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Continuous distributions. Five important continuous distributions: uniform distribution (contiuous) Normal distribution c 2 –distribution[“ki-square”] t -distribution F -distribution. A reminder. Definition: Let X: S R be a continuous random variable. - PowerPoint PPT Presentation
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lecture 51
Continuous distributions
Five important continuous distributions:
1. uniform distribution (contiuous)
2. Normal distribution
3. 2 –distribution [“ki-square”]
4. t-distribution
5. F-distribution
lecture 52
A reminder
Definition: Let X: S R be a continuous random variable.
A density function for X, f (x), is defined by:
1. f (x) 0 for all x
2.
3.
f (x)
xba
P(a<X<b)
1)( dxxf
b
a
dxxfbXaP )()(
lecture 53
Uniform distributionDefinition
Definition: Let X be a random variable. If the density function is given by
then the distribution of X is the (continuous) uniform distribution on the interval [A,B].
xA B
f (x)
BxAAB
xf
1
)(AB
1
lecture 54
Uniform distributionMean & variance
Theorem:Let X be uniformly distributed on the interval [A,B]. Then we have:
• mean of X:
• variance of X: 12
)()(
2)(
2ABXVar
BAXE
lecture 55
Normal distribution Definition
Definition:Let X be a continuous random variable. If the density function is given by
then the distribution of X is called the normal distribution with parameters and 2 (known).
My notation:
The book’s: for density function
xexf
x 222
1
22
1)(
),;(
),(~ 2
xn
NX
lecture 56
Normal distributionExamples
The normal distribution is without doubt the most important continuous distribution, since many phenomena are well described by it.
IQ among AAU students Height among AAU students
90 100 110 120 130 140 1500
0.01
0.02
0.03
0.04
160 170 180 190 2000
0.02
0.04
0.06
0.08
Plotting in Matlab:>> x=90:1:150; y=normpdf(x,120,10); plot(x,y)
lecture 57
Normal distributionMean & variance
Theorem:If X ~ N(,) then
• mean of X:
• variance of X:
Density function:
N(1,1)N(0,2)
N(0,1)
2)(
)(
XVar
XE
-4 -2 0 2 4
0.0
0.1
0.2
0.3
0.4
x
lecture 58
Normal distributionStandard normal distribution
Standard normal distribution: Z ~ N(0,1)
Density function: Distribution function:
(see Table A.3)
Notice!! Due to symmetryP(Z z) = 1P(Z z)
2
2
1
2
1)(
zezf
zz
dxezZPzF2
2
1
2
1)()(
2
1
2
1 2
2
1)21( dzeZP
z
-3 -2 -1 0 1 2 3
0.0
0.1
0.2
0.3
0.4
lecture 59
We typically have X~N() where and .
Normal distributionStandard normal distribution
Standard normal distribution, N(0,1), is the only normal distribution for which the distribution function is tabulated.
Example: X ~ N(1,4)
What is P( X 3 ) ?
-2 0 2 4
0.0
0.1
0.2
0.3
0.4
1587.0)1(1)1()2
13
2
1()3(
ZPZP
XPXP
-2 0 2 4
0.0
0.1
0.2
0.3
0.4
-2 0 2 4
0.0
0.1
0.2
0.3
0.4
lecture 510
Normal distributionStandard normal distribution
Example cont.: X ~ N(1,4). What is P(X 3) ?
Theorem: Standardise
If X ~ N(,2) then
Z ~ N(0,1)X ~ N(1,4)
Equal areas = 0.1587
Z ~ N(0,1)
10~ ,NX
Normal distributionStandard normal distribution
lecture 511
-2 0 2 4
0.0
0.1
0.2
0.3
0.4
15870841301)1(1)3(8413.0)1( ..ZPXPZP
Normal distributionStandard normal distribution
lecture 512
1587.0)3(1)3( XPXPExample cont.: X ~ N(1,4). What is P(X 3) ?
