1. 1 1 Slide Continuous Probability Distributions Chapter 6 BA 201

2. 2 2 Slide Continuous Probability Distributions n n A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. n n It is not possible to talk about the probability of the random variable assuming a particular value. n n Instead, we talk about the probability of the random variable assuming a value within a given interval.

3. 3 3 Slide Continuous Probability Distributions n n The probability of the random variable assuming a value within some given interval from x 1 to x 2 is defined to be the area under the graph of the probability density function between x 1 and x 2. f ( x ) x x Uniform x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x Normal x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 Exponential x x x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2

4. 4 4 Slide Area as a Measure of Probability n n The area under the graph of f ( x ) and probability are identical. n n This is valid for all continuous random variables. n n The probability that x takes on a value between some lower value x 1 and some higher value x 2 can be found by computing the area under the graph of f ( x ) over the interval from x 1 to x 2.

5. 5 5 Slide UNIFORM PROBABILITY DISTRIBUTION

6. 6 6 Slide Uniform Probability Distribution where: a = smallest value the variable can assume b = largest value the variable can assume f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere n n A random variable is uniformly distributed whenever the probability is proportional to the interval’s length. n n The uniform probability density function is: f ( x ) x x Uniform x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2

7. 7 7 Slide Var( x ) = ( b - a ) 2 /12 E( x ) = ( a + b )/2 Uniform Probability Distribution n Expected Value of x n Variance of x

8. 8 8 Slide Uniform Probability Distribution Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces.

9. 9 9 Slide n Uniform Probability Density Function f ( x ) = 1/10 for 5 < x < 15 = 0 elsewhere where: x = salad plate filling weight Uniform Probability Distribution

10. 10 Slide n Expected Value of x E( x ) = ( a + b )/2 = (5 + 15)/2 = 10 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = 8.33 Uniform Probability Distribution n Variance of x

11. 11 Slide n Uniform Probability Distribution for Salad Plate Filling Weight f(x)f(x) f(x)f(x) x x 1/10 Salad Weight (oz.) Uniform Probability Distribution 5 5 10 15 0 0

12. 12 Slide f(x)f(x) f(x)f(x) x x 1/10 Salad Weight (oz.) 5 5 10 15 0 0 P(12 < x < 15) = 1/10(3) =.3 What is the probability that a customer will take between 12 and 15 ounces of salad? Uniform Probability Distribution 12

13. 13 Slide UNIFORM PROBABILITY DISTRIBUTION PRACTICE

14. 14 Slide Scenario Smile time in eight week old babies ranges from zero to 23 seconds. 1. 1. What is f(x)? 2. 2. Compute E(x). 3. 3. Compute Var(x). 4. 4. What is the probability a baby smiles a. a. Less than or equal to five seconds? b. b. 10 to 18 seconds inclusive?

15. 15 Slide NORMAL PROBABILITY DISTRIBUTION

16. 16 Slide Normal Probability Distribution n n The normal probability distribution is the most important distribution for describing a continuous random variable. n n It is widely used in statistical inference. n n It has been used in a wide variety of applications including: Heights of people Rainfall amounts Test scores Scientific measurements

17. 17 Slide The distribution is symmetric; its skewness measure is zero. Normal Probability Distribution n Characteristics x

18. 18 Slide The entire family of normal probability distributions is defined by its mean  and its standard deviation . Normal Probability Distribution n Characteristics Standard Deviation  Mean  x

19. 19 Slide The highest point on the normal curve is at the mean, which is also the median and mode. Normal Probability Distribution n Characteristics x

20. 20 Slide Normal Probability Distribution n Characteristics -10025 The mean can be any numerical value: negative, zero, or positive. x

21. 21 Slide Normal Probability Distribution n Characteristics  = 15 The standard deviation determines the width of the curve: larger values result in wider, flatter curves.  = 25 x

22. 22 Slide Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (0.5 to the left of the mean and 0.5 to the right). Normal Probability Distribution n Characteristics.5.5 x

23. 23 Slide Normal Probability Distribution n Characteristics (basis for the empirical rule) x   – 1   + 1  68.26%  – 2   + 2  95.44%  – 3   + 3  99.72%

24. 24 Slide Normal Probability Distribution n Normal Probability Density Function  = mean  = standard deviation  = 3.14159 e = 2.71828 where:

