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1 Probability distribution Dr. Deshi Ye College of Computer Science, Zhejiang University

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Presentation on theme: "1 Probability distribution Dr. Deshi Ye College of Computer Science, Zhejiang University"— Presentation transcript:

1 1 Probability distribution Dr. Deshi Ye College of Computer Science, Zhejiang University yedeshi@zju.edu.cn

2 2 Outline  Random variable  The Binomial distribution  The Hypergeometric Distribution  The Mean and the Variance of the a Probability distribution.

3 3 Random variables  We concern with one number or a few number that associated with the outcomes of experiments.  EX. Inspection: number of defectives road test: average speed and average fuel consumption. In the example of tossing dice, we are interested in sum 7 and not concerned whether it is (1,6) or (2, 5) or (3, 4) or (4, 3) or (5, 2) or (6, 1).

4 4 Definition  A random variable: is any function that assigns a numerical value to each possible outcome of experiments.  Discrete random variable: only a finite or a countable infinity of values. Otherwise, continuous random variables.

5 5 Probability distribution  The probability distribution of the random variable: is the probabilities that a random variable will take on any one value within its range.  The probability distribution of a discrete random variable X is a list of possible values of X together with their probabilities The probability distribution always satisfies the conditions

6 6 Checking probability distribution x 0 1 2 3 Prob..26.5.22.02 Another 1)f(x) = (x-2)/2, for x=1,2, 3, 4 2)H(x) =x 2 /25, x=0,1,2,3,4

7 7 Probability histogram & bar chart

8 8

9 9 Bar chart

10 10 Histogram

11 11 Histogram

12 12 Cumulative distribution  F( x ): value of a random variable is less than or equal to x.

13 13 EX. x 0 1 2 3 Prob..26.76.98 1.0

14 14 Binomial Distribution  Foul Shot: 1. Min Yao (Hou).862. 2. O’Neal Shaquille.422  The Question is: what is the probability of them in two foul shots that they get 2 points, respectively?

15 15 Binomial distribution  Study the phenomenon that the probability of success in repeat trials.  Prob. of getting x “success” in n trials, otherwords, x “success” and n – x failures in n attempt.

16 16 Bernoulli trials  1. There are only two possible outcomes for each trial.  2. The probability of success is the same for each trial.  3. The outcomes from different trials are independent.  4’. There are a fixed number n of Bernoulli trials conducted.

17 17  Let X be the random variable that equals the number of success in n trials. p and 1- p are the probability of “success” and “failure”, the probability of getting x success and n- x failure is

18 18 Def. of Binomial Dist.  #number of ways in which we can select the x trials on which there is to be a success is  Hence the probability distribution of Binomial is

19 19 Expansions Binomial coefficient

20 20 Table 1  Table 1: Cumulative Binomial distribution

21 21 EX Solve  Foul shot example: Here n=2, x=2, and p=0.862 for Yao, and p=0.422 for Shaq. 0 1 2 Yao.02.24.74 Shaq..33.49.18

22 22 Bar Chats

23 23 Minitab for Binomial

24 24

25 25

26 26

27 27 Skewed distribution Positively skewed

28 28 Hypergeometric Distr.  Sampling with replacement  Sampling without replacement

29 29 Hyergeometric distr.  Sampling without replacement  The number of defectives in a sample of n units drawn without replacement from a lot containing N units, of which are defectives. Here: population is N, and are total defectives Sampling n units, what is probability of x defectives are found?

30 30 formulations  Hypergeometric distr.

31 31 Discussion  Is Hypergeometric distribution a Bernolli trial?  Answer: NO! The first drawing will yield a defect unit is a/N, but the second is (a - 1)/(N-1) or a/(N-1).

32 32 EX  A shipment of 20 digital voice recorders contains 5 that are defective. If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective?  Solution: a=5, n=10, N=20, and x=2

33 33

34 34 Expectation  Expectation: If the probability of obtaining the amounts then the mathematical expectation is

35 35 Motivations  The expected value of x is a weighted average of possible values that X can take on, each value being weighted by the probability that X assumes it.  Frequency interpretation

36 36 4.4 The mean and the variance  Mean and variance: two distributions differ in their location or variation  The mean of a probability distribution is simply the mathematical expectation of a random variable having that distribution.  Mean of discrete probability distribution Alternatively,

37 37 EX  The mean of a probability distribution measures its center in the sense of an average.  EX: Find the mean number of heads in three tosses.  Solution: The probabilities for 0, 1, 2, or 3 heads are 1/8, 3/8, 3/8, and 1/8

38 38 Mean of Binomial distribution  Contrast: please calculate the following

39 39 Mean of b()  Mean of binomial distribution: Proof.

40 40 Mean of Hypergeometric Distr. Proof. Similar proof or using the following hints:

41 41 EX  5 of 20 digital voice records were defectives,find the mean of the prob. Distribution of the number of defectives in a sample of 10 randomly chose for inspection.  Solution: n=10, a= 5, N=20. Hence

42 42 Expectation of a function of random variable  Let X denote a random variable that takes on any of the values -1,0,1 respective probabilities P{x=-1}=0.2, P{x=0}=0.5, P{x=1}=0.3  Compute E[X 2 ] Answer = 0.5

43 43 Proposition  If X is a discrete random variable that takes on one of the value of x i, with respective to probability p(x i ), then for any real-valued function g,

44 44 Variance of probability  Variance of a probability distribution f(x), or that of the random variable X which has that probability distribution, as We could also denote it as

45 45 Standard deviation  Standard deviation of probability distribution

46 46 Relation between Mean and Variance

47 47 Ex.  Find the variance of the number of heads in four tosses.  Solution:

48 48 Variance of binomial distr.  Variance of binomial distribution: Proof. Detailed proof after the section of disjoint probability distribution

49 49 Some properties of Mean  C is a constant, then E(C) = C.  X is a random variable and C is a constant E(CX) = CE(X)  X and Y are two random variables, then E(X+Y) = E(X)+E(Y) If X and Y are independent random variables E(XY) = E(X)E(Y)

50 50 Variance of hypergeometric distr.

51 51 K-th moment  K-th moment about the origin  K-th moment about the mean

52 Case study  Occupancy Problem 52

53 53 Homework  problems


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