Transcript standard normal distribution

SOME CONTINUOUS
PROBABILITY
DISTRIBUTIONS
Uniform, Normal, Exponential,
Gamma and Chi-Square
Distributions
1
Uniform Distribution
– A random variable X is said to be uniformly
distributed if its density function is
1
f ( x) 
a  x  b.
ba
– The expected value and the variance are
ab
(b  a) 2
E(X) 
V( X ) 
2
12
2
Indicator functions
• It is sometimes convenient to express the
p.m.f. or p.d.f. by using indicator functions.
This is especially true when the range of
random variable depends on a parameter.
• Ex: Uniform distribution
3
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,000 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(2500X3000) = (3000-2500)(1/3000) = .1667
1/3000
2000 2500 3000
5000
x
4
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X4000) = (5000-4000)(1/3000) = .333
1/3000
2000
4000
5000
x
5
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X=2500) = (2500-2500)(1/3000) = 0
1/3000
2000 2500
5000
x
6
Normal Distribution
• This is the most popular continuous
distribution.
– Many distributions can be approximated by a
normal distribution.
– The normal distribution is the cornerstone
distribution of statistical inference.
7
Normal Distribution
• A random variable X with mean m and
variance s2 is normally distributed if its
probability density function is given by
 xm 
 (1 / 2)

 s 
e
2
1
f (x) 
   x  ; s  0
s 2
where   3.14159... and e  2.71828...
8
The Shape of the Normal
Distribution
The normal distribution is bell shaped, and
symmetrical around m.
90
m
Why symmetrical? Let m = 100. Suppose x = 110.
f (110) 
1
s 2
 110 100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2) 
s
e
110
Now suppose x = 90
2
f (90) 
1
s 2
 90 100 
 (1 / 2)

s


e
2

1
s 2
 10 
 (1 / 2)

