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Chapter 4

Chapter 4. Continuous Random Variables and Probability Distributions. Learning Objectives. Determine probabilities from probability density functions Determine probabilities from cumulative distribution functions Calculate means and variances Standardize normal random variables

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Chapter 4

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  1. Chapter 4 Continuous Random Variables and Probability Distributions

  2. Learning Objectives • Determine probabilities from probability density functions • Determine probabilities from cumulative distribution functions • Calculate means and variances • Standardize normal random variables • Approximate probabilities for some binomial and Poisson distributions • Calculate probabilities, determine means and variances for the continuous probability distributions presented

  3. Continuous Random Variables • Discussed about discrete random variables • Continuous random variable, X, has a distinctly different distribution from the discrete random variables • Includes all values in an interval of real numbers • Thought of as a continuum

  4. Probability Distributions • Describe the probability distribution of a continuous random variable X • Probability that X is between a and b is determined as the integral of f(x) from a to b

  5. Probability Density Function • Continuous random variable X, a probability density function 1. 2. 3. area under f(x) from a to b • Zero for x values that cannot occur

  6. Important Point • f(x) is used to calculate an areathat represents the probability that X assumes a value in [a, b] • Probability at any point is zero, because every point has zero width • P(X=x)=0 • Not distinguish between inequalities such as < or  for continuous random variables • For any x1 and x2 • Not true for a discrete random variable!

  7. Example • If a random variable • Find the probability • between 1 and 3 • greater than 0.5 • Solution

  8. Class Problem • Suppose that f(x)= e-(x-4) for x>4 • Determine the following probabilities • P(1<X) • P(2X<5) • Determine such that P(X<x) =0.9

  9. Solution • Part a because f(x)=0 for x<4 • Part b • Part c • Then, x = 4 – ln(0.10) = 6.303

  10. Cumulative Distribution Function • Stated in the same way as we did for the discrete random variable • F(x) is defined for every number x by

  11. F(x) and f(x) • Probability that the random variable will take on a value on the interval from a to b is F(b) – F(a) • Fundamental theorem of integral calculus

  12. Example • Find the cumulative distribution function of the following pdf • Solution • When x=1 yields

  13. Class Problem • Suppose that f(x)=1.5 x2 for –1<x<1 • Determine the cumulative distribution function • Solution

  14. Solution • The cumulative distribution function for -1< x < 1 • Then

  15. Mean and Variance • Defined similarly to a discrete random variable • Mean or expected value of X • E[X]= • Variance • V(X)=

  16. Example • If a random variable has • Find the mean and variance of the given probability density function • Solution

  17. Uniform Distribution • Simplest continuous distribution • Probability Density • Mean & Standard Deviation • Proof f(x) x

  18. Example • Suppose X has a continuous uniform distribution over the interval [1.5, 5.5] • Determine the mean, variance, and standard deviation of X • What is P(X<2.5)? • Solution E(X) = (5.5+1.5)/2 = 3.5,

  19. Class Problem • The thickness of a flange on a aircraft component is uniformly distributed between 0.95 and 1.05 millimeters • Determine the cumulative distribution function of flange thickness • Determine the proportion of flanges that exceeds 1.02 millimeters • What thickness is exceeded by 90% of the flanges?

  20. Solution • The distribution of X is f(x) = 10 for 0.95 < x < 1.05 Now b) The probability c) If P(X > x)=0.90, then 1  F(X) = 0.90 and F(X) = 0.10. Therefore, 10x - 9.5 = 0.10 and x = 0.96.

  21. Normal Distribution • Most widely used model for the distribution of a random variable is a normal distribution • Describes many random processes or continuous phenomena • Used to approximate discrete probability distributions • Basis for classical statistical inference

  22. Mean and Variance • Random variables with different means and variances can be modeled by normal probability • E[X]= determines the center of the probability density function • V[X]=2 determines the width • Illustrates several normal probability density functions

  23. Probability Density Function • X with probability density function • Normal random variable with parameters - < < and >1 • Mean and variance • E[X]=  , V[X]= 2 • N(, 2 ) used to denote the distribution

  24. Useful Information • Total area under the curve is 1.0 • Two tails of the curve extend indefinitely • Useful results • P(-<X<+)=0.6827 • P(-2<X<+2)=0.9545 • P(-3<X<+3)=0.9973

  25. Calculating the Probabilities • Normal distributions differ by mean & standard deviation • Each distribution would require its own table • Infinite number of tables!

