Continuous Random Variables

1. Continuous Random Variables

2. Continuous Random Variable • A continuous random variable is one for which the outcome can be any value in an interval of the real number line. • Usually a measurement. • Examples • Let Y = length in mm • Let Y = time in seconds • Let Y = temperature in ºC G. Baker, Department of Statistics University of South Carolina; Slide 2

3. Continuous Random Variable • We don’t calculate P(Y = y), we calculate P(a < Y < b), where a and b are real numbers. • For a continuous random variable P(Y = y) = 0. G. Baker, Department of Statistics University of South Carolina; Slide 3

4. Continuous Random Variables • The probability density function (pdf) when plotted against the possible values of Y forms a curve. The area under an interval of the curve is equal to the probability that Y is in that interval. 0.40 f(y) a b Y G. Baker, Department of Statistics University of South Carolina; Slide 4

5. The entire area under a probability density curve for a continuous random variable • Is always greater than 1. • Is always less than 1. • Is always equal to 1. • Is undeterminable. G. Baker, Department of Statistics University of South Carolina; Slide 5

6. Time Spent Waiting for a Bus • A bus arrives at a bus stop every 30 minutes. If a person arrives at the bus stop at a random time, what is the probability that the person will have to wait less than 10 minutes for the next bus? G. Baker, Department of Statistics University of South Carolina; Slide 6

7. Identify the Random Variable • Let Y = wait time in minutes. • Since the arrival time is random, someone is as likely to arrive 1 minute before a bus arrives as 2 minutes, as 3 minutes, etc. f(y)= 1/30 1.0 G. Baker, Department of Statistics University of South Carolina; Slide 7

8. What is the probability a person will wait less than 10 minutes? 1/30 10/30 = 0.33 20/30 = 0.67 This is called a continuous uniform distribution. G. Baker, Department of Statistics University of South Carolina; Slide 8

9. What is the probability that a person will have to wait between 5 and 25 minutes? 1/30 G. Baker, Department of Statistics University of South Carolina; Slide 9

10. What is the probability that a person will have to wait more than 25 minutes? 1/30 G. Baker, Department of Statistics University of South Carolina; Slide 10

11. What is the probability that a person will have to wait 25 minutes or more? 1/30 G. Baker, Department of Statistics University of South Carolina; Slide 11

12. What is the expected waiting time? 1/30 G. Baker, Department of Statistics University of South Carolina; Slide 12

13. Properties of a Probability Density Function (pdf) • f(y) > 0 for all possible intervals of y. • If y0 is a specific value of interest, then the cumulative distribution function (cdf) is • If y1 and y2 are specific values of interest, then G. Baker, Department of Statistics University of South Carolina; Slide 13

14. Expected Value for a Continuous Random Variable • Recall Expected Value for Discrete Random Variable: • Expected value for continuous random variable: G. Baker, Department of Statistics University of South Carolina; Slide 14

15. Variance for Continuous Random Variable Recall: Variance for Discrete Random Variable: Variance for Continuous Random Variable: G. Baker, Department of Statistics University of South Carolina; Slide 15

16. G. Baker, Department of Statistics University of South Carolina; Slide 16

17. Times Between Industrial Accidents • The times between accidents for a 10-year period at a DuPont facility can be modeled by the exponential distribution. where λ is the accident rate (the expected number of accidents per day in this case) G. Baker, Department of Statistics University of South Carolina; Slide 17

18. Example of time between accidents Let Y = the number of days between two accidents. Time 12 days 35 days 5 days ● ● ● ● ● Accident Accident Accident #1 #2 #3 G. Baker, Department of Statistics University of South Carolina; Slide 18

19. Times Between Industrial Accidents • Suppose in a 1000 day period there were 50 accidents. λ = 50/1000 = 0.05 accidents per day or 1/λ = 1000/50 = 20 days between accidents G. Baker, Department of Statistics University of South Carolina; Slide 19

20. What is the probability that this facility will go less than 10 days between the next two accidents? ? f(y) = 0.05e-0.05y G. Baker, Department of Statistics University of South Carolina; Slide 20

21. ? Recall: G. Baker, Department of Statistics University of South Carolina; Slide 21