s


e
2
The Effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
How does the expected value affect the location of f(x)?
m = 10 m = 11 m = 12
10
Finding Normal Probabilities
• Two facts help calculate normal probabilities:
– The normal distribution is symmetrical.
– Any normal distribution can be transformed into
a specific normal distribution called…
“STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer
is normally distributed, with a mean of 50 minutes
and a standard deviation of 10 minutes. What is
the probability that a computer is assembled in a
time between 45 and 60 minutes?
11
STANDARD NORMAL
DISTRIBUTION
• NORMAL DISTRIBUTION WITH MEAN 0
AND VARIANCE 1.
• IF X~N(m , s2), THEN
Z
X m
s
~ N (0,1)
NOTE: Z IS KNOWN AS Z SCORES.
• “ ~ “ MEANS “DISTRIBUTED AS”
12
Finding Normal Probabilities
• Solution
– If X denotes the assembly time of a computer,
we seek the probability P(45<X<60).
– This probability can be calculated by creating a
new normal variable the standard normal
variable.
Every normal variable
with some m and s, can
be transformed into this Z.
X  mx
Z
sx
E(Z) = m = 0
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
V(Z) = s2 = 1
13
Standard normal probabilities
Copied from Walck, C (2007) Handbook on Statistical Distributions for
experimentalists
14
Standard normal table 1
15
Standard normal table 2
16
Standard normal table 3
17
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
18
Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0
z
0.0
0.1
.
.
1.0
.
.
1.2
.
.
0
0.0000
0.0398
.
.
0.3413
.
.
0.3849
.
.
0.01
0.0040
0.0438
.
.
0.3438
.
.
0.3869
.
.
…….
…….
.
.
0.05
0.0199
0.0596
.
.
0.3531
.
.
0.3944
.
.
P(0<Z<z0)
Z=0
0.06
0.0239
0.0636
.
.
0.3554
.
.
0.3962
.
.
Z = z0
19
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = -.5
z0 = 1
20
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1
P(0<Z<1)
z
0.0
0.1
.
.
1.0
.
0
0.0000
0.0398
.
.
0.3413
.
0.01
0.0040
0.0438
.
.
0.3438
.
…….
0.05
0.0199
0.0596
.
.3413
.
0.3531
.
z0 =-.5z=0 z0 = 1
0.06
0.0239
0.636
.
.
0.3554
.
21
Finding Normal Probabilities
• The symmetry of the normal distribution
makes it possible to calculate
probabilities for negative values of Z
using the table as follows:
-z0
0
+z0
P(-z0<Z<0) = P(0<Z<z0)
22
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.01
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5
23
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
.1915
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5 1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
24
Finding Normal Probabilities
• Example - continued
0
0.5000
0.5398
.
.
0.6915
.
0.01
0.5040
0.5438
.
.
….
.
…….
.3413
z
0.0
0.1
.
.
0.5
.
This table provides
probabilities from -∞ to z0
0.05
0.5199
0.5596
.
.
….
.
0.06
0.5239
0.5636
.
.
….
.
P(Z<-0.5)=1-P(Z>-0.5)=1-0.6915=0.3085
By Symmetry
P(Z<0.5)
25
Finding Normal Probabilities
• Example
– The rate of return (X) on an investment is normally
distributed with a mean of 10% and standard
deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
0%
10%
0 - 10
(i) P(X< 0 ) = P(Z<
) = P(Z< - 2)
5
X
.4772
-2
=P(Z>2) = 0.5 - P(0<Z<2) = 0.5 - .4772 = .0228
0
2
Z
26
Finding Normal Probabilities
• Example
– The rate of return (X) on an investment is normally
distributed with mean of 10% and standard
deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
X
0%
10%
0 - 10
(ii) P(X< 0 ) = P(Z<
)
10
.3413
-1
= P(Z< - 1) = P(Z>1) = 0.5 - P(0<Z<1) = 0.5 - .3413 = .1587
1
Z
27
AREAS UNDER THE STANDARD
NORMAL DENSITY
P(0<Z<1)=.3413
0
1
Z
28
AREAS UNDER THE STANDARD
NORMAL DENSITY
.3413
.4772
P(1<Z<2)=.4772-.3413=.1359
29
EXAMPLES
• P( Z < 0.94 ) = 0.5 + P( 0 < Z < 0.94 )
= 0.5 + 0.3264 = 0.8264
0.8264
0
0.94
30
EXAMPLES
• P( Z > 1.76 ) = 0.5 – P( 0 < Z < 1.76 )
= 0.5 – 0.4608 = 0.0392
0.0392
0
1.76
31
EXAMPLES
• P( -1.56 < Z < 2.13 ) =
= P( -1.56 < Z < 0 ) + P( 0 < Z < 2.13 )
Because of symmetry
P(0 < Z < 1.56)
= 0.4406 + 0.4834 = 0.9240
0.9240
-1.56
2.13
32
STANDARDIZATION
FORMULA
• If X~N(m , s2), then the standardized value Z
of any ‘X-score’ associated with calculating
probabilities for the X distribution is:
Z
X m
s
• The standardized value Z of any ‘X-score’
associated with calculating probabilities for
the X distribution is:
x  m  z.s (Converse Formula)
33
Finding Values of Z
• Sometimes we need to find the value of Z
for a given probability
• We use the notation zA to express a Z
value for which P(Z > zA) = A
A
zA
34
PERCENTILE
• The pth percentile of a set of measurements
is the value for which at most p% of the
measurements are less than that value.
• 80th percentile means P( Z < a ) = 0.80
• If Z ~ N(0,1) and A is any probability, then
P( Z > zA) = A
A
zA
35
Finding Values of Z
• Example
– Determine z exceeded by 5% of the population
– Determine z such that 5% of the population is below
• Solution
z.05 is defined as the z value for which the area on its
right under the standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
0
Z0.05
1.645
36
EXAMPLES
• Let X be rate of return on a proposed
investment. Mean is 0.30 and standard
deviation is 0.1.
a) P(X>.55)=?
Standardization formula
b) P(X<.22)=?
c) P(.25<X<.35)=?
d) 80th Percentile of X is? Converse Formula
e) 30th Percentile of X is?
37
a)P(X  0.55)  P
b)P(X  0.22)  P


ANSWERS

X - 0.3
0.55 - 0.3
=Z>
= 2.5 = 0.5 - 0.4938 = 0.0062
0.1
0.1

X - 0.3
0.22 - 0.3
=Z
= 0.8 = 0.5 - 0.2881 = 0.2119
0.1
0.1
c) P(0.25  X  0.35)  P 0.25  0.3  0.5  X - 0.3 = Z  0.35 - 0.3 = 0.5

