1. Continuous Random Variables and Probability Distributions
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.
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 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!

7. Example If a random variable Find the probability Solution
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 Solution
Determine the cumulative distribution function
Solution

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

15. Mean and Variance 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 Solution
Find the mean and variance of the given probability density function
Solution

17. Uniform Distribution f(x) x 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) = ( )/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) = and F(X) = Therefore, 10x = 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) = =
F(0.62) - F(-0.34)= – ( ) =
P(z>0.85)=1-P(z0.85)= 1-F(0.85) = =
P(z>-0.65) =1-F(-0.65)=1-[1-F(0.65)]=F(0.65) =
P(z<-0.85)= ( ) =
P(z<-4.6)=
P (z<-3.99) = =
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) =
P(3 < X < 7) = = P(0.5 < Z < 0.5)=P(Z < 0.5)  P(Z < 0.5)=
P(2 < X < 9) = =P(1.75 < Z < 1) = [P(Z < 1)  P(Z < 1.75)] =

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 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

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 =

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