22. If: G. Baker, Department of Statistics University of South Carolina; Slide 22

23. In General… G. Baker, Department of Statistics University of South Carolina; Slide 23

24. Exponential Distribution G. Baker, Department of Statistics University of South Carolina; Slide 24

25. If the time to failure for an electrical component follows an exponential distribution with a mean time to failure of 1000 hours, what is the probability that a randomly chosen component will fail before 750 hours? Hint: λ is the failure rate (expected number of failures per hour). G. Baker, Department of Statistics University of South Carolina; Slide 25

26. Mean and Variance for an Exponential Random Variable Note: Mean = Standard Deviation G. Baker, Department of Statistics University of South Carolina; Slide 26

27. The time between accidents at a factory follows an exponential distribution with a historical average of 1 accident every 900 days. What is the probability that that there will be more than 1200 days between the next two accidents? G. Baker, Department of Statistics University of South Carolina; Slide 27

28. If the time between accidents follows an exponential distribution with a mean of 900 days, what is the probability that there will be less than 900 days between the next two accidents? G. Baker, Department of Statistics University of South Carolina; Slide 28

29. Relationship between Exponential & Poisson Distributions • Recall that the Poisson distribution is used to compute the probability of a specific number of events occurring in a particular interval of time or space. • Instead of the number of events being the random variable, consider the time or space between events as the random variable. G. Baker, Department of Statistics University of South Carolina; Slide 29

30. Exponential or Poisson Distribution? • We model the number of industrial accidents occurring in one year. • We model the length of time between two industrial accidents (assuming an accident occurring is a Poisson event). • We model the time between radioactive particles passing by a counter (assuming a particle passing by is a Poisson event). • We model the number of radioactive particles passing by a counter in one hour G. Baker, Department of Statistics University of South Carolina; Slide 30

31. Recall: For a Poisson Distribution y = 0,1,2,… where λ is the mean number of events per base unit of time or space and t is the number of base units inspected. The probability that no events occur in a span of time (or space) is: G. Baker, Department of Statistics University of South Carolina; Slide 31

32. Now let T = the time (or space) until the next Poisson event. In other words, the probability that the length of time (or space) until the next event is greater than some given time (or space), t, is the same as the probability that no events will occur in time (or space) t. G. Baker, Department of Statistics University of South Carolina; Slide 32

33. Relationship between Exponential & Poisson Exponential distribution models time (or space) between Poisson events. TIME G. Baker, Department of Statistics University of South Carolina; Slide 33

34. Radioactive Particles • The arrival of radioactive particles at a counter are Poisson events. So the number of particles in an interval of time follows a Poisson distribution. Suppose we average 2 particles per millisecond. • What is the probability that no particles will pass the counter in the next 3 milliseconds? • What is the probability that more than 3 milliseconds will elapse before the next particle passes? G. Baker, Department of Statistics University of South Carolina; Slide 34

35. Machine Failures • If the number of machine failures in a given interval of time follows a Poisson distribution with an average of 1 failure per 1000 hours, what is the probability that there will be no failures during the next 2000 hours? • What is the probability that the time until the next failure is more than 2000 hours? G. Baker, Department of Statistics University of South Carolina; Slide 35

36. Number of failures in an interval of time follows a Poisson distribution. If the mean time to failure is 250 hours, what is the probability that more than 2000 hours will pass before the next failure occurs? • e-8 • 1 – e-8 • e-0.125 • 1 – e-0.125 G. Baker, Department of Statistics University of South Carolina; Slide 36

37. Common Application for the Exponential Distribution • Time between failures when a product is in constant failure rate mode. Burn In Ware Out Constant Failure Rate Mode G. Baker, Department of Statistics University of South Carolina; Slide 37

38. Constant Failure Rate Mode • What is the probability that an electrical component in constant failure rate mode with an average time to failure of 2000 hours will still be operating at 2500 hours? G. Baker, Department of Statistics University of South Carolina; Slide 38

39. If ten of these components are placed in different devices, what is the probability that at least one will still be operating at 2500 hours? G. Baker, Department of Statistics University of South Carolina; Slide 39

40. For how many hours would you guarantee an electrical component that is in constant failure rate mode if the average time to failure is 5000 hours and you want no more than 5% of the components subject to the guarantee? Hint: ln(ex) = x G. Baker, Department of Statistics University of South Carolina; Slide 40