1. Ch.4 DISCRETE PROBABILITY DISTRIBUTION Prepared by: M.S Nurzaman, S.E, MIDEc. ( deden )‏ dedenmsn@gmail.com 081310315562

2. Introduction We already know that a Discrete Random Variable is a variable which can assume only integer values, such as, 7, 9, and so on. In other words, a discrete random variable cannot take fractions as value. Things such as people, cars, or defectives are things we can count and are discrete items. There are some types of Discrete Probability Distribution: Binomial Distribution, Poisson Distribution, and Hypergeometric Distribution. In this lecture we will discussing the first type

3. A probability distribution is similar to the frequency distribution of a quantitative population because both provide a long-run frequency for outcomes. In other words, a probability distribution is listing of all the possible values that a random variable can take along with their probabilities. for example, suppose we want to find out the probability distribution for the number of heads on three tosses of a coin: First toss.........T T T T H H H H Second toss.....T T H H T T H H Third toss........T H T H T H T H Probability Distribution:

4. the probability distribution of the above experiment is as follows (columns 1, and 2 in the following table). (Column 1).....................(Column 2)..............(Column 3)‏ Number of heads...............Probability.................(1)(2)‏ X.....................................P(X)..........................(X)P(X)‏ 0......................................1/8................................0.0 1......................................3/8................................0.375 2......................................3/8................................0.75 3......................................1/8................................0.375 Total...................................................................1.5=E(X)‏

5. The equation for computing the mean, or expected value of discrete random variables is as follows: Mean = E(X) = Summation[X.P(X)] where: E(X) = expected value, X = an event, and P(X) = probability of the event Note that in the above equation, the probability of each event is used as the weight. For example, going back to the problem of tossing a coin three times, the expected value is: E(X) = [0(1/8)+1(3/8)+2(3/8)+3(1/8) = 1.5 (column 3 in the above table). Thus, on the average, the number of heads showing face up in a large number of tossing a coin is 1.5.

7. Example: Suppose a charity organization is mailing printed return-address stickers to over one million homes in the U.S. Each recipient is asked to donate either $1, $2, $5, $10, $15, or $20. Based on past experience, the amount a person donates is believed to follow the following probability distribution: X:..... $1......$2........$5......$10.........$15......$20 P(X)....0.1.....0.2.......0.3.......0.2..........0.15.....0.05 The question is, what is expected that an average donor to contribute, and what is the standard devation.

8. The solution is as follows. (1)......(2).......(3).............(4)..................(5)..........................................(6)‏ X......P(X)....X.P(X).......X - mean......[(X - mean)]squared...............(5)x(2)‏ 1.......0.1......0.1...........- 6.25...............39.06........................................3.906 2.......0.2......0.4...........- 5.25...............27.56........................................5.512 5.......0.3......1.5...........- 2.25.................5.06........................................1.518 10.....0.2......2.0.............2.75.................7.56........................................1.512 15.....0.15....2.25...........7.75...............60.06........................................9.009 20.....0.05....1.0...........12.75.............162.56.........................................8.125 Total...........7.25 = E(X)....................................................................29.585 Thus, the expected value is $7.25, and standard deviation is the square root of $29.585, which is equal to $5.55. In other words, an average donor is expected to donate $7.25 with a standard deviation of $5.55.

9. Binomial Distribution: One of the most widely known of all discrete probability distributions is the binomial distribution. Several characteristics underlie the use of the binomial distribution. Characteristics of the Binomial Distribution: 1. The experiment consists of n identical trials. 2. Each trial has only one of the two possible mutually exclusive outcomes, success or a failure. 3. The probability of each outcome does not change from trial to trial, and 4. The trials are independent, thus we must sample with replacement.

10. Note that if the sample size, n, is less than 5% of the population, the independence assumption is not of great concern. Therefore the acceptable sample size for using the binomial distribution with samples taken without replacement is [n<5% N] where n is equal to the sample size, and N stands for the size of the population. The birth of children (male or female), true-false or multiple- choice questions (correct or incorrect answers) are some examples of the binomial distribution

11. Binomial Equation: When using the binomial formula to solve problems, all that is necessary is that we be able to identify three things: the number of trials (n), the probability of a success on any one trial (p), and the number of successes desired (X). The formulas used to compute the probability, the mean, and the standard deviation of a binomial distribution are as follows

12. where: n = the sample size or the number of trials, X = the number of successes desired, p = probability of getting a success in one trial, and q = (1 - p) = the probability of getting a failure in one trial

13. Let's go back to previous lecture and solve the probability problem of defective TVs by applying the binomial equation once again. We said, suppose that 4% of all TVs made by Panasonic Company in 1995 are defective. If eight of these TVs are randomly selected from across the country and tested, what is the probability that exactly three of them are defective? Assume that each TV is made independently of the others

14. In this problem, n=8, X=3, p=0.04, and q=(1-p)=0.96. Plugging these numbers into the binomial formula (see the above equation) we get: P(X) = P(3) = 0.0003 or 0.03% which is the same answer as in lecture number four. The mean is equal to (n) x (p) = (8)(0.04)=0.32, the variance is equal to np (1 - p) = (0.32)(0.96) = 0.31, and the standard deviation is the square root of 0.31, which is equal to 0.6

15. The Binomial Table: Mathematicians constructed a set of binomial tables containing presolved probabilities The binomial tables are easy to use. Simply look up n and p, then find X (located in the first column of each table), and read the corresponding probability. The following table is the binomial probabilities for n = 6. Note that the probabilities in each column of the binomial table must add up to 1.0.

16. Binomial Probability Distribution Table (n = 6)‏ -------------------------------------------------------------------------------- Probability X.....0.1........0.2.....0.3.....0.4.....0.5.....0.6.....0.7.....0.8.....0.9 ------------------------------------------------------------------------------- 0.....0.531............0.118....................................................0.000 1.....0.354............0.303....................................................0.000 2.....0.098............0.324....................................................0.001 3.....0.015............0.185....................................................0.015 4.....0.001............0.060....................................................0.098 5.....0.000............0.010....................................................0.354 6.....0.000............0.001....................................................0.531 --------------------------------------------------------------------------------

17. Example: Suppose that an examination consists of six true and false questions, and assume that a student has no knowledge of the subject matter. The probability that the student will guess the correct answer to the first question is 30%. Likewise, the probability of guessing each of the remaining questions correctly is also 30%. What is the probability of getting more than three correct answers?

18. For the above problem, n = 6, p = 0.30, and X >3. In the above table, search along the row of p values for 0.30. The problem is to locate the P(X > 3). Thus, the answer involves summing the probabilities for X = 4, 5, and 6. These values appear in the X column at the intersection of each X value and p = 0.30, as follows: P (X > 3) = Summation of {P (X=4) + P(X=5) +P(X=6)} = (0.060)+(0.010)+(0.001) = 0.071 or 7.1% Thus, we may conclude that if 30% of the exam questions are answered by guessing, the probability is 0.071 (or 7.1%) that more than four of the questions are answered correctly by the student.

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