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Continuous Distributions
Uniform Distribution
Definition 6.1: The continuous random variable X
has a uniform distribution if its p.d.f. is equal to a
constant on its support. If the support is the interval
[a, b ], then its p.d.f. is
Special Probability Densities
1
f ( x) 
The most common parameters in most of
the distributions are the lower moments:
mean m and variance s 2.
b a
a  x  b.
,
It is usually denoted as U(a, b ).
f(x)
1
1
F(x)
0
a
b a
a
0
1
The mean, variance, and m.g.f. of a continuous
random variable X that has a uniform
distribution are:
a b
s
,
2

(b  a )
2
2
m 
,
0  10
s
 5,
2

(10  0 )
2
ta
e e

,
M (t )   t ( b  a )

1,

2
Example: Let X be U(0,10), find the mean and
the variance, and the m.g.f. of X.
12
tb
b
Uniform Distribution
Uniform Distribution
m 
b
t  0.

12
 e 10 t  1

,
M ( t )   10 t

 1,
t  0,
2
25
,
3
t  0,
t  0.
Pseudo-Random Number Generator on most computers U(0, 1)
3
Gamma Distribution
4
Graphs of Gamma Distributions
Definition 6.2: The continuous random variable X
has a Gamma distribution, Gamma( a, b ). if its
p.d.f. is
1

a 1  x / b
x e
,
for

f ( x )    (a ) b a

0 ,
elsewhere.
b=4
a=4
a = 1/4
b = 5/6
0 x
b =1
a=1
a=2
a=3
Gamma Function: (n) = (n – 1)!
* X can be the waiting time until ath success in a Poisson process.
5
0
10
b=2
b =3
a=4
20
30
0
10
b=4
20
30
6
CD - 1
Continuous Distributions
Gamma Distribution
The mean, variance, and m.g.f. of a continuous
random variable X that has an Gamma
distribution are:
m  ab ,
M (t ) 
s
1
(1  b t )
a
t 
,
2
 ab
2
,
1
b
•Gamma(1, b )  Exponential Distribution
•Gamma(2, r/2 )  Ch-square with d.f. = r.
7
8
Special Notation c2a (r)
Let X be a random that has Chi-square distribution with
degrees of freedom r,
P [ X  c a ( r )]  a
2
Tail area
Probability of X
greater than or
equal to c2a(r) is a.
Try This!
• Find c2.01(7) = ?
• Find c2.025(4) = ?
• Find the 10th percentile from a c2
distribution with degrees of freedom 6.
c2a(r)
P[ X  c
2
1 a
( r )]  a
Example: Find c2.05(3) = ?7.815
9
Exponential Distribution
Exponential Distribution
Definition 6.3 The continuous random variable X has
an exponential distribution if its p.d.f. is
 1 x /

x
f ( x)   
 0 ,
,
10
The mean, variance, and m.g.f. of a continuous
random variable X that has an exponential
distribution are:
m ,
for 0  x
elsewhere
M (t ) 
where  is the mean of the distribution.
* X can be the waiting time until next success in a Poisson process.
11
c.d.f. 
s
1
1  t
,
t 
2
 ,
2
1

0,

F ( x)  
 x /
1

e
,

  x0
0 x
12
CD - 2
Continuous Distributions
Exponential Distribution
Exponential Distribution
Example: Let X have an exponential distribution
with a mean of 30, what is the first quartile of
this distribution?
0,

F ( x)  
 x / 30
1

e
,

F (p . 25 )  1  e
 p . 25 / 
  x0
F(w) = P(W  w) = 1 – P(W > w)
= 1 – P(W > w)
= 1 – P(no success in [0,w])
= 1 – e-lw  d.f. of exponential distribution.
F (w) = f(w) = le-lw  p.d.f. of exponential distribution.
0 x
 . 25  e
 p . 25 / 
Let W be the waiting time until next success in a
Poisson process in which the average number of
success in unit interval is l, then, for w  0
 . 75
 p.25 / = ln(.75)  p.25 =   ·ln(.75)
{ = 30}  p.25 = - 30·ln(.75)  p.25 = 8.63
1
pp =  · ln(1 – p)
13
Exponential Distribution

e
 x /
14
Exponential Distribution
Let W be the waiting time until next success in
a Poisson process in which the average number
of success in unit interval is l, then, l = 1/,
where  is the average waiting time until
next success, for w  0
Suppose that number of arrivals of customers
follows a Poisson process with a mean of 10 per
hour. What is the probability that the next customer
will arrive within 15 minutes? (15 min. = 0.25
hour)
P(X  x) = F (x) = 1 – e-x/
F(w) = 1 – e-lw =1 – e-w/
f(w) =
1

