2. 12.1 Discrete Probability Distributions (Poisson Distribution)

3. LEARNING OUTCOMES At the end of this lesson, students will be able to : (f) the understand the Poisson Distribution, P o ( ) (g) Identify the mean and variance of Poisson distribution (h) use the Poisson distribution to solve related problems (i) use the Poisson distribution to approximate the Binomial distribution

4. The Poisson Distribution is used to model situations where the random variable X is the number of occurrences of a particular event over a given period of time (or space)

5. * telephone calls made to a switchboard in a given minute (time) * road accidents on a particular motorway in one day (time) * flaws in a meter of fabric (space) * typing error in one page (space) Examples where a Poisson model could be used :

7. 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 8.0 0123401234 1.0000 1.0000 1.0000 1.0000 1.0000 0.9992 0.9993 0.9993 0.9994 0.9994 0.9995 0.9995 0.9996 0.9996 0.9997 0.9933 0.9939 0.9944 0.9949 0.9953 0.9957 0.9961 0.9964 0.9967 0.9970 0.9725 0.9745 0.9764 0.9781 0.9797 0.9812 0.9826 0.9839 0.9851 0.9862 0.9233 0.9281 0.9326 0.9368 0.9409 0.9446 0.9482 0.9515 0.9547 0.9576 Poisson Probability Tables The tabulated value is P(X ≥ r), where r has a Poisson Distribution with mean For a Poisson distribution with mean 7.6, the probability of two or more events occurring is 0.9957 r

8. Poisson Cumulative Distribution

9. EXAMPLE 1: SOLUTION: or FROM CALCULATOR

10. FROM TABLE or (Turn to page 12)(r=0&1; =3.5)

12. EXAMPLE 2: The number of telephone calls made to a switchboard during an afternoon can be modeled by a Poisson distribution with a mean of eight calls per five-minute. Find the probability that in the next five minutes : (a) no calls is made (b) five calls are made (c) at least three calls are made (d) less than four calls are made (e) i. between three and five calls are made ii. between three and five calls are made (inclusive).

13. SOLUTION: X is the number of calls made in five-minute=8

15. EXAMPLE 3: The number of customers arriving at a supermarket checkout in a 10 minute interval may be modeled by a Poisson Distribution with mean 4. Find the probability that the number of customers arriving in a specific 10-minute interval is : (a) two or fewer (b) less than seven (c) exactly three (d) between three and seven inclusive (e) at least five / five or more

16. SOLUTION: (a) P(X ≤ 2) = 1 – P(X ≥ 3) = 1 – 0.7619 = 0.2381 (b) P(X < 7) = 1 – P(X ≥ 7) = 1 – 0.1107 = 0.8893 (c) P(X = 3) = P(X ≥ 3) – P(X ≥ 4) = 0.7619 - 0.5665 = 0.1954 (d) P(3 ≤ X ≤ 7) = P(X ≥ 3) - P(X ≥ 8) = 0.7619 – 0.0511 = 0.7108 (e) P(X ≥ 5) = 0.3712

17. The number of industrial injuries per working week in a particular factory is known to follow a Poisson distribution with mean 0.5. Find the probability that (a) in a particular week there will be (i) less than 2 accidents, (ii) more than 2 accidents; (b) t here will be no accidents in a three week period.... EXAMPLE 4:

18. SOLUTION: (a) (i) P(X < 2) = 1 – P(X ≥ 2) = 1- 0.0902 = 0.9098 (ii) P(X >2) = P(X ≥ 3) = 0.0144 therefore, P(X = 0) = 1 – P(X ≥ 1) = 1 – 0.7769 = 0.2231

19. Mean And Variance Of A Poisson Distribution * Mean of the Poisson distribution: * Variance of the Poisson distribution:

20. EXAMPLE 5: If the random variable X follows a Poisson distribution with mean 9, find that (a) E(X) (b) Var(X) (c) P ( X < 4) SOLUTION: (a) E(X) = 9 (b) Var(X) = 9 (c) P ( X < 4 ) = 1 – P ( X  4 ) = 1 – 0.9788 = 0.0212

21. EXAMPLE 6: The number of phone calls received by a household each day over a period of 300 days was recorded and the results were as follows: Number of calls0123456 Number of days1211105016300 Find the mean and variance of the following frequency table.

