Continuous Random Variables and Probability Distributions 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 Approximate probabilities for some binomial and Poisson distributions Calculate probabilities, determine means and variances for the continuous probability distributions presented

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

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

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

Important Point f(x) is used to calculate an area that 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!

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

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

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

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

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

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

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

Solution The cumulative distribution function for -1< x < 1 Then

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

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

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

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,

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?

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.

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

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

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

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

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

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

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

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)

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

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

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)

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

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

Approximation is approximately a standard normal random variable 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

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

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

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

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

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?

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.

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

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?

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

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

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