Chap3b.continuous Distrib
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Transcript of Chap3b.continuous Distrib
Chapter Topics
Gamma Distribution Weibull Distribution Exponential Distribution The normal distribution The standardized normal distribution Evaluating the normality assumption
Continuous Probability Distributions
Continuous random variable Values from interval of numbers Absence of gaps
Continuous probability distribution Distribution of continuous random variable
Most important continuous probability distribution The normal distribution
Let X be a random variable with range [0,2] and pdf defined by f(x)=1/2 for all x between 0 and 2 and f(x)=0 for all other values of x. Note that since the integral of zero is zero we get
22
00
1( ) 1/ 2 1 0 1
2f x dx dx x
Example
That is, as with all continuous pdfs, the total area under the curve is 1. We might use this random variable to model the position at which a two-meter with length of rope breaks when put under tension, assuming “every point is equally likely”. Then the probability the break occurs in the last half-meter of the rope is
22 2
3/ 2 3/ 23/ 2
1(3/ 2 2) ( ) 1/ 2 1/ 4
2P X f x dx dx x
Example
Let Y be a random variable whose range is the nonnegative reals and whose pdf is defined by
for nonnegative values of x (and 0 for negative values of x). Then
/ 7501( )
750xf x e
/ 750 / 750
0 0
/ 750 0 750/
0
1( ) lim
750
lim lim 1 0 1
tx x
t
tx t
t t
f x dx e dx e dx
e e e
Cumulative Distribution Functions
In the second example above, F(x)=0 if x is negative and for nonnegative x we have
Thus the probability of a light bulb lasting between 500 and 1000 hours is
/ 750 / 750 / 750 / 750
00
1( ) 1 1
750
x xt t x xF x e dt e e e
1000/ 750 500/ 750 2/3 4 /3(1000) (500) (1 ) (1 ) 0.250F F e e e e
The random variable Y might be a reasonable choice to model the lifetime in hours of a standard light bulb with average life 750 hours. To find the probability a bulb lasts under 500 hours, you calculate
500 500/ 750 / 750 2/3
00
1(0 500) 1 0.487
750x xP Y e dx e e
Gamma Random Variables
A continuous r.v. whose density is given by
Gamma function
It is easy to show by induction that for general n
0,
0,0
0,),|(
1
x
xxe
xf
x
0
1dxxe x
!1 nn
Mean and Variance
Mean Similarly we have that
k
dx
dxex)X(E
kex
k
k
/xk
k
k
/xk
k
k
k
1
01
1
0
111
1
111
2
1
02
2
0
1212
1
2
122
kk
dx
dxex)X(E
kex
k
k
/xk
k
k
/xk
k
k
k
Moment Gf
Mgf The rth derivative
k
k
ext
t
tk
kX
t
xtd
dxex)t(M
k
xtkk
k
x
k
1
1
1
0
1
1
0
111
111
r
k)rk(r
rkrk
)rk(
rkr
rX
XE
t
t
)rk)....(k(ktM
1
1
11
Exponential Distributions
arrival time 1
: any value of continuous random variable
: the population average number of
arrivals per unit of time
1/ : average time between arrivals
2.71828
XP X e
X
e
e.g.: Drivers Arriving at a Toll Bridge; Customers Arriving at an ATM Machine
Exponential Distributions
Describes time or distance between events Used for queues
Density function
Parameters
(continued)
f(X)
X
= 0.5
= 2.0
1 x
f x e
The CDF of X is No Memory Property
/x
x/t
x/t
/tx
e
e
/tde
dte)x(F
1
0
0
0
1
)tX(P
)aX|taX(P
a
ta
ee
)aX(P)taX(P
)aX(P)aXandtaX(P
Example
e.g.: Customers arrive at the check out line of a supermarket at the rate of 30 per hour. What is the probability that the arrival time between consecutive customers to be greater than five minutes?