Solution in Matlab: >> 1 – normcdf(3,1,2) ans =
0.1587
Cumulative distribution functionnormcdf(x,,)
Solution in R: > 1 - pnorm(3,1,2) [1] 0.1586553
Cumulative distribution functionpnorm(x,,)
lecture 513
Normal distributionExample
Problem:The lifetime of a light bulb is normal distributed with mean 800 hours and standard deviation 40 hours:
1. Find the probability that the lifetime of a bulb is between 750-850 hours.
2. Find the number of hours b, such that the probability of a bulb having a lifetime longer than b is 90%.
3. Find a time period symmetric around the mean so that the probability of a lifetime in this interval has probability 95%
650 700 750 800 850 900 950
0.000
0.002
0.004
0.006
0.008
0.010
Normal distributionExample
lecture 514
Solution in Matlab:1. P(750 < X < 850) = ?
>> normcdf(850,800,40)-normcdf(750,800,40)2. ?3. ...
Solution in R:1. P(750 < X < 850) = ?
> pnorm(850,800,40)-pnorm(750,800,40)2. P(X > b) = 0.90 P(X b) = 0.10
> qnorm(0.1,800,40)3. ...
lecture 515
Normal distributionRelation to the binomial distribution
)1( pnp
npXZ
If X is binomial distributed with parameters n and p, then
is approximately normal distributed.
Rule of thumb:If np5 and n(p)5, then the approximation is good
0 5 10 15 20
0.00
0.05
0.10
0.15
lecture 516
Normal distributionLinear combinations
Theorem: linear combinationsIf X1, X2,..., Xn are independent random variables, where
and a1,a2,...,an are constant, then the linear combination
where
, for ,n,,i,σμNX iii 21),(~ 2
,( ),~ 22211 YYnn NXaXaXaY
2222
22
21
21
2
2211
nnY
nnY
aaa
aaa
lecture 517
The distributionDefinition
Definition: (alternative to Walpole, Myers, Myers & Ye)If Z1, Z2,..., Zn are independent random variables, where
Zi ~ N(0,1), for i =1,2,…,n,
then the distribution of
is the -distribution with n degrees of freedom.
Notation: Critical values: Table A.5
n
iin ZZZZY
1
2222
21
)(~ 2 nY
0 2 4 6 8 10
0.0
0.1
0.2
0.3
0.4
0.5
The distributionDefinition
lecture 518
Definition: A continuous random variable X follows a -distribution with n degrees of freedom if it has density function
0
)2(2
1)( 212
2
xex
nxf xn
n for
Y ~ 2(1)
Y ~ 2 (3)
Y ~ 2 (5)
Antag Y ~ 2(n)
E(Y) = nVar(Y) = 2n
E(Y/n) = 1Var(Y/n) = 2/n
lecture 519
t-distributionDefinition
Definition: Let Z ~ N(0,1) and V ~ (n) be two independent random variables. Then the distribution of
is called the t-distribution with n degrees of freedom.
Notation: T ~ t(n) Critical values: Table A.4
nV
ZT
t-distributionCompared to standard normal
lecture 520
X ~ N(0,1)
T ~ T(3)
-3 -2 -1 0 1 2 3
0.0
0.1
0.2
0.3
0.4
T ~ T(1)
•The t-distribution is symmetric around 0•The t-distribution is more flat than the standard normal•The more degrees of freedom the more the t-distribution looks like a standard normal
lecture 521
F-distribution Definition
Definition: Let U ~ 2(n1) and V ~ 2(n2) be two independent random variables. Then the distribution of
is called the F-distribution with n1 and n2 degrees of freedom.
Notation: F ~ F(n1 , n2) Critical values: Table A.6
2
1
nV
nU
F
0 1 2 3 4
0.0
0.2
0.4
0.6
0.8
1.0
F-distribution Example
lecture 522
F ~ F(10,30)
F ~ F(6,10)
F ~ F(20,50)