= 2.* (0.1915)  0.3830
0.1
0.1
0.1

d) 80th Percentile of X is x  m  s .z0.20  .3+(.85)*(.1)=.385
e) 30th Percentile of X is x  m  s .z0.70  .3+(-.53)*(.1)=.247
38
The Normal Approximation to the
Binomial Distribution
• The normal distribution provides a close
approximation to the Binomial distribution
when n (number of trials) is large and p
(success probability) is close to 0.5.
• The approximation is used only when
np  5 and
n(1-p)  5
39
The Normal Approximation to the
Binomial Distribution
• If the assumptions are satisfied, the Binomial random
variable X can be approximated by normal distribution
with mean m = np and s2 = np(1-p).
• In probability calculations, the continuity correction
improves the results. For example, if X is Binomial
random variable, then
P(X  a) ≈ P(X<a+0.5)
P(X  a) ≈ P(X>a-0.5)
40
EXAMPLE
• Let X ~ Binomial(25,0.6), and want to find P(X ≤ 13).
• Exact binomial calculation:
P(X  13) 
13  25 
x
25  x


(
0
.
6
)
(
0
.
4
)
 0.267
 x 
x  0 
• Normal approximation (w/o correction): Y~N(15,2.45²)
13  15
P(X  13)  P(Y  13)  P( Z 
)  P( Z  0.82)  0.206
2.45
Normal approximation is good, but not great!
41
EXAMPLE, cont.
Bars – Bin(25,0.6); line – N(15, 2.45²)
Copied from Casella & Berger (1990)
42
EXAMPLE, cont.
• Normal approximation (w correction):
Y~N(15,2.45²)
P(X  13)  P(X  13.5)  P(Y  13.5)  P( Z  0.61)  0.271
Much better approximation to the exact value:
0.267
43
Exponential Distribution
• The exponential distribution can be used
to model
– the length of time between telephone calls
– the length of time between arrivals at a
service station
– the lifetime of electronic components.
• When the number of occurrences of an
event follows the Poisson distribution, the
time between occurrences follows the
exponential distribution.
44
Exponential Distribution
A random variable is exponentially
distributed if its probability density function
is given by
f
1
X
 x  e
 x/
, x  0,   0
 = e-x,
f(x)
x>=0.
 is a
distribution
parameter.
is the
distribution parameter
>0).
E(X) = 
V(X) = 2
The cumulative distribution function is
F(x) =1e-x/, x0
45
Exponential distribution for 1 = .5, 1, 2
2.5
f(x) = 2e-2x
2
f(x) = 1e-1x
1.5
f(x) = .5e-.5x
1
0.5
0
0
1
2
3
4
5
2.5
2
1.5
1
P(a<X<b) = e-a/ e-b/
0.5
0
a
b
46
Exponential Distribution
• Finding exponential probabilities is
relatively easy:
– P(X < a) = P(X ≤ a)=F(a)=1 – e –a/ 
– P(X > a) = e–a/ 
– P(a< X < b) = e – a/ – e – b/ 
47
Exponential Distribution
• Example
– The lifetime of an alkaline battery is
exponentially distributed with mean 20 hours.
– Find the following probabilities:
• The battery will last between 10 and 15 hours.
• The battery will last for more than 20 hours.
48
Exponential Distribution
• Solution
– The mean = standard deviation =
  20 hours.
– Let X denote the lifetime.
• P(10<X<15) = e-.05(10) – e-.05(15) = .1341
• P(X > 20) = e-.05(20) = .3679
49
Exponential Distribution
• Example
The service rate at a supermarket checkout
is 6 customers per hour.
– If the service time is exponential, find the
following probabilities:
• A service is completed in 5 minutes,
• A customer leaves the counter more than 10
minutes after arriving
• A service is completed between 5 and 8 minutes.
50
Exponential Distribution
• Solution
– A service rate of 6 per hour =
A service rate of .1 per minute (1 = .1/minute).
– P(X < 5) = 1-e-.lx = 1 – e-.1(5) = .3935
– P(X >10) = e-.lx = e-.1(10) = .3679
– P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572
51
Exponential Distribution
• If pdf of lifetime of fluorescent lamp is
exponential with mean 0.10, find the life
for 95% reliability?
The reliability function = R(t) = 1F(t) = e-t/
R(t) = 0.95  t = ?
52
GAMMA DISTRIBUTION
• X~ Gamma(,)
1
 1  x / 
f  x 
x
e , x  0,   0,   0

   
E  X    and Var  X   
M  t   1   t  , t 

2
1

53
GAMMA DISTRIBUTION
• Gamma Function:

     x e dx
 1
x
0
where  is a positive integer.
Properties:
   1     ,  0
  n    n  1! for any integer n  1
1

   
 2
54
GAMMA DISTRIBUTION
(  1)  ( )


