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# Discrete random variable.

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Name                                       Shakeel Nouman
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### Discrete random variable.

1. 1. Discrete Random Variables Slide 1 Shakeel Nouman M.Phil Statistics Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
2. 2. 3        Slide 2 Random Variables Using Statistics Expected Values of Discrete Random Variables The Binomial Distribution Other Discrete Probability Distributions Continuous Random Variables Using the Computer Summary and Review of Terms Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
3. 3. 3-1 Using Statistics Slide 3 Consider the different possible orderings of boy (B) and girl (G) in four sequential births. There are 2*2*2*2=24 = 16 possibilities, so the sample space is: BBBB BBBG BBGB BBGG BGBB BGBG BGGB BGGG GBBB GBBG GBGB GBGG GGBB GGBG GGGB GGGG If girl and boy are each equally likely [P(G)=P(B) = 1/2], and the gender of each child is independent of that of the previous child, then the probability of each of these 16 possibilities is: (1/2)(1/2)(1/2)(1/2) = 1/16. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
4. 4. Random Variables Slide 4 Now count the number of girls in each set of four sequential births: BBBB BBBG BBGB BBGG • • • (0) (1) (1) (2) BGBB BGBG BGGB BGGG (1) (2) (2) (3) GBBB GBBG GBGB GBGG (1) (2) (2) (3) GGBB GGBG GGGB GGGG (2) (3) (3) (4) Notice that: each possible outcome is assigned a single numeric value, all outcomes are assigned a numeric value, and the value assigned varies over the outcomes. The count of the number of girls is a random variable: A random variable, X, is a function that assigns a single, but variable, value to each element of a sample space. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
5. 5. Random Variables (Continued) Slide 5 0 BBBB BGBB GBBB BBBG BBGB GGBB GBBG BGBG BGGB GBGB BBGG BGGG GBGG GGGB GGBG GGGG 1 X 2 3 4 Sample Space Points on the Real Line Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
6. 6. Random Variables (Continued) Slide 6 Since the random variable X = 3 when any of the four outcomes BGGG, GBGG, GGBG, or GGGB occurs, P(X = 3) = P(BGGG) + P(GBGG) + P(GGBG) + P(GGGB) = 4/16 The probability distribution of a random variable is a table that lists the possible values of the random variables and their associated probabilities. P(x) 1/16 4/16 6/16 4/16 1/16 16/16=1 P r o b a b ility D is trib u tio n o f t h e N u m b e r o f G irls in F o u r B ir th s 0 .4 0 .3 7 5 0 0 .3 0 .2 5 0 0 P(x) x 0 1 2 3 4 0 .2 5 0 0 0 .2 0 .1 0 .0 6 2 5 0 0 .0 6 2 5 1 2 3 4 N u m b e r o f g irls , x Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
7. 7. Example 3-1 Slide 7 Consider the experiment of tossing two six-sided dice. There are 36 possible outcomes. Let the random variable X represent the sum of the numbers on the two dice: 3 1,3 2,3 3,3 4,3 5,3 6,3 4 1,4 2,4 3,4 4,4 5,4 6,4 5 1,5 2,5 3,5 4,5 5,5 6,5 6 1,6 2,6 3,6 4,6 5,6 6,6 7 8 9 10 11 12 P(x)* 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 1 Po a tyDstrib tio u w ic r b bili i u nofSmofT oD e 0.17 0.12 p(x) 1,1 2,1 3,1 4,1 5,1 6,1 2 1,2 2,2 3,2 4,2 5,2 6,2 x 2 3 4 5 6 7 8 9 10 11 12 0.07 0.02 2 3 4 5 6 7 8 9 10 x * Note that: P(x)  (6  (7  x) 2 ) / 36 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 11 12
8. 8. Example 3-2 Slide 8 Probability Distribution of the Number of Switches P(x) 0.1 0.2 0.3 0.2 0.1 0.1 1 TePr ba yD b o fth u berofS itc es h o bilit istri uti no eNm w h 0.4 0.3 P(x) x 0 1 2 3 4 5 0.2 0.1 0.0 0 1 2 3 4 5 x Probability of more than 2 switches: P(X > 2) = P(3) + P(4) + P(5) = 0.2 + 0.1 + 0.1 = 0.4 Probability of at least 1 switch: P(X ³ 1) = 1 - P(0) = 1 - 0.1 = .9 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
9. 9. Discrete and Continuous Random Variables   Slide 9 A discrete random variable:  has a countable number of possible values  has discrete jumps (or gaps) between successive values has measurable probability associated with individual values  counts A continuous random variable:  has an uncountably infinite number of possible values  moves continuously from value to value  has no measurable probability associated with each value measures (e.g.: height, weight, speed, value, duration, length) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