P(X  0.25) = F (0.25) = 1 – e-0.25/
 d.f. of exponential
distribution.
l = 10   = 0.1
F (0.25) = 1 – e-0.25/0.1 = .918
ew/  p.d.f. of exponential distribution.
15
16
Beta Distribution
Beta Distribution
The mean and variance of a continuous
random variable X that has an Beta
distribution are:
Definition 6.5: The continuous random variable X
has a Beta distribution, Beta( a, b ). if its p.d.f. is
  (a  b ) a  1
b 1
x (1  x ) ,

f ( x )    (a )  ( b )
 0 ,
for
0 x 1
m 
elsewhere.
a
a b
s
2

ab
(a  b ) (a  b  1 )
2
for a > 0 and b > 0.
𝐵𝑒𝑡𝑎 function :
Γ(𝛼)Γ(𝛽)
Γ(𝛼+𝛽)
=
1 𝛼−1
𝑥
(1 −
0
𝑥)𝛽−1 𝑑𝑥
17
18
CD - 3
Continuous Distributions
Normal Probability
Density Function
Definition 6.6: The continuous
random variable X has a normal
distribution if its p.d.f. is
f ( x) 
1
s
2p
e
1  xm 
 

2 s 
Normal Distribution
s
The mean, variance, and m.g.f. of a
continuous random variable X that has a
normal distribution are:
m
2
E[ X ]  m ,
 < x < 
Notation: N (m, s 2)  A normal distribution with
mean m and standard deviation s
M (t )  e
2
m t  s t / 2
2 2
19
20
m.g.f. of Normal Distribution
Normal Distribution
Example: If the m.g.f. of a random variable X is
M(t) = exp(2t + 16t 2), what is the distribution of
this random variable? What are the mean and
standard deviation of this distribution and what
is the p.d.f. of this random variable?
Distribution  ?
Mean  ?
Standard deviation  ?
p.d.f.  ?
Var [ X ]  s ,
1. “Bell-Shaped” &
Symmetrical
2. Mean, Median,
Mode Are Equal
3. Random Variable
Has Infinite Range
<x<
f(X)
X
Mean
Median
Mode
21
22
Effect of Varying Parameters (m & s)
Example
N (72, 5)  A normal distribution with mean 72
and variance 5.
Possible situations: Test scores, pulse
rates, …
f(x)
B
N (130, 24)  A normal distribution with mean
A
130 and variance 24.
Possible situations: Weight, Cholesterol levels, …
C
x
23
24
CD - 4
Continuous Distributions
Infinite Number
of Tables
Standard Normal Distribution
Standard Normal Distribution:
Normal distributions differ by mean &
standard deviation.
Each distribution would require
its own table.
f(X)
Definition 6.7: A normal distribution with mean = 0
and standard deviation = 1, is referred to as the
Standard Normal Distribution.
s=1
Notation:
Z ~ N (m = 0, s2 = 1)
X
That’s an infinite number!
25
Cap letter Z
Z
0
26
Area under Standard Normal
Curve
Z
0
z
How to find the proportion of the are
under the standard normal curve below
z or say P ( Z < z ) = ?
Use Standard Normal Table!!!
27
Standard Normal Distribution
28
Standard Normal Distribution
P(Z > 0.32) = Area above .32 = 1  .6255 = .3745
P(Z < 0.32) = F (0.32)  Area below .32 = 0.6255
?
Areas in the upper tail of the standard normal distribution
0 .32
0 .32
29
30
CD - 5
Continuous Distributions
Standard Normal Distribution
P(0 < Z < 0.32) = Area between 0 and .32 = ?
= 0.6255 – 0.5 = 0.1255
P ( -1.00 < Z < 1.00 ) = __?___
.6826
.3413
-1.00
0 .32
32
2.00
-3.00
…
.4987
.4987
.4772
0
1.00
.9974
P ( 3.00 < Z < 3.00 ) = _____
.9544
P ( -2.00 < Z < 2.00 ) = _____
-2.00
0
0.8413  0.5
31
.4772
.3413
0
3.00
.9544 = .4772 + .4772
33
34
Normal Distribution
s=1
P ( 1.40 < Z < 2.33 ) = .9093
____
Standard Normal
.4192
.4901
m=0
.0099
.0808
- 1.40
0
s=3
s=9
…
2.33
m = 50
35
m = 70
36
CD - 6
Continuous Distributions
Standardize the
Normal Distribution
Z
Theorem 6.7: If X has a normal distribution
with mean m and variance s2, then Z =
X m
has a standard normal distribution.
s
Normal
Distribution
Standardized Normal
Distribution
s
s =1
𝑃 𝑥1 < 𝑋 < 𝑥2 =
𝑥1 − 𝜇
𝜎
𝑥2 − 𝜇
𝑧2 =
𝜎
𝑧1 =
m
m= 0
X
Z
One table!
Standardize the
Normal Distribution
N ( m, s)
2𝜋
𝑧2
𝑧1
𝑥1
𝑧2
1
2
1
2
𝑑𝑥
𝑒 −2 𝑧 𝑑𝑥
𝑧1
1
2𝜋
𝑒 −2 𝑧 𝑑𝑥
38
Standard Normal
Distribution
Z
m
s = 10
b
s
bm 
am
P (a  X  b)  P 
Z 