  26. Definition • Normal random variable with=0 and 2=1 • Called a standard normal random variable • Denoted as Z • Cumulative distribution function of a standard normal random variable is denoted as • Appendix Table II provides cumulative probability values

  27. Standardize the Normal Distribution • Use the following random variable z to standardize a normal distribution into standard normal distribution • Calculate the probabilities Standardized Normal Distribution Normal distribution

  28. Working with Table • Table II provides values of (z) for values of Z • Suppose Z=1.5 • Note that P(a<X<b)=F(a) – F(b)

  29. Example • Find probabilities that a random variable having the standard normal distribution will take on a value • between 0.87 and 1.28 • between –0.34 and 0.62 • greater than 0.85 • greater than –0.65 • less than –0.85 • less than –4.6

  30. Solution • F(1.28) – F(0.85) = 0.8997-0.8078 = 0.0919 • F(0.62) - F(-0.34)= 0.7324 – (1-0.6331) = 0.3655 • P(z>0.85)=1-P(z0.85)= 1-F(0.85) = 1-0.85023 = 0.1977 • P(z>-0.65) =1-F(-0.65)=1-[1-F(0.65)]=F(0.65) = 0.7422 • P(z<-0.85)= (1-0.8551) = 0.1949 • P(z<-4.6)= • P (z<-3.99) = 1-0.99967 = 0.000033 • P(z<-4.6)< P(z<-3.99) = ~ 0

  31. Class Problem • Assume X is normally distributed with a mean of 5 and a standard deviation of 4 • Determine the following • P(X<11) • P(X>0) • P(3<X<7) • P(2 < X < 9)

  32. Solution • P(X < 11) = = P(Z < 1.5) = 0.93319 • P(X > 0) = P(Z > (5/4)) = P(Z > 1.25) = 1  P(Z < 1.25) = 0.89435 • P(3 < X < 7) = = P(0.5 < Z < 0.5)=P(Z < 0.5)  P(Z < 0.5)= 0.38292 • P(2 < X < 9) = =P(1.75 < Z < 1) = [P(Z < 1)  P(Z < 1.75)] = 0.80128

  33. Normal Approximation of Binomial Distribution • Difficult to calculate probabilities when n is large • Used to approximate binomial probabilities for cases in which n is large • Gives approximate probability only

  34. Approximation • If X is a binomial random variable • is approximately a standard normal random variable • Good for np>5 and n(1-p)>5 • Accuracy of approximation • Calculate the interval: • If Interval lies in range 0 to n, normal approximation can be used

  35. Normal Approximation of Poisson Distribution • If X is a Poisson random variable with E(X)= and V(X)=  • Approximated a standard normal random variable • Good for  >5

  36. Exponential Distribution • Poisson distribution defined a random variable to be the number events during a given time interval or in a specified regions • Time or distance between the events is another random variable that is often of interest • Probability density function • Mean and variance

  37. Example • Suppose that the log-ons to a computer network follow a Poisson process with an average of 3 counts per minute • What is the mean time between counts? • What is the standard deviation of the time between counts? • Determine x such that the probability that at least one count occurs before time x minutes is 0.95

  38. Solution • E(X) = 1/ =1/3 = 0.333 minutes • V(X) = 1/2 = 1/32 = 0.111,  = 0.33 • Value of x • Thus, x = 0.9986

  39. Class Problem • The time between the arrival of electronic messages at your computer is exponentially distributed with a mean of two hours • (a) What is the probability that you do not receive a message during a two-hour period? • (b) What is the expected time between your fifth and sixth messages?

  40. Solution • Let X denote the time until a message is received. Then, X is an exponential random variable and • P(X > 2) = • E(X) = 2 hours.

  41. Erlang Distribution • Describes the length until the first count is obtained in a Poisson process • Generalization of the exponential distribution is the length until r counts occur in a Poisson process. • Random variable that equals the interval length until r counts occur in a Poisson process • PDF • Mean and Variance • E(X)=r/ and V(X)= r/2

  42. Example • Errors caused by contamination on optical disks occur at the rate of one error every bits. Assume the errors follow a Poisson distribution. • What is the mean number of bits until five errors occur? • What is the standard deviation of the number of bits until five errors occur? • The error-correcting code might be ineffective if there are three or more errors within 105 bits. What is the probability of this event?

  43. Solution • X denote the number of bits until five errors occur • Then, X has an Erlang distribution with r = 5 and =10-5 error per bit • E(X) = • V(X) = • Y denote the number of errors in 105 bits. Then, Y is a Poisson random variable with 10-5 error per bit which equals 1 error per 105 bits

  44. Gamma Function • Generalization of factorial function leads • Definition of gamma function for r > 0 • Integral (r) is finite • Using integration by parts it can be shown • (r)=(r-1) (r-1) • if r is a positive integer • (r)=(r-1)! • Used in development of the gamma distribution

  45. Gamma Distribution • X with the following PDF • x > 0 • Gamma random variable with parameters >0 and r>0 • If r is an integer, X has an Erlang distribution • Erlang is a special case of the gamma distribution • Mean and Variance • E(X)=r/ and V(X)= r/2

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