10. 10. Rules of Discrete Probability Distributions Slide 10 The probability distribution of a discrete random variable X must satisfy the following two conditions. 1. P(x)  0 for all values of x. 2.  P(x)  1 all x Corollary: 0  P( X )  1 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
11. 11. Cumulative Distribution Function Slide 11 The cumulative distribution function, F(x), of a discrete random variable X is: F(x)  P( X  x)   P(i) all i  x P(x) 0.1 0.2 0.3 0.2 0.1 0.1 F(x) 0.1 0.3 0.6 0.8 0.9 1.0 1 Cumulative Prob ability Distribution of the Numb er of Switche s 1 .0 0 .9 0 .8 0 .7 F(x) x 0 1 2 3 4 5 0 .6 0 .5 0 .4 0 .3 0 .2 0 .1 0 .0 0 1 2 3 4 5 x Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
12. 12. Cumulative Distribution Function Slide 12 The probability that at most three switches will occur: The P ro b a bility Tha t at Mo s t T hre e S witc he s W ill O c c ur P(x) 0.1 0.2 0.3 0.2 0.1 0.1 F(x) 0.1 0.3 0.6 0.8 0.9 1.0 0 .4 P( X  3)  F(3) 0 .3 P (x) x 0 1 2 3 4 5 0 .2 0 .1 1 0 .0 0 1 2 3 4 5 x Note: P(X < 3) = F(3) = 0.8 = P(0) + P(1) + P(2) + P(3) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
13. 13. Using Cumulative Probability Distributions (Figure 3-8) Slide 13 The probability that more than one switch will occur: The Probability That More than One Switch Will Occur P(x) 0.1 0.2 0.3 0.2 0.1 0.1 F(x) 0.1 0.3 0.6 0.8 0.9 1.0 0 .4 0 .3 P ( x) x 0 1 2 3 4 5 P( X  1  1 F(1 ) ) F(1 ) 0 .2 0 .1 1 0 .0 0 1 2 3 4 5 x Note: P(X > 1) = P(X > 2) = 1 – P(X < 1) = 1 – F(1) = 1 – 0.3 = 0.7 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
14. 14. Using Cumulative Probability Distributions (Figure 3-9) Slide 14 The probability that anywhere from one to three switches will occur: The Probability That Anywhere from One to Three Switches Will Occur P(x) 0.1 0.2 0.3 0.2 0.1 0.1 F(x) 0.1 0.3 0.6 0.8 0.9 1.0 1 0 .4 F(1 X  3)  F(3)  F(0) F(3) 0 .3 P (x) x 0 1 2 3 4 5 0 .2 F(0) 0 .1 0 .0 0 1 2 3 4 5 x Note: P(1 < X < 3) = P(X < 3) – P(X < 0) = F(3) – F(0) = 0.8 – 0.1 = 0.7 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
15. 15. 3-2 Expected Values of Discrete Random Variables The mean of a probability distribution is a measure of its centrality or location, as is the mean or average of a frequency distribution. It is a weighted average, with the values of the random variable weighted by their probabilities. 0 1 2 3 Slide 15 4 5 2.3 The mean is also known as the expected value (or expectation) of a random variable, because it is the value that is expected to occur, on average. The expected value of a discrete random variable X is equal to the sum of each value of the random variable multiplied by its probability. m  E ( X )   xP ( x ) x P(x) xP(x) 0 0.1 0.0 1 0.2 0.2 2 0.3 0.6 3 0.2 0.6 4 0.1 0.4 5 0.1 0.5 1.0 2.3 = E(X) = m all x Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
16. 16. A Fair Game Slide 16 Suppose you are playing a coin toss game in which you are paid \$1 if the coin turns up heads and you lose \$1 when the coin turns up tails. The expected value of this game is E(X) = 0. A game of chance with an expected payoff of 0 is called a fair game. x P(x) xP(x) -1 0.5 -0.50 1 0.5 0.50 1.0 0.00 = E(X)=m -1 0 1 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
17. 17. Expected Value of a Function of a Discrete Random Variables Slide 17 The expected value of a function of a discrete random variable X is: E [ h ( X )]   h ( x ) P ( x ) all x Example 3-3: Monthly sales of a certain product are believed to follow the given probability distribution. Suppose the company has a fixed monthly production cost of \$8000 and that each item brings \$2. Find the expected monthly profit h(X), from product sales. E [ h ( X )]   h ( x ) P ( x )  5400 all x of items, x 5000 6000 7000 8000 9000 Number P(x) xP(x) h(x) h(x)P(x) 0.2 1000 2000 400 0.3 1800 4000 1200 0.2 1400 6000 1200 0.2 1600 8000 1600 0.1 900 10000 1000 1.0 6700 5400 Note: h(X) = 2X – 8000 where X = # of items sold The expected value of a linear function of a random variable is: E(aX+b)=aE(X)+b In this case: E(2X8000)2E(X)8000(2)(6700)80005400 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