30 5/ 60
30 5 / 60 hours
arrival time > 1 arrival time
1 1
.0821
X
P X P X
e
Exponential Distribution in PHStat
PHStat | probability & prob. Distributions | exponential
Example in excel spreadsheet
Microsoft Excel Worksheet
Exponential Random Variables
X: exponential RV with parameter λ
Y: exponential RV with parameter μ
X, Y: independentThen:1. min{X, Y}: exponential
RV with parameter λ+μ
2. P{X<Y} = λ/(λ+μ)
Proof:
0 0
0 0
0 0
0
( )
0 0
{ } ( , )
(1 )
( )
1
y
XY
yx y
yy x
y y
y y
P X Y f x y dx dy
e e dx dy
e e dx dy
e e dy
e dy e dy
( )
( )
{min{ , } } { , }{ } { }
{min{ , } } 1
t t t
t
P X Y t P X t Y tP X t P Y te e e
P X Y t e
Weibull Distribution
A continuous r.v. X is said to have the Weibull distribution with parameters ß,Ø>0 if it has a pdf of the form
It follows that the 100 x pth percentile has the form
01,,
0,,
/
/1
xexF
xexxf
x
x
.1ln /1 px
pxF
p
p
The Mean and Variance
The Mean The Variance
/11
;/
)(
0
1)/11(
/1
0
/1)1(
/1
0
dtet
txxt
dxex
dxexxXE
t
x
x
/21
;/
)(
2
0
1)/21(2
/1
0
/1)2(
/1
0
22
dtet
txxt
dxex
dxexxXE
t
x
x
The Normal Distribution
“Bell shaped” Symmetrical Mean, median and
mode are equal Interquartile range
equals 1.33 Random variable
has infinite range
Mean Median Mode
X
f(X)
The Mathematical Model
21
2
2
1
2
: density of random variable
3.14159; 2.71828
: population mean
: population standard deviation
: value of random variable
X
f X e
f X X
e
X X
Many Normal Distributions
By varying the parameters and , we obtain different normal distributions
There are an infinite number of normal distributions
Which Table to Use?
An infinite number of normal distributions means an infinite number of tables to look
up!
Solution: The Cumulative Standardized Normal
Distribution
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.5478.02
0.1 .5478
Cumulative Standardized Normal Distribution Table (Portion)
Probabilities
Shaded Area Exaggerated
Only One Table is Needed
0 1Z Z
Z = 0.12
0
Standardizing Example
6.2 50.12
10
XZ
Normal Distribution
Standardized Normal
Distribution
Shaded Area Exaggerated
10 1Z
5 6.2 X Z0Z
0.12
Example:
Normal Distribution
Standardized Normal
Distribution
Shaded Area Exaggerated
10 1Z
5 7.1 X Z0Z
0.21
2.9 5 7.1 5.21 .21
10 10
X XZ Z
2.9 0.21
.0832
2.9 7.1 .1664P X
.0832
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.5832.02
0.1 .5478
Cumulative Standardized Normal Distribution Table (Portion)
Shaded Area Exaggerated
0 1Z Z
Z = 0.21
Example: 2.9 7.1 .1664P X
(continued)
0
Z .00 .01
-03 .3821 .3783 .3745
.4207 .4168
-0.1.4602 .4562 .4522
0.0 .5000 .4960 .4920
.4168.02
-02 .4129
Cumulative Standardized Normal Distribution Table (Portion)
Shaded Area Exaggerated
0 1Z Z
Z = -0.21
Example: 2.9 7.1 .1664P X
(continued)
0
Normal Distribution in PHStat
PHStat | probability & prob. Distributions | normal …
Example in excel spreadsheet
Microsoft Excel Worksheet
Example: 8 .3821P X
Normal Distribution
Standardized Normal
Distribution
Shaded Area Exaggerated
10 1Z
5 8 X Z0Z
0.30
8 5.30
10
XZ
.3821
Example: 8 .3821P X
(continued)
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.6179.02
0.1 .5478
Cumulative Standardized Normal Distribution Table (Portion)
Shaded Area Exaggerated
0 1Z Z
Z = 0.30
0
.6217
Finding Z Values for Known Probabilities
Z .00 0.2
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
.6179 .6255
.01
0.3
Cumulative Standardized Normal Distribution Table
(Portion)
What is Z Given Probability = 0.1217 ?
Shaded Area Exaggerated
.6217
0 1Z Z
.31Z 0
Recovering X Values for Known Probabilities
5 .30 10 8X Z
Normal Distribution
Standardized Normal
Distribution10 1Z
5 ? X Z0Z 0.30
.3821.1179
Assessing Normality
Not all continuous random variables are normally distributed
It is important to evaluate how well the data set seems to be adequately approximated by a normal distribution
Assessing Normality Construct charts
For small- or moderate-sized data sets, do stem-and-leaf display and box-and-whisker plot look symmetric?
For large data sets, does the histogram or polygon appear bell-shaped?
Compute descriptive summary measures Do the mean, median and mode have similar
values? Is the interquartile range approximately 1.33 ?
Is the range approximately 6 ?
(continued)
Assessing Normality
Observe the distribution of the data set Do approximately 2/3 of the observations lie
between mean 1 standard deviation? Do approximately 4/5 of the observations lie
between mean 1.28 standard deviations? Do approximately 19/20 of the observations
lie between mean 2 standard deviations? Evaluate normal probability plot
Do the points lie on or close to a straight line with positive slope?
(continued)
Assessing Normality
Normal probability plot Arrange data into ordered array Find corresponding standardized normal
quantile values Plot the pairs of points with observed data
values on the vertical axis and the standardized normal quantile values on the horizontal axis
Evaluate the plot for evidence of linearity
(continued)
Assessing Normality
Normal Probability Plot for Normal Distribution
Look for Straight Line!
30
60
90
-2 -1 0 1 2
Z
X
(continued)