1. 1

2. 2 At the end of the lesson, students will be able to (c)Understand the Binomial distribution B(n,p) (d) find the mean and variance of Binomial distribution.

3.  Bernoulli trial is a trial that has 2 possible outcomes 3 or

4.  The Bernoulli trial is repeated n times and each trial is independent. The probability of a success stays the same for each trial.  The number of successes (X) are recorded as a random variable. 4 Intro Bernoulli

5.  Probability distribution for X is a binomial distribution  Denoted by X ~ B( n, p ) where n = the number of independent trials p = the probability of success X~B(n,p) read as ‘X is binomially distributed where n is the number of trials and p is the probability of success’ 5

6.  The binomial probability distribution is given by 6 x = 0, 1, 2, 3, …, n Where x = the number of success n = the number of trials p = the probability of success q = 1 – p Binomial Example

7. A fair coin is tossed three times. Find the probability of getting (a) no head, (b) exactly one head, (c) exactly two heads, (d) exactly three heads. 7

8. If X represents the number of heads obtained, then X~B(3, 0.5) P(X=x) = 8 (a) P(X = 0) = (b) P(X = 1) =

9. (c) P(X = 2) = 9 (d) P(X = 3) =

10. In a box, 20% of bulbs are damaged. If a random sample of 10 bulbs is taken, find the probability that (a) exactly 3 bulbs are damaged. (b) all bulbs are good. (c) not more three bulbs are damaged. 10

11. If X represents the number of damaged bulbs, then X~B(10, 0.2) 11 (a) P(X = 3) = (b) P(X = 0) =

12. (c) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 12 = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791

13. The probability that students watch a certain TV program is 0.4. If eight students are selected at random, find the probability that (a) 5 students watch the TV program. (b) less than 4 students watch the TV program. (c) at least one student watches the TV program. 13

14. If X represents the number of students watch a certain TV program, then X~B(8, 0.4) and 14 (a) P(X = 5) = = 0.1239

15. (b) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 15 = 0.01679 + 0.08957 + 0.20901 + 0.27869 = 0.5941

16. (c) P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) 16 = 0.9832

17. If X ~ B(n, p), then (i) mean, μ = E(X) = np, (ii) variances, σ 2 = Var(X) = npq *Notes: q = 1 – p 17

18. A fair die is rolled 300 times. Let X represents the number of times the number ‘6’ is observed. Find the mean and standard deviation. 18

19. If X represents the number of times the number 6 is observed, then 19 Mean, μ = np =

20. Variance, σ 2 = npq = 20 Standard Deviation, σ

21. A study revealed that two of five families have a video recorder. (a) If there are 4 families, find the probability that two or more families have video recorders. (b) If there are 900 families, calculate the mean and standard deviation of the numbers of families who have video recorders 21

22. If X represents the number of families who have video recorders, then X~B(4, 0.4) and 22 (a) P(X ≥ 2) = 0.5248

23. (b) μ = np = (900)(0.4) = 360 σ = 23 X~B(900, 0.4)

24. 24 P(X ≥ r) is ‘Probability of X is larger and Equals to r. To find probability for the value of p from 0.001 to 0.5

25. 25 Binomial Cumulative Distribution

26. Let X be a random such that X~B(5, 0.3). By using the binomial table, find (a) P(X ≥ 3) (b) P(X > 3) (c) P(X ≤ 3) (d) P(X < 3) (e) P(X = 3) 26

27. (a) P(X ≥ 3) = 0.1631 nprpr 0.10 0.15 0.200.25 0.30 0.35 50123401234 1.000 1.000 1.000 0.4095 0.5563 0.6723 0.0815 0.1648 0.2627 0.0086 0.0266 0.0579 0.0005 0.0022 0.0067 1.000 1.000 1.000 0.7627 0.8319 0.8840 0.3672 0.4718 0.5716 0.1035 0.1631 0.2352 0.0156 0.0308 0.0540 27 (Turn to page 5 from Statistical Tables) Check Table

28. (b) P(X > 3) = P(X ≥ 4) = 28 0.0308 (c) P(X ≤ 3) = 1 – P(X ≥ 4) =0.9692 (d) P(X < 3) = 1 – P(X ≥ 3) =0.8369 (e) P(X = 3) = P(X ≥ 3) – P(X ≥ 4) = 0.1631 – 0.0308 = 0.1323 Check Table

29. 29

30. Let X be a random variable such that X~B(10, 0.8). By using the binomial table, find (a) P(X ≥ 4) (b) P(X > 4) (c) P(X ≤ 4) (d) P(X < 4) 30

31. (a) P(X ≥ 4)  X ~ B(10,0.8) P(X ≤ 6)  X ~ B(10,0.2) = 1 - P(X ≥ 7) = 1 - 0.0009 = 31 0.9991 (b) P (X > 4) = P(X ≥ 5)  X ~ B(10,0.8) P (X ≤ 5)  X ~ B(10,0.2) = 1 - P(X ≥ 6) = 1 - 0.0064 = 0.9936

32. 32 (d) P(X < 4) = P (X ≤ 3 )  X ~ B(10,0.8) P(X ≥ 7)  X ~ B(10,0.2) = 0.0009 (c) P(X ≤ 4)  X ~ B(10,0.8) P(X ≥ 6)  X ~ B(10,0.2) = 0.0064

33. For a random variable X with a binomial distribution B(10, 0.45), find a when (a) P(X ≥ a) = 0.4956 (b) P(X < a) = 0.89801 (c) P(X = a) = 0.23836 33

34. (a) P(X ≥ a) = 0.4956 from the binomial probability tables, 34 P(X ≥ 5) = 0.4956 Thus, a = 5

35. 35 (b) P(X < a) = 0.89801 1 – P(X ≥ a) = 0.89801 P(X ≥ a) = 1 – 0.89801 = 0.1020 From the binomial probability tables, P(X ≥ 7) = 0.1020 Thus, a = 7

36. (c) P(X = a) = 0.23836 36 P(X ≥ a) – P(X ≥ a + 1) = 0.23836 From the binomial probability tables, P(X ≥ 4) – P(X ≥ 5) = 0.23836 Thus, a = 4

37.  The binomial probability distribution is given by 37 x = 0, 1, 2, 3, …, n To find probability for the value of p > 0.5