18. 18. Variance and Standard Deviation of a Random Variable Slide 18 The variance of a random variable is the expected squared deviation from the mean: s 2  V ( X )  E [( X  m ) 2 ]   (x  m ) 2 P(x) a ll x      E ( X 2 )  [ E ( X )] 2    x 2 P ( x )     xP ( x )   a ll x   a ll x  2 The standard deviation of a random variable is the square root of its variance: s  SD( X )  V ( X ) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
19. 19. Variance and Standard Deviation of a Slide 19 Random Variable – using Example 3-2 s 2 = V ( X ) = E[( X - m)2 ] Table 3-8 Switches, x 0 1 2 3 4 5 m Recall: P(x) 0.1 0.2 0.3 0.2 0.1 0.1 Number of xP(x) (x-m) (x-m)2 P(x-m)2 0.0 -2.3 5.29 0.529 0.2 -1.3 1.69 0.338 0.6 -0.3 0.09 0.027 0.6 0.7 0.49 0.098 0.4 1.7 2.89 0.289 0.5 2.7 7.29 0.729 2.3 2.010 = 2.3. x2P(x) 0.0 0.2 1.2 1.8 1.6 2.5 7.3 = å ( x - m)2 P( x) = 2.01 all x = E( X 2 ) - [ E( X )]2 2 é ù é ù = ê å x2 P( x)ú - ê å xP( x)ú ëall x û ëall x û = 7.3 - 2.32 = 2.01 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
20. 20. Variance of a Linear Function of a Random Variable Slide 20 The variance of a linear function of a random variable is: V(a X b)  a2V( X)  a2s2 Example 3-3: Number of items, x P(x) xP(x) x2 P(x) 5000 0.2 1000 5000000 6000 0.3 1800 10800000 7000 0.2 1400 9800000 8000 0.2 1600 12800000 9000 0.1 900 8100000 1.0 6700 46500000 s2 V(X)  E ( X 2 )  [ E ( X )]2 2     2   x P( x )    xP( x ) all x  all x   46500000  ( 67002 )  1610000 s  SD( X )  1610000  1268.86 V (2 X  8000)  (2 2 )V ( X )  ( 4)(1610000)  6440000 s ( 2 x  8000)  SD(2 x  8000)  2s x  (2)(1268.86)  2537.72 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
21. 21. Some Properties of Means and Variances of Random Variables Slide 21 The mean or expected value of the sum of random variables is the sum of their means or expected values: m( XY)  E( X Y)  E( X)  E(Y)  mX  mY For example: E(X) = \$350 and E(Y) = \$200 E(X+Y) = \$350 + \$200 = \$550 The variance of the sum of independent random variables is the sum of their variances: s 2 ( X Y )  V ( X  Y)  V ( X ) V (Y)  s 2 X  s 2 Y if and only if X and Y are independent. For example: V(X) = 84 and V(Y) = 60 V(X+Y) = 144 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
22. 22. Chebyshev’s Theorem Applied to Probability Distributions Slide 22 Chebyshev’s Theorem applies to probability distributions just as it applies to frequency distributions. For a random variable X with mean m, standard deviation s, and for any number k > 1: 1 P( X  m  ks)  1 2 k 1 At least 1 1 3  1    75% 2 4 4 2 1 1 8 1  2  1    89% 9 9 3 1 1 15 1 2  1   94% 16 16 4 2 Lie within 3 4 Standard deviations of the mean Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
23. 23. Using the Template to Calculate statistics of h(x) Slide 23 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
24. 24. 3-3 Bernoulli Random Variable Slide 24 • If an experiment consists of a single trial and the outcome of the trial can only be either a success* or a failure, then the trial is called a Bernoulli trial. • The number of success X in one Bernoulli trial, which can be 1 or 0, is a Bernoulli random variable. • Note: If p is the probability of success in a Bernoulli experiment, the E(X) = p and V(X) = p(1 – p). * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
25. 25. The Binomial Random Variable Slide 25 Consider a Bernoulli Process in which we have a sequence of n identical trials satisfying the following conditions: 1. Each trial has two possible outcomes, called success *and failure. The two outcomes are mutually exclusive and exhaustive. 2. The probability of success, denoted by p, remains constant from trial to trial. The probability of failure is denoted by q, where q = 1-p. 3. The n trials are independent. That is, the outcome of any trial does not affect the outcomes of the other trials. A random variable, X, that counts the number of successes in n Bernoulli trials, where p is the probability of success* in any given trial, is said to follow the binomial probability distribution with parameters n (number of trials) and p (probability of success). We call X the binomial random variable. * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