(  1)   x e dx   xx 1e  x dx
0
 x
0
( )   1  x

xx e dx  ( )

( ) 0
since the last integral is the expectation of a
Gamma distribution with parameters alpha and 1.
55
GAMMA DISTRIBUTION
• Let X1,X2,…,Xn be independent rvs with
Xi~Gamma(i, ). Then,
 X ~ Gamma   , 

n
i 1
i
n
i 1
i

•Let X be an rv with X~Gamma(, ). Then,
cX ~ Gamma  , c  where c is positive constant.
• Let X1,X2,…,Xn be a random sample with
Xi~Gamma(, ). Then,
n
X


X
~ Gamma  n , 
n
n

i 1
i
56
GAMMA DISTRIBUTION
• Special cases: Suppose X~Gamma(α,β)
– If α=1, then X~ Exponential(β)
– If α=p/2, β=2, then X~ 2 (p) (will come back in a min.)
– If Y=1/X, then Y ~ inverted gamma.
• Gamma approximates to Normal
distribution on the limit as alpha goes to
infinity.
57
Integral tricks
• Recall: h(t) is an odd function if h(-t)=-h(t);
It is an even function if h(-t)=h(t).
• Ex: h(x)=x exp{-x²/2} odd;
h(x)= exp{-x²/2-x} neither odd nor even

•
 odd function =0


•

 even function = 2* 0 even function

58
CHI-SQUARE DISTRIBUTION
Chi-square with  degrees of freedom
• X~ 2()= Gamma(/2,2)
1
f ( x)   / 2
x / 21e  x / 2 , x  0,   1,2,...
2 ( / 2)
E  X    and Var  X   2
M (t )  (1  2t )  / 2 t  1 / 2
59
DEGREES OF FREEDOM
• In statistics, the phrase degrees of freedom is
used to describe the number of values in the
final calculation of a statistic that are free to vary.
• The number of independent pieces of
information that go into the estimate of a
parameter is called the degrees of freedom (df) .
• How many components need to be known
before the vector is fully determined?
60
CHI-SQUARE DISTRIBUTION
• Chi-square (2) ≡ Exponential (2)
61
CHI-SQUARE DISTRIBUTION
• If rv X has Gamma(,) distribution, then
Y=2X/ has Gamma(,2) distribution. If 2
2
is positive integer, then Y has  2
distribution.
•Let X be an rv with X~N(0, 1). Then,
2
2
X ~ 1
•Let X1,X2,…,Xn be a r.s. with Xi~N(0,1). Then,
X ~ 
n
i 1
2
2
i
n
62
BETA DISTRIBUTION
• The Beta family of distributions is a
continuous family on (0,1) and often used
to model proportions.
1
 1
 1
f  x 
x 1  x  , 0  x  1,   0,   0.
B  ,  
where
1
B( ,  )   x

0
 1
(1  x)
 1
( )(  )
dx 
(   )

EX  
and Var  X  
 
        1 63
2
WEIBULL DISTRIBUTION
• To model the failure time data or hazard
functions.
P( t  T  t   | T  t )
h(t )  l i m
• Hazard function:

 0
Rate of change in prob that subject survives a little
past time t, given that subject survives upto time t.
• If X~Exp(), then Y=X1/ has Weibull(, )
distribution.

f  y  y e

 1
Y
 y /
, y  0,   0,   0
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CAUCHY DISTRIBUTION
• It is a symmetric and bell-shaped
distribution on (,) with pdf
f (x) 
1
s 1 (
1
x 
s
)2
,s  0
Since E X  , the mean does not exist.
• The mgf does not exist.
•  measures the center of the distribution
and it is the median.
• If X and Y have N(0,1) distribution, then Z=X/Y
has a Cauchy distribution with =0 and σ=1.
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CAUCHY DISTRIBUTION
• When studying hypothesis tests that assume
normality, seeing how the tests perform on data
from a Cauchy distribution is a good indicator of
how sensitive the tests are to heavy-tail
departures from normality.
• Likewise, it is a good check for robust
techniques that are designed to work well under
a wide variety of distributional assumptions.
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LOG-NORMAL DISTRIBUTION
• An rv X is said to have the lognormal
distribution, with parameters µ and s2, if
Y=ln(X) has the N(µ, s2) distribution.
•The lognormal distribution is used to model
continuous random quantities when the
distribution is believed to be skewed, such as
certain income and lifetime variables.
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STUDENT’S T DISTRIBUTION
• This distribution will arise in the study of
population mean when the underlying
distribution is normal.
• Let Z be a standard normal rv and let U be
a chi-square distributed rv independent of
Z, with  degrees of freedom. Then,
Z
X
~ t
U /
When n, XN(0,1).
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F DISTRIBUTION
• Let U and V be independent rvs with chisquare distributions with 1 and 2
degrees of freedom. Then,
U /
X
~ F 
V /
1
1, 2
2
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MULTIVARIATE
DISTRIBUTIONS
70
BIVARIATE NORMAL
DISTRIBUTION
• A pair of continuous rvs X and Y is said to
have a bivariate normal distribution if it has
a joint pdf of the form

