Uniform Distribution Uniform Distribution Uniform Distribution ...

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

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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


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

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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




e

 x /

14


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

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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  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




s=1 .4772

0 2 2.0

Z 48

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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%


.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


.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

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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  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

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