POISSON DISTRIBUTIONS

1. POISSON DISTRIBUTIONS

2. POISSON

3. POISSON 1781-1840

4. DEFINITION • The Poisson distribution is a discrete probability distribution • It expresses the probability of a number of events occurring in a fixed time if these events occur with a known average rate and are independent of the time since the last event.

5. EXAMPLES NOTICE HOW ALL OF THESE INVOLVE RATES • The number of emergency calls received by an ambulance control in an hour. • The number of vehicles approaching a motorway toll bridge in a five-minute interval. • The number of flaws in a meter length of material.

6. CONDITIONS • Events occur at random • The probability of the event happening in a given interval of time or space is approximately proportional to the size of that interval. • Events occur singly in a given interval of time or space. • The probability of the event happening in one interval is independent of the probability of the event happening in any other non-overlapping interval.

8. PROBABILITY

9. EXAMPLE 1 • A student finds that the average number of amoebas in 10 ml of pond water from a particular pond is four. Assuming that the number of amoebas follows a Poisson distribution. • Find the probability that in a 10 ml sample • a) there are exactly five amoebas. • b) There are no amoebas • c) There are fewer than three amoebas.

10. a) Find the probability that in a 10 ml sample there are exactly five amoebas

11. b) Find the probability that in a 10 ml sample there are no amoebas

12. c) Find the probability that in a 10ml sample, there are fewer than three amoebas.

13. Graphics Calc Poisson Dist Stats Mode from Calc For point dist select Ppd For cumulative select Pcd after Poisn F5 Distribution Then F6 next F1 Poisn Select Data: Variable For P(X=1) X:1 For P(X<3), For P(X=1) = 0.1494 The calcuator uses mu instead of lambda for the mean Since tables only go to 4 dp, round to 4dp

14. ON AVERAGE THERE ARE THREE BABIES BORN A DAY WITH HAIRY BACKS Find the probability that in one day two babies are born hairy. Find the probability that in one day no babies are born hairy. EXAMPLE 2

15. USING FORMULA SHEET • Find the probability that in one day two babies are born hairy. • Probability = 0.2240

16. USING FORMULA SHEET • Find the probability that in one day no babies are born hairy. • Probability = 0.0498

17. SEE: ‘POISSON DISTRIBUTION… REAL SLOW’ DOCUMENT NOW

18. QUEUE THEORY

19. Example: Suppose a bank knows that on average 60 customers arrive between 10 A.M. and 11 A.M. daily. • Find the probability that exactly two customers arrive in a given one-minute time interval between 10 and 11 A.M.

20. Example: there are on average 3 flaws per square metre of material at CDA TEXTILES. • Find the probability that there are no flaws in half a square metre of material

21. Example: A call centre receives, on average, 8 calls every 10 minute period. • Find the probability of less than 10 calls being received in a twenty minute interval

22. Example: A call centre receives, on average, 8 calls every 10 minute period. • Find the probability of less than 2 calls in an 8 minute interval

23. MEAN AND STANDARD DEVIATION OF A POISSON DISTRIBUTION

24. Note the following: • The binomial distribution is affected by the sample size and the probability of success while the Poisson distribution is ONLY affected by the mean. • The binomial distribution has values from x = 0 to n but the Poisson distribution has values from x = 0 to infinity.

25. INVERSE POISSON

26. Whenever working with an inverse problem. Try to get P(X=0) This will simplify the equation But we know P(X=0) =0.4 So we can set the 2 equal to each other. = 0.4 Solve… =3.22 When we used P(X=0) our X was defined in a 15 minute time interval. Naturally the mean found will be in the same interval (15 mins) Adjust lambda and find simple probability from there

27. Students often struggle with the concept of finding lambda. For example, when they use P(X=0)=0.04 (the probability of scoring 0 goals being 0.04) to find lambda=3.22 (on average 3.22 goals are scored), they think that that average is linked to the fact that X=0. NO! That lambda was found using X=0, but it applies for all X. The idea of the Poisson distribution is: Look…on average we score 3.22 goals in a game. BUT obviously other things can happen, I might only score one goal…I wonder what the probability of that happening is? Etc. Obviously, when I score 3.22 goals on average, scoring only one goal would be a relatively high probability [ P(X=1)=0.129 ] but the probability of scoring 10 goals, though possible, would come out as a very low probability [ P(X=10)=0.001 ] as it would have to be a very out of the ordinary performance to score 10 goals by a team who on average only scores 3.22 goals per game.

28. In Mr Chidas restaurant, The probability that none of the customers order a vegetarian dish throughout the entire working day is 4%. Estimate, and hence find probability that a randomly selected sample will eat 2 vege dishes in a working day lambda= 3.2188 P(X=2)=0.207 Part of Lebron Chida’s morning basketball workout is shooting three pointers (shots from a distance aimed at going through the basket) for 5 minutes, 98% of the time he makes some shots. Estimate the mean number of shots in his morning session? What is the probability Lebron Chida makes 6 shots in those 5 minutes? lambda=3.91 P(x=6)=0.099