s 
 s
m= 5 6.2 X
39
Example
X m

40
Example
55
0
10
X m 6.25
Z

.12
s
10
Standardized Normal
s
For a normal distribution that has
a mean = 5 and s.d. = 10, what
percentage of the distribution is
between 5 and 6.2?
am 0 bm
s
s = 10
2𝜋𝜎
1
1 𝑥−𝜇 2
𝜎
𝑒 −2
Example
Normal
Distribution
Normal
Distribution
=
𝑥2
1
= 𝑃 𝑧1 < 𝑍 < 𝑧2
X
Z
=
37
N ( 0 , 1)
Normal
Distribution
a
𝑋−𝜇
𝜎
P(5  X  6.2)
 P(0  Z  .12)
Distribution
Standardized Normal
Distribution Table
s =1
s =1
m= 0 .12
m= 5 6.2 X
m= 0 .12
Z
41
Z
Area = .5478 - .5 = .047842
CD - 7
Continuous Distributions
Example
P(3.8  X  5)
Example
Z
Z
Normal
Distribution
s = 10
X m
s
X m
s

55
0
10
6.25

.12
10
Standardized Normal
P(5  X  6.2)
 P(0  Z  .12)
 0.0478  4.78%
Distribution
s =1
Z
X m
s

3.85
.12
10
P(3.8  X  5)
=P(.12  Z  0)
.0478
Normal
Distribution
s = 10
Standardized Normal
Distribution
s =1
.0478
m= 0 .12
m= 5 6.2 X
Z
3.8 m = 5
-.12 m = 0
X
43
Example
P(X > 8)
Z
Normal
Distribution
s = 10
X m
s
Z
44
Area = .0478
Example
P(X > 8)
85

.30
10
P(X > 8)
=P(Z > .30)
=.3821
Standardized Normal
Distribution
s =1
Normal
Distribution
s = 10
62%
38%
.3821
Value 8 is the 62nd percentile
m =5
8
X
m =0
.30 Z
Area = .3821
m =5
8
X
45
46
More on Normal Distribution
P(42  X  60) = ?
Z
The work hours per week for residents in Ohio has a
normal distribution with m = 42 hours & s = 9 hours.
Find the percentage of Ohio residents whose work
hours are
A. between 42 & 60 hours.
P(42  X  60) =?
B. less than 20 hours.
P(X  20) = ?
Normal
Distribution
s = 200
9
Z
X m
s
X m
s
4242
0
9
6042