26. 26. Binomial Probabilities (Introduction) Slide 26 Suppose we toss a single fair and balanced coin five times in succession, and let X represent the number of heads. There are 25 = 32 possible sequences of H and T (S and F) in the sample space for this experiment. Of these, there are 10 in which there are exactly 2 heads (X=2): HHTTT HTHTH HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH The probability of each of these 10 outcomes is p3q3 = (1/2)3(1/2)2=(1/32), so the probability of 2 heads in 5 tosses of a fair and balanced coin is: P(X = 2) = 10 * (1/32) = (10/32) = .3125 10 (1/32) Number of outcomes with 2 heads Probability of each outcome with 2 heads Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
27. 27. Binomial Probabilities (continued) Slide 27 P(X=2) = 10 * (1/32) = (10/32) = .3125 Notice that this probability has two parts: 10 (1/32) Number of outcomes with 2 heads Probability of each outcome with 2 heads In general: 1. The probability of a given sequence of x successes out of n trials with probability of success p and probability of failure q is equal to: pxq(n-x) 2. The number of different sequences of n trials that result in exactly x successes is equal to the number of choices of x elements out of a total of n elements. This number is denoted: n!  n nCx      x x!( n  x)! Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
28. 28. The Binomial Probability Distribution The binomial probability distribution: n!  n x ( n  x ) P( x)    p q  px q ( n x) x!( n  x)!  x where : p is the probability of success in a single trial, q = 1-p, n is the number of trials, and x is the number of successes. N u m b er o f su ccesses, x 0 1 2 3  n Slide 28 P ro b ab ility P (x ) n! p 0 q (n 0) 0 !( n  0 ) ! n! p 1 q ( n  1) 1 !( n  1 ) ! n! p 2 q (n 2) 2 !( n  2 ) ! n! p 3 q (n 3) 3 !( n  3 ) !  n! p n q (n n) n !( n  n ) ! 1 .0 0 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
29. 29. The Cumulative Binomial Probability Table (Table 1, Appendix C) Slide 29 n=5 p x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99 0 .951 .774 .590 .328 .168 .078 .031 .010 .002 .000 .000 .000 .000 1 .999 .977 .919 .737 .528 .337 .187 .087 .031 .007 .000 .000 .000 2 1.000 .999 .991 .942 .837 .683 .500 .317 .163 .058 .009 .001 .000 3 1.000 1.000 1.000 .993 .969 .913 .813 .663 .472 .263 .081 .023 .001 4 1.000 1.000 1.000 1.000 .998 .990 .969 .922 .832 .672 .410 .226 .049 h F(h) P(h) 0 0.031 0.031 1 0.187 0.156 2 0.500 0.313 3 0.813 0.313 4 0.969 0.156 5 1.000 0.031 1.000 Cumulative Binomial Probability Distribution and Binomial Probability Distribution of H,the Number of Heads Appearing in Five Tosses of a Fair Coin Deriving Individual Probabilities from Cumulative Probabilities F (x )  P ( X  x )   P(i ) all i  x P(X) = F(x) - F(x - 1) For example: P (3)  F (3)  F (2)  .813 .500  .313 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
30. 30. Calculating Binomial Probabilities Example Slide 30 60% of Brooke shares are owned by LeBow. A random sample of 15 shares is chosen. What is the probability that at most three of them will be found to be owned by LeBow? n=15 0 1 2 3 4 ... .50 .000 .000 .004 .018 .059 ... p .60 .000 .000 .000 .002 .009 ... .70 .000 .000 .000 .000 .001 ... F (x )  P ( X  x )   P (i ) all i  x F ( 3)  P ( X  3)  .002 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
31. 31. Mean, Variance, and Standard Slide 31 Deviation of the Binomial Distribution Mean of a binomial distribution: m  E ( X )  np For example, if H counts the number of heads in five tosses of a fair coin : Variance of a binomial distribution: m  E ( H )  (5)(.5)  2.5 H s  V ( X )  npq 2 s  V ( H )  (5)(.5)(.5)  1.25 2 H Standard deviation of a binomial distribution: s = SD(X) = npq s  SD( H )  1.25  1.118 H Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
32. 32. Calculating Binomial Probabilities using the Template Slide 32 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
