Statistical Experiment A statistical experiment is any process by which measurements are obtained.

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1 (التوزيعات الا حتمالية ( Distributions Probability Statistical Experiment A statistical experiment is any process by which measurements are obtained. Examples of Statistical Experiments Counting the number of trees in a field. Counting the number of mistakes on a page. Measuring the amount of rainfall during the month of January. Random Variable A random variable is a numerical value determined by the outcome of an experiment that varies from trial to trial. Example Consider a random experiment in which a coin is tossed three times. Let x be the number of heads. Let H represent the outcome of a head and T the outcome of a tail. The possible outcomes for such an experiment will be: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH. Thus the possible values of x (number of heads) are x = 0: TTT x = 1: TTH, THT, HTT x = 2: THH, HTH, HHT x = 3: HHH From the definition of a random variable, x as defined in this experiment, is a random variable. Classes of random variables 1. Discrete Random Variable A discrete random variable is a quantitative random variable that can be obtained by counting, and can take on only a finite number of values or a countable whole number (0, 1, 2, 3,.., etc.) of values (Poisson random variable is exception ( )). Examples The number of children per family (0, 1, 2, 3, 4, 5,.,etc.) The number of cavities a patient has in a year (0, 1, 2, 3, 4, 5,.,etc.). The number of bacteria which survive treatment with some antibiotic. The number of times a person had a cold in Gaza Strip. 2. Continuous Random Variable A continuous random variable is a quantitative random variable that can take infinite number of values within an interval. 42

2 Examples The amount of rainfall in your city during the month of January. The distance students travel to class. The time it takes an executive to drive to work. The length of an afternoon nap. The length of time of a particular phone call. Probability Distribution It is a table, graph, formula, or other device used to specify all possible values of a random variable along with their respective probabilities. Depending on the variable, the probability distribution can be classified into: Discrete probability distribution Continuous probability distribution Discrete probability distribution It is the probability distribution of a discrete random variable, include Binomial Distribution Poisson Distribution Other Distribution Example 1 Experiment: Toss 2 coins. Count number of tails. Probability distribution Probability Histogram Value of (x) Probabilities P(x) 0 1/4 = /4 = /4 = 0.25 Fig. p(x) x Graphical representation of the Probability distribution of number of tails of two coins tossed Features of a Probability Distribution Probabilities must be between zero and one (0 < P < 1) ΣP(x) =1 43

3 ) توزيع ذات حدين ( Distribution Binomial The binomial distribution is used to determine the probability of yes/no events the number of times a given outcome occurs in a given number of attempts. Examples If a certain drug is known to cause a side effect 10% of the time and if five patients are given this drug, what is the probability that four or more experience the side effect? The number of patients out of n that respond to treatment. The number of people in a class of n students that are asthmatic. Features of a Binomial Experiment 1. The experiment consists of n repeated trials. 2. Each trial has only two possible outcomes-success or failure, yes or no, girl or boy, sick or well, dead or alive, at risk or not at risk survived not survived, infected not infected, white nonwhite, or simply positive negative. 3. The outcome of each trial is independent of the outcomes of any other trial; that is, the outcome of one trial has no effect on the outcome of any other trial. 4. The probability of success, denoted by p, remains constant from trial to trial. The probability of a failure, 1- p, is denoted by q. Since each trial results in success or failure, p + q = 1 and q = 1 p. 6. Our interest is in the number of successes r occurring in the n trials. Binomial Probability Formula The probability of an event consisting of r successes out of n trials is P(r successes) = Where "!" represents the factorial function. n! = (n)(n - 1) (n - 2)(n - 3)... (1). For example, 4! = (4)(3)(2)(1) = 24 By definition, 0! = 1 n = the number of trials in an experiment r = the number of successes n - r = the number of failures p = the probability of success q = 1 - p, the probability of failure Example 1 n! p r!( n r)! r r q n What is the probability of having two girls and one boy, 3 boys, and at least one boy in a threechild family if the probability of having a boy is 0.5? Solution n! From the calculations in equation p r!( n r)! r r q n, it can be seen that 44