22. SOLUTION:

23. Using the Poisson Distribution as an Approximation to the Binomial distribution  It is appropriate to use the Poisson distribution as an approximation to the binomial when (i) n is large ( n > 50 ) or / and (ii) p is small ( p < 0. 1 )  In fact, when p = 0.1, n 30, both Poisson and binomial distribution are almost identical.  This particular approximation is more accurate when p 0 and n  ( for Poisson ) = np ( for binomial)

24. Example 1 Let X ~ B( 100, 0.1 ). By using the Poisson distribution as an approximation to the binomial distribution, determine (a) P( X > 5 ) (b) P( X ≤ 3 ) (c) P( X = 10 )

25. Solution X ~ P o ( 10) (a) P( X > 5 )= P( X ≥ 6 ) = 0.9329 using table (b) P( X ≤ 3 ) = 1 – P( X ≥ 4) = 1 – 0.9897 = 0.0103

26. = P( X ≥ 10 ) - P( X ≥ 11)(c) P( X = 10 ) = 0.5421 – 0.4170 = 0.1251 Using the table Or P( X = 10 ) Using definition = 0.1251 ( 4 d.p.)

27. Example 2 Suppose each computer contains 1000 electrical components. The probability that one of these components fails to work in one year is 0.0022. Consider that each function of the components is independent of the functions of the other components. Estimate the probability in one year (a)3 components fail to work (b)between 1 and 3 components fail to work (inclusive)

28. Solution Let X be the number of components that fail to work So X ~ B( n, p ) p = 0.0022 50 can use Poisson distribution Now X ~ P o ( 2.2 )

29. (a) P( X = 3 ) = Using definition = 0.1966 (b) P( 1 ≤ X ≤ 3 )= P( X ≥ 1 ) – P( X ≥ 4 ) Using the table = 0.8892 - 0. 1806 = 0. 7086

30. Example 3 On average one in 200 cars break down on a certain stretch of roads per day. Find the probability that on a certain day (a)none of 250 cars breaks down (b)more than 2 of 300 cars break down Solution (a)Let X be the number of cars( out of 250 cars) that breaks down So X ~ B( n, p )

31. thus,we can use Poisson distribution Now X ~ P o ( ) P( X = 0 ) = = 0.287

32. (b)Let X be the number of cars( out of 300 cars) that breaks down So X ~ B( n, p ) So X ~ P o ( ) P( X > 2)= P( X ≥ 3 ) = 0.1912 Using the table

33. Example 4 500 eggs were placed in a box. On average 0.6% of the eggs break when they arrive at a supermarket. Find the probability that a box containing 500 eggs (a)exactly 3 eggs were broken. (b)less than 2 eggs were broken. (c)at least 3 eggs were broken. (d)not more than 3 eggs were broken.

34. Solution Let X be the number of eggs were broken, So X ~ B( 500, 0.006 ) Poisson distribution can be used as an approximation to binomial distribution since n > 50 and p < 0.01 Now X ~ P o ( 3 )

35. (a)P( X = 3 ) = (b)P( X < 2 )= 1 – P( X ≥ 2 ) Using the table = 1 – 0.8009 = 0. 1991 = 0.22404 ( 5 d.p.)

36. (c)P( X ≥ 3 ) = 0.5768 using table (d)P( X ≤ 3 )= 1 – P( X ≥ 4) = 1 – 0.3528 = 0.6472 using table

37. Poisson Distribution

38. Poisson distribution can be used as an approximation to binomial distribution when n and p in binomial distribution are respectively large and small, i.e. when n > 50 or / and p < 0.1 ( in Poisson ) = np ( in binomial )