1
1

f x, y  
exp 
2
 2 1  
2s1s 2 1   2


2 
2

y

m
y

m







y
y 
 x  m x   2   x  m x 


 








 s1 
s1  s 2   s 2   



   x  ,  y  , s1  0, s 2  0, 1    1.
71
If
BIVARIATE NORMAL
DISTRIBUTION
2
2
 X ,Y  ~ BVN m x ,m y ,s 1 ,s 2 ,   , then
2
2
X ~ N m x ,s 1  and Y ~ N m y ,s 2 
and  is the correlation coefficient btw X and Y.
1. Conditional on X=x,





s2
Y x ~ N m y   (x  mX ), s 22 1   2
s1

2. Conditional on Y=y,

s1
X y ~ N m x  
( y  mY ), s12 1   2
s2







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Mixture of distributions
• Let f1(y) and f2(y) be density functions, and
let a be a constant such that 0≤ a ≤1.
Consider the function
f(y)=a f1(y) + (1-a) f2(y)
Such a density function is often called a
mixture distribution. Here are some
examples of real life applications for such
distributions.
73
Mixture of distributions
• Example 1: Financial returns often
behave differently in normal situations and
during crisis times. A mixture of two
normal distributions with different means
and variances can be assumed for returns,
one for the returns during normal
situations, and another for during crises.
74
Mixture of distributions
• Example 2: The prices of houses in a
particular neighborhood will tend to be
similar, while prices in a different
neighborhood may be extremely different.
For instance, if we are to collect prices in
both Cukurambar and Sincan, we will
need a mixture of two distributions to
describe these prices.
75
Mixture of distributions
• Note that f(y) is a valid density function.
• Suppose that Y1 is a random variable with density
2
function f1(y), and E(Y1)=μ1 and Var(Y1)=s 1.
Similarly, suppose that Y2 is a random variable with
2
s
density function f2(y), and E(Y2)=μ2 and Var(Y2)= 2 .
Assume that Y is random variable whose density is
a mixture of the densities corresponding to Y1 and
Y2.
i) It can be shown that E(Y)= a μ1 + (1-a) μ2
ii) It can be shown that
2
2
Var(Y)= a s 1 + (1-a) s 2 + a(1-a) (μ1 - μ2)2
76
Problems
1. Mensa (from the Latin word for “mind”) is
an international society devoted to
intellectual pursuits. Any person who has
an IQ in the upper 2% of the general
population is eligible to join. If we assume
that IQs are normally distributed with μ =
100 and σ = 16, what is the lowest IQ that
will qualify a person for membership?
77
Problems
2. Suppose that numerical grades in a statistics
class are values of a r.v. X which is Normally
distributed with mean μ = 65 and standard
deviation 15. Suppose that letter grades are
assigned according to the following rule: student
receives an A if X ≥ 85; B if 70 ≤ X < 85; C if 55
≤ X < 70; D if 45 ≤ X < 55; and F if X ≤ 45. If a
student is chosen at random from this class,
calculate the probability that the student will earn
i) A; ii) B; iii) F.
78
Problems
3. Economic conditions cause fluctuations in
the prices of raw commodities as well as in
finished products. Let X denote the price
paid for a barrel of crude oil by the initial
carrier, and let Y denote the price paid by
the refinery purchasing the product from
the carrier. Assume that the joint density
for (X,Y) is given by
f(x,y)=c 20< x < y < 40
79
Problems
a) Find the value of c that makes this a joint
density for a two-dimensional random
variable.
b) Find the probability that the carrier will
pay at least $25 per barrel and the refinery
will pay at most $30 per barrel for the oil.
c) Find the probability that the price paid by
the refinery exceeds that of the carrier by
at least $10 per barrel.
80
Problems
d) Are X and Y independent? Explain.
e) Find E(Y-X). Interpret this expectation in a
practical sense.
81