chapter 4 continuous random variables and probability distributions
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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
0)( xf
1)( dxxf
b
a
dxxfbXaP )()(
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! )( bXaP )( bXaP )( bXaP )( bXaP
Example
• If a random variable
• Find the probability– between 1 and 3– greater than 0.5
• Solution
0for 0
0for x 2)(
2 xexf
135.02)31( 623
1
2 eedxeXP x
368.02)5(5.0
12
edxeXP x
Class Problem• Suppose that f(x)= e-(x-4) for x>4 • Determine the following probabilities
– P(1<X)– P(2X<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
1)1( 4)4(
4
)4(
xx edxeXP
6321.01)52( 15
4
54
)4()4( eedxeXP xx
90.01)( )4(4
)4(
4
)4( xxxx
x eedxexXP
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
x
duufxXPxF )()()( xfor
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
)()(
xfdx
xdF
Example
• Find the cumulative distribution function of the following pdf
• Solution
• When x=1 yields
0for 0
0for x 2)(
2xexf
0for x 0
0for x e-1dt 2)(
2x-2
0
tx
exF
865.01)1( 2 eF
Class Problem
• Suppose that f(x)=1.5 x2 for –1<x<1– Determine the cumulative distribution function
• Solution
Solution
• The cumulative distribution function
for -1< x < 1• Then
5.05.05.05.1)( 31
3
1
2 xxdxxxF xx
x
x1 1,
1x1- 5.00.5x
-1x ,0
)( 3xF
Mean and Variance
• Defined similarly to a discrete random variable• Mean or expected value of X
– E[X]=
• Variance
– V(X)=
dxxxf )(
dxxfx )()( 2
Example• If a random variable has
– Find the mean and variance of the given probability density function
• Solution
0for x 0
0for x 2)(
2 xexf
5.02.)(][0
2
dxexdxxxfxE x
4/1
2)2/1(
)()(][
0
22
2
dxex
dxxfxYV
x
Uniform Distribution• Simplest continuous distribution
• Probability Density
• Mean & Standard Deviation
• Proof
abxf
1
)(ab
xf
1)(
12 and
2
abba
12 and
2
abba
x
f(x)
12
)(
)(3
)2
()2
(()(
2/)(5.0
)(
232
2
ab
ab
bax
dxab
bax
XV
baab
Xdxab
xXE
ba
b
a
ba
b
a
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)?
• SolutionE(X) = (5.5+1.5)/2 = 3.5,
866.043
4/312
2)5.15.5()(
x
XV
25.05.25.125.0
5.2
5.125.0)5.2(
x
dxXP
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
a) 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.
F x
x
x x
x
( )
, .
. , . .
, .
0 0 95
10 9 5 0 95 105
1 105
P X P X F( . ) ( . ) ( . ) . 102 1 102 1 102 0 3
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
2
2
2
)(
2
1)(
x
exf x
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!
X
f(X)
X
f(X)
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
)()( zZPz
Standardize the Normal Distribution
• Use the following random variable z to standardize a normal distribution into standard normal distribution
• Calculate the probabilities
X
X
= 0
= 1
Z = 0
= 1
Z
Normal distribution Standardized Normal Distribution
)()()( zZPxX
PxXP
Xz
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(z0.85)= 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
4
511ZP
4
574
53 ZP
4
594
52 ZP
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
• 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
pnpnpXz 1
133 pnpnp 133 pnpnp
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
Xz
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
xexf )( xexf )(
121 ][ ,][ XVXE 121 ][ ,][ XVXE
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
Solutiona) E(X) = 1/ =1/3 = 0.333 minutesb) V(X) = 1/2 = 1/32 = 0.111, = 0.33c) Value of x
• Thus, x = 0.9986
95.01
3)(
3
0
3
0
3
x
xt
xt
ee
dtexXP
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
a) P(X > 2) =
b) E(X) = 2 hours.
12
2 2
2 2
1 0 3679e dx e ex x
/ / .
1 1 2/ ( ) /E X
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
• Mean and Variance– E(X)=r/ and V(X)= r/2
... 1,2,3,r and 0for x ,)!1(
)(1
r
exxf
xrr
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?
Solutiona) X denote the number of bits until five errors occur
b) Then, X has an Erlang distribution with r = 5 and =10-5 error per bit
– E(X) = – V(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
r
5 105
r
2105 10 X 5 10 22360710
0803.0
1
)2(1
)3(
!21
!11
!01 211101
eeeYP
YP
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
,)(0
1 dxexr xr
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