4 For two girls and one boy (n = 3, r = 2, p = 1/2, q = 1/2) 2 1 3! 1 1 P(2G, 1B) = = 3(.125) =.375 2!(3 2)! 2 2 For 3 boys considering n = 3, r = 3, i.e. boy is the success 3 0 3! 1 1 P(3B) = = 1 X (.125)X 1 =.125 3!(3 3)! 2 2 For at least 1 boy P (3B) ) + P(2B, 1G) + P(1B, 2G) = = Example 2 Ten individuals are treated surgically. For each individual there is a 70% chance of successful surgery. Among these 10 people, the number of successful surgeries follows a binomial distribution with n =10, and p = 0.7. What is the probability of exactly 5 successful surgeries? Solution n = 10, r = 5, p = 70% = 0.7, q = = ! ( 5) p r = = 0.3 5!(10 5)! = (252)(0.7 5 )(0.3 5 ) = Table for Binomial Probability The probabilities of a binomial term can be obtained by reading them directly from the binomial probability table usually found in the Appendix at the end of statistic textbooks. A small portion of this table is reproduced in the Table below. Table Using the Binomial Probability Table Portion of Binomial Probability Table p n r / Find the section labeled with your value of n. Find the entry in the column headed with your value of P and row labeled with the r value of interest. Entering Probability Table above for example with n = 3, P =.5, and r = 2, in two girls and one boy example we find that P(2) =

5 Example 3 The probability that a patient recovers from a rare blood disease is 0.4. If 10 people are known to have contracted this disease, what is the probability that (a) exactly 3 survive, (b) at least 8 survive, and (c) from 2 to 5 survive? Solution Let x be the number of people that survive. Consider the binomial probability table with n = 10, and P = 0.4. We find that (a) P (r = 3) = (b) P (r 8) = P (x = 8) + P (x = 9) + P (x = 10) = = (c) P(2 r 5) = P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) = = Mean and Standard Deviation of a Binomial Distribution The mean (expected value or expectation) is found by : The variance is found by: σ = np (1 p) = Example Chemotherapy provides a 5-year survival rate of 80% for a particular type of cancer. In a group of 20 cancer patients receiving chemotherapy for this type of cancer, the mean number surviving after 5 years is µ = = 16 and the standard deviation is σ = = 1.8. npq Common English expression and corresponding inequalities µ = np Table Common English expression and corresponding inequalities (consider a binomial experiment with n trials and r success) Expression Inequalities Four or more successes r 4 At least four successes That is, r = 4, 5, 6,., n No fewer than four successes Not less than four successes Four or fewer successes r 4 At most four successes That is, r = 0, 1, 2, 3, or 4 No more than four successes The number of successes does not exceed four More than four successes r > 4 The number of successes exceeds four That is r = 5, 6, 7,., n Fewer than four successes r < 4 The number of successes is not as large as four That is, r = 0, 1, 2, 3 46

6 Poisson Probability Distribution The binomial random variable is the number of successes in a fixed number of trials of some random experiment. Many processes however generate only a single type of outcome and not within a confined number of trials, but rather along a continuum of time or space. The Poisson probability distribution is often used to describe the occurrences of random and independent event within a specific physical environment i.e. with respect to (per unit) a time interval, a length interval, a fixed area or a particular volume. This distribution has been used extensively in health science. Our interest is in the Number of events that occur in Time interval, length, area, or volume Examples The number of patient in awaiting room in an hour. The number of serious injuries in a particular factory in a year. The number of bacterial colonies growing in a certain volume of broth medium. The number of blood cells in a certain number of squares in haemocytometer. The number of birds seen in a 10 min period. The number of times a three-old child has an ear infection in a year. Note: the Poisson distribution would not be used with an infectious disease because the events (people with the disease) would not be independent of each other The formula for the Poisson probability distribution The Poisson distribution can be described mathematically using the formula: r λ λ e P( r) = for r = 0, 1, 2,. r! Where λ (Greek letter lambda) is the mean number of events in a particular interval of time (λ > 0), e is the base of natural logs = , and r is the number of events (r = 0, 1, 2, 3, ) in corresponding interval of time, volume, area, and so forth. From the above formula, the only piece of information needed to characterize any given Poisson distribution is λ. Example 1 ( λ determination) A count was made of the red blood corpuscles in each of the 64 compartments of a haemocytometer with the following results: Number of corpuscles Frequency In this case, the given area is the hematocytometer compartment and the event is the number of RBCs. The total number of RBCs is =

