1. STATISTIC & INFORMATION THEORY (CSNB134) MODULE 7A PROBABILITY DISTRIBUTIONS FOR RANDOM VARIABLES (BINOMIAL DISTRIBUTION)

2. Overview In Module 7, we will learn three types of distributions for random variables, which are: - Binomial distribution- Module 7A - Poisson distribution- Module 7B - Normal distribution- Module 7C This is a Sub-Module 7A, which includes lecture slides on Binomial Distribution.

3. The Binomial Random Variable Example of a binomial random variable: coin-tossing experiment - P(Head) = ½. Toss a fair coin n = 3 times, x = number of heads. We have seen in Module 6 that the probability distribution for the coin-tossing experiment, where a coin is toss for 3 times and x = number of heads is as follows. xp(x) 01/8 13/8 2 31/8

4. The Binomial Random Variable (cont.) Many situations in real life resemble the coin toss, but with a coin that might not be fair, i.e. P(H)  1/2. Example: A lecturer samples 10 people and counts the number who of students who are from KL. The analogy to coin tossing is as follows: Student Coin: Head / H: Tail / T: Number of tosses: P(H): From KL Not from KL n = 10 P(From KL) = proportion of students in the population who are from KL.

5. The Binomial Random Variable (cont.) Other examples: 1. Proportion of COIT students who are female. 2. Proportion of Malaysian population who watch Akademi Fantasia. 3. Proportion of people in Bangi who subscribe to ASTRO. Female/Male Watch/Don’t Watch Subscribe/Don’t Subscribe P(F) =.55 / P(M) =.45 P(W) =.35 / P(DW) =.65 P(S) =.85 / P(DS) =.15 All exhibit the common characteristic of Binomial experiment!

6. Characteristic of a Binomial Experiment 1. The experiment consists of n identical trials. 2. Each trial results in one of two outcomes, success (S) or failure (F) i.e: mutually exclusive. 1. The probability of success on a single trial is p and remains constant from trial to trial. The probability of failure is q = 1 – p. 2. The trials are independent (Note = please refer notes on Module 5 on ‘Independent Event’). 3. We are interested in x, the number of successes in n trials.

7. Binomial or Not? Example: Select two people from the U.S. population (say 300 million), and suppose that 15% (45 million) of the population has the Alzheimer’s gene. For the first person: p = P(gene) = 45,000,000/300,000,000 =.15 For the second person: p = P(gene) = 44,999,999/299,999,999 .15, even though one person has been removed from the population. We assume that the removal of one person from the population have negligible effect.

8. The Binomial Probability Distribution For a binomial experiment with n trials and probability p of success on a given trial, the probability of k successes in n trials is: And the measures of center and spread are:

9. Exercise 1  A fair coin is flipped 6 times. (1) What is the probability of obtaining exactly 3 heads?  For this problem, N =6, k=3, and p =.5, thus:  (2) Determine the probability of obtaining 3 or more successes.  P(x >=3) = P(3) + P(4) + P(5) + P(6) = pls find this out…..

10. Exercise 2 A rifle shooter hits a target 80% of the time. He fires five shots at the target. Assuming x = number of hits, what is the mean and standard deviation for x?

11. Exercise 2 (cont.) Would it be unusual to find that none of the shots hit the target? The value x = 0 lies more than 4 standard deviations below the mean. Very unusual!! (Note: Refer to z-score in Module 3)

12. n = p =x = success = Exercise 2 (cont.) What is the probability that exactly 3 shots hit the target? 5.8hit# of hits

13. Exercise 2 (cont.) What is the probability that more than 3 shots hit the target?

14. Example 2 (Cont.) For the same rifle shooter, if he fires 20 shots at the target. What is the probability that more than 5 shots hit the target?

15. Effect of p Two binomial distributions are shown below. Here, π is the same as p Notice that for p =.5, the distribution is symmetric whereas for p =.3, the distribution is skewed right.

16. Probability Tables  We can use the binomial probability tables to find probabilities for selected binomial distributions.  They can either be  Probability tables for each x = k or  Cumulative probability tables

17. Probability Tables (cont.) Probability table for each x = k (part of)

18. Cumulative Probability table (part of) (for n= 5) Probability table for each x = k (for n = 5) P(x<=3) = 0.263 (from the cumulative table) P(x<=3)= 0.000 + 0.006 + 0.051 + 0.205 (from the x=k table) = 0.262 Probability Tables (cont.)

19. Cumulative Probability Tables We can use the cumulative probability tables to find probabilities for selected binomial distributions.  Find the table for the correct value of n.  Find the column for the correct value of p.  The row marked “k” gives the cumulative probability, P(x  k) = P(x = 0) +…+ P(x = k)  Find the table for the correct value of n.  Find the column for the correct value of p.  The row marked “k” gives the cumulative probability, P(x  k) = P(x = 0) +…+ P(x = k)

20. Exercise 3 A rifle shooter hits a target 80% of the time. He fires five shots at the target. What is the probability that exactly 3 shots hit the target?  Find the table for the correct value of n.  Find the column for the correct value of p.  Row “k” gives the cumulative probability, P(x  k) = P(x = 0) +…+ P(x = k)

21. Exercise 3 (cont.) kp =.80 0.000 1.007 2.058 3.263 4.672 51.000 P(x = 3) P(x = 3) = P(x  3) – P(x  2) =.263 -.058 =.205 P(x = 3) P(x = 3) = P(x  3) – P(x  2) =.263 -.058 =.205 Check from formula: P(x = 3) =.2048

22. Exercise 2 (cont.) kp =.80 0.000 1.007 2.058 3.263 4.672 51.000 What is the probability that more than 3 shots hit the target? P(x > 3) P(x > 3) = 1 - P(x  3) = 1 -.263 =.737 P(x > 3) P(x > 3) = 1 - P(x  3) = 1 -.263 =.737 Check from formula: P(x > 3) =.7373

23. STATISTIC & INFORMATION THEORY (CSNB134) PROBABILITY DISTRIBUTIONS OF RANDOM VARIABLES (BINOMIAL DISTRIBUTIONS) --END--