25. 25 Slide NORMAL PROBABILITY DISTRIBUTION PRACTICE

26. 26 Slide Scenario Mean75pi3.14159 Std Dev8.95e2.71828 Compute f(75), f(66), and f(84)

27. 27 Slide STANDARD NORMAL PROBABILITY DISTRIBUTION

28. 28 Slide Standard Normal Probability Distribution A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. n Characteristics

29. 29 Slide  0 z The letter z is used to designate the standard normal random variable. Standard Normal Probability Distribution n Characteristics

30. 30 Slide n Converting to the Standard Normal Distribution Standard Normal Probability Distribution We can think of z as a measure of the number of standard deviations x is from .

31. 31 Slide Standard Normal Probability Distribution 0 z P(z ≤ 0)=0.500 z=0 P(z ≤ -1)= 0.159 z=-1 P(z ≤ -2)=0.023 z=-2

32. 32 Slide Standard Normal Probability Distribution 0 z P(z ≤ 1)= 0.841 z=1

33. 33 Slide Standard Normal Probability Distribution 0 z P(z ≤ 1)= 0.841 z=1 P(z ≤ -1)= 0.159 z=-1 P(-1 ≤ z ≤ +1) P(z ≤ 1) P(z ≤ -1) -0.841-0.159== 0.682=

34. 34 Slide Standard Normal Probability Distribution Pep Zone Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order.

35. 35 Slide It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. Standard Normal Probability Distribution Pep Zone The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? P ( x > 20) = ?

36. 36 Slide z = ( x -  )/  = (20 - 15)/6 = 0.83 n Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. Step 2: Find the area under the standard normal curve to the left of z = 0.83. Standard Normal Probability Distribution

37. 37 Slide n Cumulative Probability Table for the Standard Normal Distribution Standard Normal Probability Distribution z.00.01.02.03.04.05.06.07.08.09............5.6915.6950.6985.7019.7054.7088.7123.7157.7190.7224.6.7257.7291.7324.7357.7389.7422.7454.7486.7517.7549.7.7580.7611.7642.7673.7704.7734.7764.7794.7823.7852.8.7881.7910.7939.7967.7995.8023.8051.8078.8106.8133.9.8159.8186.8212.8238.8264.8289.8315.8340.8365.8389........... P ( z < 0.83) Appendix B Table 1

38. 38 Slide P ( z > 0.83) = 1 – P ( z < 0.83) = 1- 0.7967 = 0.2033 n Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z = 0.83. Probability of a stockout P ( x > 20) Standard Normal Probability Distribution

39. 39 Slide n Solving for the Stockout Probability 0 0.83 Area = 0.7967 Area = 1 - 0.7967 = 0.2033 z Standard Normal Probability Distribution

40. 40 Slide n Standard Normal Probability Distribution Standard Normal Probability Distribution If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than 0.05, what should the reorder point be? --------------------------------------------------------------- (Hint: Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.)

41. 41 Slide n Solving for the Reorder Point 0 Area = 0.9500 Area = 0.0500 z z 0.05 Standard Normal Probability Distribution

42. 42 Slide n Solving for the Reorder Point Step 1: Find the z -value that cuts off an area of 0.05 in the right tail of the standard normal distribution. We look up the complement of the tail area (1 - 0.05 = 0.95) Standard Normal Probability Distribution

43. 43 Slide n Solving for the Reorder Point Step 2: Convert z 0.05 to the corresponding value of x. x =  + z 0.05   = 15 + 1.645(6) = 24.87 or 25 A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than) 0.05. Standard Normal Probability Distribution

44. 44 Slide Normal Probability Distribution n Solving for the Reorder Point 15 x 24.87 Probability of a stockout during replenishment lead-time = 0.05 Probability of no stockout during replenishment lead-time = 0.95

45. 45 Slide n Solving for the Reorder Point By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about.20 to.05. This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer’s desire to make a purchase. Standard Normal Probability Distribution

46. 46 Slide STANDARD NORMAL PROBABILITY DISTRIBUTION PRACTICE

47. 47 Slide Scenario (6-20)

48. 48 Slide a) P(x<=50)

49. 49 Slide Scenario (6-22)

50. 50 Slide