7 The number of compartments is = 64 So the mean number of RBCs per compartment is = Since λ = 7, the probabilities of any number of corpuscles in a randomly selected compartment can be easily determined Example 2 An Urgent Care facility specializes in caring for minor injuries, colds, and flu. For the evening hours of 6-10 PM the mean number of arrivals is 4.0 per hour. What is the probability that in a randomly chosen hour, the number of arrival will be 2? Example 3 Suppose we are interested in the number of snake bite cases in a particular hospital in a year. Assume that the number of snake bite cases at the hospital has a Poisson (6) (i.e., λ = 6) distribution. a. What is the probability that in a randomly chosen year, the number of snake bite cases will be 7? b. What is the probability that the number of such cases will be less than 2? c. What is the expected number of snake bite cases in a year? a. Find the probability that 7 bites will occur in a randomly selected year. 7-6 (6). e pr= ( 7) = = = ! 5040 b. The second question asks us to find p(r <2). (6). e (6). e pr ( < 2) = pr ( = 1) + pr ( = 0) = + 1! 0! = e -6 + (6).e -6 = c. The expected number of snake bite cases in a year at this hospital is µ = λ = 6 cases Example 4 r λ e p( r = 2) = r! λ 2 4 e = 2! =.1465 Assume that the probability of dying from Leukaemia is described by a Poisson distribution with a mean of 2.3 deaths per children. What is the probability of observing a local death rate of 6 deaths per (assuming that nothing is acting to increase the rate)? (2.3). e pr ( = 6) = = = = 2.1% 6!

8 Mean and Variance of a Poisson Random Variable If x is a Poisson random variable with parameter µ, then Mean (expected number) µ = λ An interesting feature of the Poisson distribution is the fact that the mean and variance are equal. 2 2 Variance σ = λ and Standard Deviation σ = σ = λ Use of Tables A table of the Poisson probability distribution for selected values of λ and the number of successes r is often available in the Appendix of statistics books. Example: To find the value of p (r = 2) when λ = 0.3, look in the column labeled by 0.3 and the row labeled by 2. r λ Example 5 Bacteria are distributed independently of each other in a solution and it is known that the number of bacteria per millilitre follows a Poisson distribution with mean 2. Find the probability that a sample of 1 ml of solution contains a. One or fewer bacteria. By using the table we see that when λ = 2, p(r 1) = p(r = 1) + p(r = 0) = = b. More than five bacteria. By using the table we see that when λ = 2, p(r > 5) = 1 - p(r 5) = 1 [p (r = 5) + p(r = 4) + p(r = 3) + p(r = 2) + p(r = 1) + p(r = 0)] =1 [ = =

9 Poisson approximation of the Binomial Probability Distribution The Poisson distribution can be used instead of binomial distribution to approximate the probabilities of a discrete random variable x which has a Binomial (n, p) distribution when a binomial experiment has a quite low p and high n Sources are not in agreement with respect to what values of n and p are appropriate for employing the Poisson approximation of the binomial distribution. Hogg and Tanis (1997) recommend the approximation if n 20 and p 0.05 Daniel and Terrell (1995) state the approximation is usually very good when n 100, p 0.1 and np 10. Rosner (1995) states the more conservative criterion that n 100 and p Example 6 Assume that there is a.03 probability of a specific microorganism growing in a culture. What is the probability that in a batch of 200 cultures five of the cultures will contain the microorganism? Compute the binomial probability for n = 200, r = 5, p =.03, and q =.97. The probability that in a batch of 200 cultures 5 will contain the microorganism. 200! pr ( = 5) = !(200-5)! = The Poisson estimate of the binomial probability is determined as follows. First, we compute the expected value µ = λ = np = = 6. We then calculate the probability that in a batch of 200 cultures 5 will contain the microorganism. r λ 5 6 λ e 6 e P( r = 5) = = =.1606 r! 5! Note that the value p(r = 5) =.1606 is very close to the exact binomial probability of p(r = 5) = Example 7 Suppose that the probability that a person making hajj suffers from heat exhaustion on the Day of Arafat (when the temperature is 38 C or more) is 1 in 2500, that is 1/2500 = If we can assume that one person s suffering from heat exhaustion is independent from another s and we observe 1000 people making hajj on such day, what is the probability that at least one of these is suffering heat exhaustion? What is the expected number to be suffering from heat exhaustion? Since n = 1000 is large and p = is small, binomial probabilities are difficult to calculate so we use the Poisson approximation. We are asked to find p(r 1) e (0.4) 0.4 p(r 1) = 1 p (r < 1) = 1 p (r = 0) = 1 = 1 e = ! The expected number of heat exhaustion suffers in 1000 is λ = np =