1. Chapter 4 Continuous Random Variables and their Probability Distributions Another one of life’s great adventures is about to begin. Chapter 4A

2. The Road Ahead today + special bonus – the triangular distribution!

3. Continuous Random Variables Experiments and tests can result in values that are spread over a continuum. Even if our measurement device has discrete values, it is often impractical to use a discrete distribution because the number of allowed values is so large. We gain modeling flexibility by expanding the distributions available to us. Measured values can be represented as R.V.’s Range of values is an interval of real numbers An ‘infinite’ number of outcomes are possible A Probability Density Function (pdf), f(x), is used to describe the probability distribution of a continuous R.V., X.

4. Continuous Random Variables - Examples T = a continuous random variable, time to failure X = a continuous random variable, the distance in miles between cable defects Z = a continuous random variable, the monthly consumption of power in watts T = a continuous random variable, the repair time of a failed machine Y = a continuous random variable, the time between arrivals of customers at City National Bank X = a continuous random variable, the procurement lead-time for a critical part

5. 4-2 Probability Distributions and Probability Density Functions Definition

6. Probability Density Function Fig 4.2 – Probability is determined from area under f(x). area = 1

7. 4-2 Probability Distributions and Probability Density Functions Figure 4-3 Histogram approximates a probability density function.

8. Continuous Random Variables Special Message Box: If X is a continuous R.V., P(X = x) = 0 (very important idea) There are an infinite number of points on the X-axis under the curve, the probability that the R.V., X, takes on any particular value, P(X = x), is zero.

9. Our First Example Example 4-2

10. 4-2 Probability Distributions and Probability Density Functions Figure 4-5 Probability density function for Example 4-2.

11. 4-2 Probability Distributions and Probability Density Functions Example 4-2 (continued)

12. Our Very next Example Given the PDF:

13. More of Our Very next Example Given the PDF: 1.P(X > 1) 2.P(1 < X < 2.5) 3.P(X = 3) 4.P(X < 4)

14. 4-3 Cumulative Distribution Functions Definition

15. 4-3 Cumulative Distribution Functions Example 4-4

16. 4-3 Cumulative Distribution Functions Figure 4-7 Cumulative distribution function for Example 4-4.

17. Given the CDF, find the PDF I can do this.

18. Cumulative Distribution Functions Figures 4-4 and 4-6 the PDF and CDF The relation of these shapes ought to become intuitive to you.

19. Problem 4-11 Suppose F(x) =.2x for 0 5. F(x) = 0 otherwise. Determine the following: a.P(X < 2.8) b.P(X > 1.5) c.P(X < -2) d.P(X > 6) = F(2.8) = (.2)(2.8) =.56 = 1 - F(1.5) = 1 - (.2)(1.5) =.7 = F(-2) = 0 = 1 – F(6) = 1 – 1 = 0

20. A CDF Problem Determine the CDF for f(x) =1.5x 2, for –1 < x < 1

21. PDF & CDF f(x) – PDF –> 1.5x 2 F(x) – CDF ->.5x 3 +.5

22. 4-4 Mean and Variance of a Continuous Random Variable Definition

23. A Complete Example Let X = a continuous random variable, the time to complete a complex task in hours. The CDF is given by the following where b, the distribution parameter, is in hours: then

24. Let’s Graph the PDF and CDF b = 10

25. Let’s find some probabilities

26. More of a complete example

27. The Median Define the median such that Pr{X  median} =.5 or F(median) =.5 For b = 10 median = 2.9289

28. 4-4 Mean and Variance of a Continuous Random Variable Expected Value of a Function of a Continuous Random Variable Note for continuous RV: E[a + bX] = a + b E[X] Var[a + bX] = b 2 Var[X]

29. More of the Example The cost for completing the task in the previous example is $50 times the square root of the task time. What is the expected cost of completing the task if b = 10 hours.

30. Problem 4-29 The thickness of a conductive coating in micrometers has a density function of 600x -2 for 100  m < x < 120  m (a) Find the mean and variance (b) If the coating cost $.50 per  m of thickness on each part, what is the average cost of the coating per part? (b) E[cost] =.50 (109.34) = 54.70

31. Begin the Bonus Round He is really going to do it - discuss the triangular distribution!

32. The Right Triangular Distribution A continuous random variable is said to have a right triangular distribution if its density function is given by: x f(x) b Find the value of k that makes f(x) a PDF:

33. More Right Triangular Distribution Now find the CDF and the mean of a right triangular distribution.

34. More Right Triangular Distribution Now find the median of a right triangular distribution.

35. The Left Triangular Distribution Big bonus 35 a t 2/a Please professor. Can we students work this one?

36. The General Triangular Distribution Often used as a “rough” model in the absence of data. a – optimistic value b – pessimistic value c – most likely value (mode) Bigger bonus ac b f(x) a < c < b

37. The General Triangular Distribution It says here that the overachieving student will explore these distributions in detail by finding the CDF, mean, and median. An overachieving student

38. Today’s discussion on continuous random variables has concluded Say it isn’t so.