33. 33. Shape of the Binomial Distribution p  0.1 Slide 33 p  0.3 Binomial Probability: n=4 p=0.1 p  0.5 Binomial Probability: n=4 p=0.3 Binomial Probability: n=4 p=0.5 0.6 0.5 0.5 0.4 0.4 0.4 0.3 P(x) 0.6 0.5 P(x) 0.7 0.6 n4 0.7 P(x) 0.7 0.3 0.2 0.2 0.1 0.1 0.0 0.2 0.1 0.0 0 1 2 3 0.3 0.0 4 0 1 2 x 3 4 0 Binomial Probability: n=10 p=0.1 Binomial Probability: n=10 p=0.3 4 0.4 0.4 0.3 0.3 0.3 P(x) 0.5 P(x) P(x) 3 Binomial Probability: n=10 p=0.5 0.5 0.4 0.2 0.2 0.2 0.1 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 0.0 0 1 2 3 4 5 6 7 8 0 9 10 x Binomial Probability: n=20 p=0.3 1 2 x Binomial Probability: n=20 p=0.1 0.1 4 5 x 6 7 8 9 10 P(x) 0.2 0.1 0.0 3 Binomial Probability: n=20 p=0.5 0.2 P(x) 0.2 P(x) n  20 2 x 0.5 n  10 1 x 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 1011121314151617181920 0 1 2 3 4 5 6 7 8 9 1011121314151617181920 0 1 2 3 4 5 6 7 8 9 1011121314151617181920 x x x Binomial distributions become more symmetric as n increases and as p .5. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
34. 34. 3-5 Negative Binomial Distribution Slide 34 The negative binomial distribution is useful for determining the probability of the number of trials made until the desired number of successes are achieved in a sequence of Bernoulli trials. It counts the number of trials X to achieve the number of successes s with p being the probability of success on each trial. Negative Binomial Distribution : P( X  x)  ( ) x 1 s 1 s p (1  p ) ( x  s) The mean is : m  s p The variance is : s 2  s (1  p ) p2 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
35. 35. Negative Binomial Distribution Example Example: Suppose that the probability of a manufacturing process producing a defective item is 0.05. Suppose further that the quality of any one item is independent of the quality of any other item produced. If a quality control officer selects items at random from the production line, what is the probability that the first defective item is the eight item selected. Slide 35 Heres = 1, x = 8, and p = 0.05. Thus,  8  1 P( X  8)   0.05 (1  0.05) 1  1   0.0349 1 ( 81 ) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
36. 36. Calculating Negative Binomial Probabilities using the Template Slide 36 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
37. 37. 3-6 The Geometric Distribution Slide 37 Within the context of a binomial experiment, in which the outcome of each of n independent trials can be classified as a success (S) or a failure (F), the geometric random variable counts the number of trials until the first success.. Geometric distribution: x1 P ( x )  pq where x = 1,2,3, . . . and p and q are the binomial parameters. The mean and variance of the geometric distribution are: m 1 p s 2  q 2 p Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
38. 38. The Geometric Distribution Example Slide 38 Example: A recent study indicates that Pepsi-Cola has a market share of 33.2% (versus 40.9% for Coca-Cola). A marketing research firm wants to conduct a new taste test for which it needs Pepsi P(1)  (.332)(.668)(11)  0332 . drinkers. Potential participants for the ( 21) test are selected by random screening of P(2)  (.332)(.668)  0222 . soft drink users to find Pepsi drinkers. P(3)  (.332)(.668)( 31)  0148 . What is the probability that the first ( 41)  0.099 randomly selected drinker qualifies? P( 4)  (.332)(.668) What’s the probability that two soft drink users will have to be interviewed to find the first Pepsi drinker? Three? Four? Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
39. 39. Calculating Geometric Distribution Probabilities using the Template Slide 39 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
40. 40. 3-7 The Hypergeometric Distribution Slide 40 The hypergeometric probability distribution is useful for determining the probability of a number of occurrences when sampling without replacement. It counts the number of successes (x) in n selections, without replacement, from a population of N elements, S of which are successes and (N-S) of which are failures. H ypergeom etric D istribution: The mean of the hypergeometric distribution is: m  np , where p        X n  x   N    n S P(x)  N  S The variance is: s 2  N  n  npq     N  1 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer S N
