y < k) Multinomial Probability Distribution: ▪ ▪ An extension of the binomial experiment, Consists of n identical independent trials, with k possible outcomes. Properties of the Multinomial Experiment: 1. The experiment consists of n identical, independent trials 2. There are only k possible outcomes on each trial 3. The probabilities of k outcomes, denoted by p1, p2, ... pk, remain the same from trial to trial; Note: p1+p2+ ... + pk=1 4. The random variables of interest are the counts y1, y2, ...yk in each of the k categories. The Multinomial Probability Distribution: n! P( y1, y2 ,.., yk ) = ( p1) y1 ( p2 ) y2 ....( pk ) yk y1! y2!..yk ! where: pi = probability of outcome i on a single trial ▪ p1 + p2 + ... + pk = 1 ▪ n= y1 + y2 + ... + yk = number of trials ▪ yi = # of occurrences of i in n trials ▪ mean  i = npi, and ▪ variance i2= npi (1-pi) Example: For a multinomial distribution with; n=40, k=3, p1=0.1, p2=0.3, and p3=0.6, find: P( y1=10,y2=10,y3=20) P( y1=10,y2=10,y3=20) 40! = ________ 10! 10! 20! = 0.00055 (0.1)10 (0.3)10 (0.6)20 Note: • No tabulation possible. • The binomial is a multinomial with k = 2.

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Sksantana8
School: UC Berkeley

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