Special Continuous Probability Distributions Normal Distributions Lognormal Distributions

1. Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS • Special Continuous Probability Distributions • Normal Distributions • Lognormal Distributions Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08

2. f(x) x Normal Distribution A random variable X is said to have a normal (or Gaussian) distribution with parameters  and , where - < <  and  > 0, with probability density function for - < x < 

3. Properties of the Normal Model • the effects of  and 

4. Normal Distribution • Mean or expected value of • Mean = E(X) =  • Median value of • X0.5 =  • Standard deviation

5. Normal Distribution • Standard Normal Distribution • If ~ N(, ) • and if • then Z ~ N(0, 1). • A normal distribution with  = 0 and  = 1, is called • the standard normal distribution.

6. Normal Distribution P (X<x’) = P (Z<z’) f(z) f(x) x z 0  Z’ X’

7. Standard Normal Distribution Table of Probabilities • http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html • Enter table with • and find the • value of  • Excel f(z)  z 0 z Normal Distribution

8. Normal Distribution - Example The following example illustrates every possible case of application of the normal distribution. Let ~ N(100, 10) Find: (a) P(X < 105.3) (b) P(X  91.7) (c) P(87.1 <  115.7) (d) the value of x for which P(  x) = 0.05

9. a. P( < 105.3) = = P( < 0.53) = F(0.53) = 0.7019 Normal Distribution – Example Solution f(x) f(z) x z 105.3 100 0.53 0

10. b. P(  91.7) = = P( -0.83) = 1 - P( < -0.83) = 1- F(-0.83) = 1 - 0.2033 = 0.7967 Normal Distribution – Example Solution f(x) f(z) x z 0 100 91.7 -0.83

11. c. P(87.1 <  115.7) = F(115.7) - F(87.1) = P(-1.29 < Z < 1.57) = F(1.57) - F(-1.29) = 0.9418 - 0.0985 = 0.8433 Normal Distribution – Example Solution f(x) f(x) x x 87.1 100 115.7 -1.29 0 1.57

12. Normal Distribution – Example Solution f(x) f(z) 0.05 0.05 x z 1.64 116.4 100 0

13. Normal Distribution – Example Solution (d) P(  x) = 0.05 P(  z) = 0.05 implies that z = 1.64 P(  x) = therefore x - 100 = 16.4 x = 116.4

14. Normal Distribution – Example Solution The time it takes a driver to react to the brake lights on a decelerating vehicle is critical in helping to avoid rear-end collisions. The article ‘Fast-Rise Brake Lamp as a Collision-Prevention Device’ suggests that reaction time for an in-traffic response to a brake signal from standard brake lights can be modeled with a normal distribution having mean value 1.25 sec and standard deviation 0.46 sec. What is the probability that reaction time is between 1.00 and 1.75 seconds? If we view 2 seconds as a critically long reaction time, what is the probability that actual reaction time will exceed this value?

15. Normal Distribution – Example Solution

16. Normal Distribution – Example Solution

17. An Application of Probability & Statistics Statistical Quality Control Statistical Quality Control is an application of probabilitistic and statistical techniques to quality control

18. Statistical Quality Control - Elements Analysis of process capability Statistical process control Process improvement Acceptance sampling

19. Statistical Tolerancing - Convention Normal Probability Distribution 0.00135 0.9973 0.00135 Nominal LTL UTL +3 -3 

20. Statistical Tolerancing - Concept x LTL Nominal UTL

21. Caution For a normal distribution, the natural tolerance limits include 99.73% of the variable, or put another way, only 0.27% of the process output will fall outside the natural tolerance limits. Two points should be remembered: 1. 0.27% outside the natural tolerances sounds small, but this corresponds to 2700 nonconforming parts per million. 2. If the distribution of process output is non normal, then the percentage of output falling outside   3 may differ considerably from 0.27%.

22. Normal Distribution - Example The diameter of a metal shaft used in a disk-drive unit is normally distributed with mean 0.2508 inches and standard deviation 0.0005 inches. The specifications on the shaft have been established as 0.2500  0.0015 inches. We wish to determine what fraction of the shafts produced conform to specifications.

23. Normal Distribution - Example Solution f(x) 0.2500 nominal 0.2508 x 0.2485 LSL 0.2515 USL

24. Normal Distribution - Example Solution Thus, we would expect the process yield to be approximately 91.92%; that is, about 91.92% of the shafts produced conform to specifications. Note that almost all of the nonconforming shafts are too large, because the process mean is located very near to the upper specification limit. Suppose we can recenter the manufacturing process, perhaps by adjusting the machine, so that the process mean is exactly equal to the nominal value of 0.2500. Then we have

25. Normal Distribution - Example Solution f(x) nominal 0.2485 LSL 0.2515 USL x 0.2500

30. Using a normal probability distribution as a model for a quality characteristic with the specification limits at three standard deviations on either side of the mean. Now it turns out that in this situation the probability of producing a product within these specifications is 0.9973, which corresponds to 2700 parts per million (ppm) defective. This is referred to as three-sigma quality performance, and it actually sounds pretty good. However, suppose we have a product that consists of an assembly of 100 components or parts and all 100 parts must be non-defective for the product to function satisfactorily. Normal Distribution - Example

31. The probability that any specific unit of product is non-defective is 0.9973 x 0.9973 x . . . x 0.9973 = (0.9973)100 = 0.7631 That is, about 23.7% of the products produced under three sigma quality will be defective. This is not an acceptable situation, because many high technology products are made up of thousands of components. An automobile has about 200,000 components and an airplane has several million! Normal Distribution - Example

32. Lognormal Distribution

33. f(x) x 0 Lognormal Distribution • Definition - A random variable is said to have the • Lognormal Distribution with parameters  and , • where > 0 and  > 0, if the probability density • function of X is: • , for x >0 • , for x 0

34. Lognormal Distribution - Properties • Rule: If ~ LN(,), • then = ln ( ) ~ N(,) • Probability Distribution Function • where F(z) is the cumulative probability distribution • function of N(0,1)

35. Lognormal Distribution - Properties Mean or Expected Value • Median • Standard Deviation

36. Lognormal Distribution - Example A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. Suppose the parameters are  = 5 and  = 0.1 (a) Compute E( ) and Var( ) (b) Compute P( > 120) (c) Compute P(110   130) (d) What is the value of median ductile strength? (e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120? (f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

37. Lognormal Distribution –Example Solution a) b)

38. Lognormal Distribution –Example Solution c) d)

39. Lognormal Distribution –Example Solution e) Let Y=number of items tested that have strength of at least 120 Y=0,1,2,…,10

40. Lognormal Distribution –Example Solution f) The value of x, say xms, for which is determined as follows: