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Chapter 6 Continuous Probability Distributions Outline 6.1. Continuous Uniform Distribution a. Definition (function) & Thm 6.1 (mean, & variance) b. Ex 6.1 Conference Room booking time (check limits of integration) 6.2. Normal Distribution a. Introduction – special distribution (wide use): chapter 1 Empirical Rule b. Definition (function, mean, & variance) c. Properties of the Normal Curve 6.3. Areas under the Normal Curve a. Integrating pdf vs Standard Normal Curve (Table Appendix A.3) Have X (or Z) find P(Z) i. Ex 6.2 ii. Ex 6.4 iii. Ex 6.5 iv. ER 2 b. Using the Normal Curve in Reverse Have P(Z) find X = + z i. Ex 6.3 ii. Ex 6.6 6.4. Applications of the Normal Distribution a. Ex 6.7 Battery Storage problem (Have X (or Z) find P(Z)) b. Ex 6.10 Gauge Limits (Have P(Z) find X = + z) c. Ex 6.13 A/B cut-points problem (Have P(Z) find X = + z) ● Selected textbook problem Q5 p157. What kind of normal distribution problem? 6.5. Normal Approximation to the Binomial Distribution a. When to approximate (as n : approx good when np > 5 and nq > 5) b. Ex 6.15 Recovery from rare blood disease ● Selected textbook problem Q10 p165. What kind of normal distribution problem? Why? 6.6. Exponential Distributions a. Gamma Function b. Gamma Distribution function ( & Thm 6.3 for mean & variance) c. Exponential Distribution: A special case of the Gamma Distribution function (& Corollary 1 of Thm 6.3 for mean & variance) d. Relationship to Poisson Process (Waiting Time Distribution where =1/) 6.7. Application of the Exponential Distribution (Time to arrival or Time to Poisson event problems) a. Ex 6.17 Component failure time (reliability Theory - Time to failure (or Time to Poisson event) problem) b. Ex 6.18 Switchboard: Time until 2 telephone calls (Gamma) 6.8. Chi-squared Distribution a. Definition (function, Mean, & variance) b. Lab Manual Example c. Further discussions in section 8.6 Sec. 6.4. Applications of the normal distribution. Example 6.10. Gauges are used to reject all components where a certain dimension is not within the specification 5 d . It is known that this measurement is normally distributed with mean 1.5 and standard deviation 0.2. Determine the value of d such that the specifications "cover" 95% of measurements. Example 6.11. A certain machine makes electrical resistors, resistance of which is normally distributed with mean 40 ohms and standard deviation 2 ohms. What percentage of resistors will have a resistance exceeding 43 ohms? Example 6.13. The average grade for an exam is 74 and standard deviation is 7. If 12% of the class are given "A" and grades follow the normal distribution, what is the lowest possible "A" and highest possible "B"? Sec. 6.5. Normal approximation to the binomial. Objective: To use the normal dist to approximate the Binomial dist. Thm: If X is Binomial r.v. with mean n p and variance 2 n p q , then the limiting form of the distribution of Z X n p n pq as n is the standard normal distribution N(0,1) { another notation n(z;0,1). Example: Consider X Bin (4,0.5) then The graph of f(x) is like the normal curve. p(X=x) 0 1 0.0625 0.25 2 3 0.375 0.25 4 0.0625 Bin(4,0.5) 0.4 P(x=x) x 0.3 0.2 p(X=x) 0.1 0 0 1 2 3 4 X Bin (10,0.5) 4 0.205078 5 0.246094 6 7 0.205078 0.117188 8 9 0.043945 0.009766 10 0.000977 Now consider X Bin (100,0.5) p(X=x) X 10 0.043945 0.117188 8 2 3 0.3 0.25 0.2 0.15 0.1 0.05 0 6 0.000977 0.009766 4 0 1 Bin(10,0.5) 2 p(X=x) 0 x P(x=x) Now consider X 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 96 90 84 78 72 66 60 54 48 42 36 30 24 18 12 6 p(X=x) 0 P(x=x) Bin(100,0.5) X consider X Bin (100,0.1) Bin(100,0.1) 0.14 0.12 0.08 p(X=x) 0.06 0.04 0.02 X The approximation: 96 88 80 72 64 56 48 40 32 24 8 16 0 0 P(x=x) 0.1 Probabilities of events related to X can be approximated by a normal distribution with mean np and variance np1 p if the conditions np 5 and n 1 p 5 are satisfied. Continuity Correction It is the adjustment made to an integer-valued discrete random variable when it is approximated by a continuous random variable. For a binomial random variable, we inflate the events by adding or subtracting 0.5 to the event as follows: {X {X {X {X {X :X :X :X :X :X 4} { X : 3.5 X 4.5} 4} { X : X 3}={X : X 3.5} 4{ { X : X 4.5} 4} { X : X 5} {X : X 4.5} 4} { X : X 3.5} . The continuity correction should be applied anytime a discrete random variable is being approximated by a continuous random variable. Example (6.15 page 163): The probability that a patient recovers from a rare blood disease is 0.4. If 100 people are known to have contracted this disease, what is the probability that less than 30 survive? Sol. Let X= # (patient survive) X Bin (100,0.4) n p 100(0.4) 40 >5, nq=60>5 and n p q (100)(0.4)(0.6) 4.899 29 100 p (X 30) x 0 (0.4) x (0.6)100 x 0.014775318 x Using normal approximation p (X 30) p (X 29) p (X 29.5) p ( Z 29.5 40 ) p ( Z 2.14) 0.0162 4.899 Example (6.16/163): A multiple-choice quiz has 200 questions each with 4 possible answers of which only 1 is the correct answer. What is the probability that sheer guess-work yields from 25 to 30 correct answers for 80 of the 200 problems about which the student has no knowledge? Sol.: Let X= #(correct answers), p=p(correct answer)=1/4 = 0.25, n=80 X Bin (80,0.25) mean= 20 , Std= 3.873 Using normal approximation p (25 X 30) 24.5 20 30.5 20 Z ) 3.873 3.873 p (1.16 Z 2.71) .1196 p (24.5 X 30.5) p ( the exact answer is: 30 80 p (25 X 30) x 25 (0.25) x (0.75)100 x 0.119270502 x 6.6: Exponential Distribution Objectives: To introduce: The Exponential Distribution (and the Waiting time). Relationship to Poisson process. Applications. Def.: if X is a continuous r.v. has an exponential dist. with parameter then its density function is given by: x 1e , x 0 f (x ) 0 , O .W . where The mean and variance are respectively. 0 and 2 , Note: The Gamma Distribution is given by: x 1 x 1e , x 0 f (x ) ( ) , O .W . 0 The mean and variance are and 2 , respectively. If 1 then we have the special case of the Gamma which is the Exponential distribution. Relation to Poisson distribution: The exponential distribution describe the time between Poisson events Time between arrivals (Exponential) Arrival (Poisson) mean for Poisson is and 1 But for Exponential is (compare the Exp pdf with the Poisson pdf at x=0) EXAMPLE 6.18. Phone calls at a particular switchboard arrive on average of 5 calls per minute. What is the probability that a call arrive in one minute? Example: If on the average three trucks arrive per hour to be unloaded at a warehouse. Find the probability that the time between the arrivals of successive trucks will be less than 5 minutes. Sol.: The trucks arrive with average 3 per hour ( Poisson with 3/ hr ) The question is P( time between successive arrivals < 5 min) X= time between arrivals 1 ) 3 p( x 5min) P( x 5 / 60 hours) Exp (B Then X 5/ 60 0 1 e x 5/ 60 dx 1 60 3 e 3 x dx [ e 3 x ]5/ 0 0 1 e15/ 60 0.221199 See example 6.18 page 169. 6.8: Chi square Distribution: It is a special case of the Gamma dist. when 2 , 2 where v is called degrees of freedom. Def.: if x is Chi-square dist. with v degrees of freedom ( 2 ) then x 1 1 x 2 e 2, x 0 f (x ) ( ) 2 2 2 0 , O .W . The mean and the variance are v and 2v, respectively.