University:
California State University, Northridge
Course:
MATH 340 | Introduction to Probability and Statistics
Academic year:

2024
Views:

37

Pages:

45
Author:

Reese Pittman

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. f ( x) ≥ 0 ∞ 2. ∫ f ( x )dx = 1 −∞ b 3. P( a ≤ X ≤ b) = ∫ f ( x )dx = area under f(x) from a to b a • 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 P(a ≤ X ≤ b) = P(a < X ≤ b) =P(a ≤ X < b) = P(a < X < b) • Not true for a discrete random variable! Example • If a random variable 2e for x  0 f ( x) =  0 for ≤ 0 −2 x • Find the probability – between 1 and 3 – greater than 0.5 • Solution 3 P(1 ≤ X ≤ 3) = ∫ 2e − 2 x dx = e − 2 − e −6 = 0.135 1 P ( X ≥ 5) = ∞ −2 x −1 2 e dx = e = 0.368 ∫ 0.5 Class Problem • Suppose that f(x)= e-(x-4) for x>4 • Determine the following probabilities – P(1 0 f ( x) =  0 for ≤ 0 • Solution x − 2t - 2x ∫ 2e dt = 1 - e for x > 0 F ( x) =  0 0 for x ≤ 0  • When x=1 yields F (1) = 1 − e −2 = 0.865 Class Problem • Suppose that f(x)=1.5 x2 for –1 102 . ) = 1 − P( X ≤ 102 . ) = 1 − F(102 . ) = 0.3 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]=µ determines the center of the probability density function • V[X]=σ2 determines the width • Illustrates several normal probability density functions Probability Density Function • X with probability density function 1 f ( x) = =e 2πσ − ( x − µ )2 2σ 2 −∞< x <∞ • Normal random variable with parameters -∞ <µ <∞ and σ>1 • Mean and variance – E[X]= µ , V[X]= σ2 • N(µ, σ2 ) 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(µ-σ0.85)=1-P(z≤0.85)= 1-F(0.85) = 10.85023 = 0.1977 • P(z>-0.65) =1-F(-0.65)=1-[1F(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 0) = P(Z > (−5/4)) = P(Z > −1.25) = 1 − P(Z < −1.25) = 0.89435  3−5 < Z < 7 −5 P =  4 4  • 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 − 2 − 5 < Z < 9 − 5  4 4     =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 • If X is a binomial random variable z = [ X − np ] np(1 − p ) • is approximately a standard normal random variable • Good for np>5 and n(1-p)>5 • Accuracy of approximation – Calculate the interval: µ ± 3σ = np ± 3 np(1 − p ) – 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)=λ and V(X)= λ z = [X − λ ] λ • Approximated a standard normal random variable • Good for λ >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 f ( x) = λe − λx • Mean and variance µ = E[ X ] = λ1 , σ 2 = V [ X ] = 1 λ Example • Suppose that the log-ons to a computer network follow a Poisson process with an average of 3 counts per minute a) What is the mean time between counts? b) What is the standard deviation of the time between counts? c) Determine x such that the probability that at least one count occurs before time x minutes is 0.95 Solution a) E(X) = 1/λ =1/3 = 0.333 minutes b) V(X) = 1/λ2 = 1/32 = 0.111, σ = 0.33 c) Value of x P( X < x) = x − 3t e dt 3 ∫ 0 = −e • − 3t x 0 = 1 − e − 3 x = 0.95 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 λ = 1 / E( X ) = 1 / 2 a) P(X > 2) = ∞ ∞ ∫ 21 e 2 −x/ 2 b) E(X) = 2 hours. dx = − e − x / 2 2 = e −1 = 0.3679 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 f ( x) = λr x r −1e − λx ( r − 1)! • Mean and Variance , for x > 0 and r = 1,2,3, ... – E(X)=r/λ and V(X)= r/λ2 Example • Errors caused by contamination on optical disks occur at the rate of one error every bits. Assume the errors follow a Poisson distribution. a) What is the mean number of bits until five errors occur? b) What is the standard deviation of the number of bits until five errors occur? c) The error-correcting code might be ineffective if there are three or more errors within 105 bits. What is the probability of this event? a) b) Solution X denote the number of bits until five errors occur Then, X has an Erlang distribution with r = 5 and λ=10-5 error per bit r 5 – E(X) = λ = 5 × 10 r = 5 × 10 10 σ = 5 × 10 10 = 223607 2 – V(X) = λ X c) 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 P (Y ≥ 3) = 1 − P (Y ≤ 2) = 1 − [e + e + e −1 0! 10 = 0.0803 −1 1! 11 −1 2! 12 ] Gamma Function • Generalization of factorial function leads • Definition of gamma function ∞ r −1 − x for r > 0 Γ( r ) = ∫ x e dx, 0 • Integral Γ(r) is finite • Using integration by parts it can be shown – Γ(r)=(r-1) Γ(r-1) – if r is a positive integer – Γ(r)=(r-1)! • Used in development of the gamma distribution Gamma Distribution • X with the following PDF • • • • • f ( x) = r −1 − λx λx e r Γ( r ) for x>0 Gamma random variable with parameters λ>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/λ and V(X)= r/λ2

Continuous Random Variables and Probability Distributions

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