41. 41. The Hypergeometric Distribution Example Example: Suppose that automobiles arrive at a dealership in lots of 10 and that for time and resource considerations, only 5 out of each 10 are inspected for safety. The 5 cars are randomly chosen from the 10 on the lot. If 2 out of the 10 cars on the lot are below standards for safety, what is the probability that at least 1 out of the 5 cars to be inspected will be found not meeting safety standards? () ( )( ) () ()  ( ) ( ) ( ) ( )( ) () ()  (10  2)   1  (5  1)  2 P (1)  2 4  10  5 2 10  2 1 5 2 2 2! 3  10  10 5 1! 1! 4 ! 4 ! 5   0.556 9 5! 5! 8 1 8! 10 ! 10 5 P( 2)  2! 8 1 Slide 41 5 8! 1! 1! 3 ! 5! 10 !  2 9 5! 5! Thus, P(1) + P(2) = 0.556 + 0.222 = 0.778. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer  0.222
42. 42. Slide 42 Calculating Hypergeometric Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
43. 43. 3-8 The Poisson Distribution Slide 43 The Poisson probability distribution is useful for determining the probability of a number of occurrences over a given period of time or within a given area or volume. That is, the Poisson random variable counts occurrences over a continuous interval of time or space. It can also be used to calculate approximate binomial probabilities when the probability of success is small (p£0.05) and the number of trials is large (n³20). Poisson D istribution : m xe m P( x)  for x = 1,2,3,... x! where m is the mean of the distribution (which also happens to be the variance) and e is the base of natural logarithms (e=2.71828...). Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
44. 44. The Poisson Distribution Example Example 3-5: Telephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 200 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers? Slide 44 .2 0 e - .2 P ( 0) = = 0.8187 0! .2 .21 e P (1) = = 0.1637 1! - .2 .2 2 e P (2) = = 0.0164 2 !- .2 .2 3 e P ( 3) = = 0.0011 3! n = 200 m = np = (200)(0.001) = 0.2 p = 1/1000 = 0.001 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
45. 45. Slide 45 Calculating Poisson Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
46. 46. The Poisson Distribution (continued) • Slide 46 Poisson assumptions: The probability that an event will occur in a short interval of time or space is proportional to the size of the interval. In a very small interval, the probability that two events will occur is close to zero. The probability that any number of events will occur in a given interval is independent of where the interval begins. The probability of any number of events occurring over a given interval is independent of the number of events that occurred prior to the interval. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
47. 47. The Poisson Distribution (continued) m =1.5 m =1.0 0.4 0.3 0.3 ) P( x 0.4 P(x) Slide 47 0.2 0.1 0.2 0.1 0.0 0.0 0 1 2 3 4 0 1 2 3 4 X m =4 5 6 7 X m =10 0.2 0.15 P (x) P(x) 0.10 0.1 0.05 0.0 0.00 0 1 2 3 4 5 X 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 1011121314151617181920 X Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
48. 48. Discrete and Continuous Random Variables - Revisited • A continuous random variable: • A discrete random variable: – measures (e.g.: height, weight, – – possible values has discrete jumps between successive values has measurable probability associated with individual values probability is height For example: Binomial n=3 p=.5 P(x) 0.125 0.375 0.375 0.125 1.000 0.4 0.3 P(x ) x 0 1 2 3 Binomial: n=3 p=.5 0.2 0.1 0.0 0 1 2 C1 3 – – – – speed, value, duration, length) has an uncountably infinite number of possible values moves continuously from value to value has no measurable probability associated with individual values probability is area For example: In this case, the shaded area epresents the probability that the task takes between 2 and 3 minutes. Minute s to Co mplete Tas k 0.3 0.2 P(x) – counts occurrences – has a countable number of – Slide 48 0.1 0.0 1 2 3 4 Minutes Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 5 6