2
9
P(42  Z  60)
 P(0  Z  2)
 .4772(47.72%)
2400 X
42 60
47

Standardized Normal
Distribution
s=1
.4772
0 2 2.0
Z
48
CD - 8
Continuous Distributions
Finding Z Values
for Known Probabilities
P(X  20) = ?
Z
Normal
Distribution
s = 200
9
X m
s

2042
What is z given
P(Z < z) = .80 ?
 2.44
9
P(X  20)
= P(Z  2.44)
= 0.0073 = 0.73%
Standardized Normal
Distribution
.80
.20
s=1
0
0.0073
20
42 2400 X
Standardized Normal Distribution
Table
z = .84
z.20 = .84
-2.44 0 2.0
Z
49
Finding X Values
for Known Probabilities
P(Z < za) = 1 – a
50
Finding X Values
for Known Probabilities
Example: The weight of new born
infants is normally distributed with a
mean 7 lb and standard deviation of
1.2 lb. Find the 80th percentile.
Normal Distribution
Standardized Normal Distribution
.80
.80
Area to the left of 80th percentile is 0.800.
In the table, there is a area value 0.7995
(close to 0.800) corresponding to a z-score
of .84.
80th percentile = 7 + .84 x 1.2 = 8.008 lb
P(Z ≥ za) = a ;
Def. za :
7
X = 8.008
0
80th
z = .84
percentile
X  m Z s 7(.84)(1.2) 8.008
51
52
6.6 Normal Approximation for
Binomial Probability
More Examples
• The pulse rates for a certain population follow a
normal distribution with a mean of 70 per minute
and s.d. 5. What percent of this distribution that
is in between 60 to 80 per minute?
m = np = 3.5
s2= np(1p) = 1.75
P ( X b  3)  ?
• The weights of a population follow a normal
distribution with a mean 130 and s.d. 10. What
percent of this population that is in between 110
and 150 lbs?
0
53
1
2
3
4
5
6
7
54
CD - 9
Continuous Distributions
Normal Approximation for
Binomial Probability
6.7 The Bivariate Normal Distribution
m = np = 3.5
s2 = np(1p) = 1.75
P ( X b  3)  P ( X N > 2 .5 )
2 .5  3 .5
 P (Z >
)
1 . 75
Definition 6.8: The pair of random variables X and Y
have a bivariate normal distribution if and only if
their joint probability density function is given by
𝒇 𝒙, 𝒚
 P ( Z >  . 76 )
−
 . 7764
Binomial Table:
P(Xb ≥ 3) = .7734
0
1
2
3
4
5
6
7
55
𝟏
𝒙−𝝁𝟏 𝟐
−𝟐𝝆
𝒙−𝝁𝟏
𝒚−𝝁𝟐
+
𝒚−𝝁𝟐 𝟐
𝝈𝟏
𝝈𝟐
𝝈𝟐
𝒆 𝟐 𝟏−𝝆 𝟐 𝝈𝟏
=
for  < x <  and 𝟐𝝅𝝈
 <𝟏 y𝝈𝟐< ,
where
𝟏−
𝝆𝟐 s1 > 0, s2 > 0,
and 1  r < 1.
for  < x <  and  < y < , where s1 > 0, s2 > 0,
and 1  r < 1.
56
cov(X, Y)  r s1s2
r  cov(X, Y) / (s1s2) is the correlation coefficient.
Theorem 6.9: If X and Y have a bivariate normal
distribution, the conditional density of Y given X = x is
a normal distribution with
Theorem 6.10: If X and Y have a bivariate normal
distribution, they are independent if and only if r = 0.
𝜎
𝜇𝑌|𝑥 =𝜇2 + 𝜌 𝜎2 (𝑥 − 𝜇1 )
1
𝜎𝑌|𝑥 2 = 𝜎2 2 (1 − 𝜌 2 )
𝜎
𝜇𝑋|𝑦 =𝜇1 + 𝜌 𝜎1 (𝑦 − 𝜇2 )
2
𝜎𝑋|𝑦 2 = 𝜎1 2 (1 − 𝜌 2 )
57
58
CD - 10