49. 49. From a Discrete to a Continuous Distribution Slide 49 The time it takes to complete a task can be subdivided into: Half-Minute Intervals Eighth-Minute Intervals Quarter-Minute Intervals M inutes to Complete Task: Fourths of a Minute Minute s to Complete Task: By Half-Minute s M inutes to Complete Task: Eighths of a M inute 0.15 P(x) P(x) P(x) 0.10 0.05 0.00 0.01.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 . 0 M inutes 1 2 3 4 5 6 7 0 1 2 Minute s 3 4 5 6 7 M inutes Or even infinitesimally small intervals: f(z) Minutes to Complete Task: Probability Density Function 0 1 2 3 4 5 6 When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability density function corresponding to that interval. In this example, the shaded area represents P(2 £ X £ 3). 7 Minutes Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
50. 50. 3-9 Continuous Random Variables Slide 50 A continuous random variable is a random variable that can take on any value in an interval of numbers. The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x), has the following properties. 1. 2. 3. f(x) ³ 0 for all x. The probability that X will be between two numbers a and b is equal to the area under f(x) between a and b. The total area under the curve of f(x) is equal to 1.00. The cumulative distribution function of a continuous random variable: F(x) = P(X £ x) =Area under f(x) between the smallest possible value of X (often -¥) and the point x. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
51. 51. Probability Density Function and Cumulative Distribution Function F(x) 1 F(b) F(a) 0 f(x) } a b Slide 51 P(a £ X £ b)=F(b) - F(a) x P(a £ X £ b) = Area under f(x) between a and b = F(b) - F(a) 0 a b x Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
52. 52. 3-10 Uniform Distribution Slide 52 The uniform [a,b] density: { 1/(a – b) for a £ X £ b f(x)= 0 otherwise E(X) = (a + b)/2; V(X) = (b – a)2/12 f(x) Uniform [a, b] Distribution The entire area under f(x) = 1/(b – a) * (b – a) = 1.00 The area under f(x) from a1 to b1 = P(a1£X£ b1) a a 1 x b 1 b = (b1 – a1)/(b – a) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
53. 53. Uniform Distribution (continued) Slide 53 The uniform [0,5] density: { 1/5 for 0 £ X £ 5 f(x)= 0 otherwise E(X) = 2.5 The entire area under f(x) = 1/5 * 5 = 1.00 Uniform [0,5] Distribution 0.5 0.4 f(x) 0.3 The area under f(x) from 1 to 3 = P(1£X£3) 0.2 0.1 .0.0 -1 0 1 2 3 4 5 6 x = (1/5)2 = 2/5 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
54. 54. Slide 54 Calculating Uniform Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
55. 55. 3-11 Exponential Distribution Slide 55 E xp o n e ntial Dis trib utio n: l = 2 The exponential random variable measures the time between two occurrences that have a Poisson distribution. 2 Exponential distribution: f(x) The density function is: f (x)  lelx for x  0, l  0 1 1 The mean and standard deviation are both equal to . l The cumulative distribution function is: F(x)  1  elx for x  0. 0 0 1 2 3 Time Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
56. 56. Exponential Distribution Example Slide 56 Example The time a particular machine operates before breaking down (time between breakdowns) is known to have an exponential distribution with parameter l = 2. Time is measured in hours. What is the probability that the machine will work continuously for at least one hour? What is the average time between breakdowns? F (x ) = 1 - e - l x Þ P( X ³ x ) = e - l x P ( X ³ 1) = e ( - 2 )(1) =.1353 E(X ) = 1 1 = =.5 l 2 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
57. 57. Slide 57 Calculating Exponential Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
58. 58. Slide 58 Name Religion Domicile Contact # E.Mail M.Phil (Statistics) Shakeel Nouman Christian Punjab (Lahore) 0332-4462527. 0321-9898767 sn_gcu@yahoo.com sn_gcu@hotmail.com GC University, . (Degree awarded by GC University) M.Sc (Statistics) Statitical Officer (BS-17) (Economics & Marketing Division) GC University, . (Degree awarded by GC University) Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development Department, Govt. of Punjab Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer