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 Special Probability Densities [a, b], then its p.d.f. is

1 f ( x)  , a  x  b . The most common parameters in most of b  a the distributions are the lower moments: It is usually denoted as U(a, b). mean m and variance s 2. f(x) 1 F(x) 1 b  a 1 0 a b 0 a b 2

Uniform Distribution Uniform Distribution

The mean, variance, and m.g.f. of a continuous Example: Let X be U(0,10), find the mean and random variable X that has a uniform distribution are: the variance, and the m.g.f. of X.

2 2 a  b 2 (b  a ) 0  10 (10  0) 25 m  , s  , m   5, s 2   , 2 12 2 12 3 tb ta  e  e 10 t  , t  0,  e  1 M (t )   t (b  a )  , t  0,  M (t )   10 t  1, t  0.  1, t  0. Pseudo-Random Number Generator on most computers U(0, 1) 3 4

Gamma Distribution Graphs of Gamma Distributions

Definition 6.2: The continuous random variable X b = 4 a = 4 has a Gamma distribution, Gamma( a, b ). if its p.d.f. is a = 1/4  1 b = 5/6  x a 1e  x / b , for 0  x a b = 1 f ( x)    (a )b a = 1  b = 2  0 , elsewhere. a = 2 a = 3 b = 3 a = 4 b = 4 Gamma Function: (n) = (n – 1)! 0 10 20 30 0 10 20 30 * X can be the waiting time until ath success in a Poisson process. 5 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 , s 2  ab 2 , 1 1 M (t )  , t  (1  b t ) a b

•Gamma(1, b )  Exponential Distribution •Gamma(2, r/2 )  Ch-square with d.f. = r. 7 8

2 r Special Notation c a ( ) Try This! Let X be a random that has Chi-square distribution with degrees of freedom r, 2 • Find c .01(7) = ? P[ X  c 2 (r )]  a a 2 • Find c .025(4) = ? Probability of X Tail area th 2 greater than or • Find the 10 percentile from a c 2 equal to c a(r) is a. distribution with degrees of freedom 6.

2 c a(r) 2 P[ X  c 1a (r )]  a

2 Example: Find c .05(3) = ?7.815 9 10

Exponential Distribution Exponential Distribution The mean, variance, and m.g.f. of a continuous Definition 6.3 The continuous random variable X has random variable X that has an exponential an exponential distribution if its p.d.f. is distribution are:

 1 2 2  x  x /  , for 0  x m   , s   , f ( x)    1 1  0 , elsewhere M (t )  , t  1  t  where  is the mean of the distribution.

 0,    x  0 * X can be the waiting time until next success in a Poisson process. c.d.f.  F ( x)   1  e  x / , 0  x   11  12

CD - 2 Continuous Distributions

Exponential Distribution Exponential Distribution

Example: Let X have an exponential distribution Let W be the waiting time until next success in a with a mean of 30, what is the first quartile of Poisson process in which the average number of this distribution? success in unit interval is l, then, for w  0  0,    x  0 F ( x)    x / 30 F(w) = P(W  w) = 1 – P(W > w) 1  e , 0  x   = 1 – P(W > w)  p /  p / F (p )  1  e .25  .25  e .25  .75 .25 = 1 – P(no success in [0,w]) -lw  p.25 / = ln(.75)  p.25 =   ·ln(.75) = 1 – e  d.f. of exponential distribution. -lw { = 30}  p.25 = - 30·ln(.75)  p.25 = 8.63 F (w) = f(w) = le  p.d.f. of exponential distribution. 1  x / p =  · ln(1 – p) 13 e 14 p 

Exponential Distribution Exponential Distribution Let W be the waiting time until next success in Suppose that number of arrivals of customers a Poisson process in which the average number follows a Poisson process with a mean of 10 per of success in unit interval is l, then, l = 1/, hour. What is the probability that the next customer where  is the average waiting time until will arrive within 15 minutes? (15 min. = 0.25 next success, for w  0 hour)

P(X  x) = F (x) = 1 – e-x/ F(w) = 1 – e-lw =1 – e-w/  d.f. of exponential P(X  0.25) = F (0.25) = 1 – e-0.25/ distribution. 1 w/ l = 10   = 0.1 f(w) = e  p.d.f. of exponential distribution.  F (0.25) = 1 – e-0.25/0.1 = .918 15 16

Beta Distribution Beta Distribution The mean and variance of a continuous Definition 6.5: The continuous random variable X random variable X that has an Beta has a Beta distribution, Beta( a, b ). if its p.d.f. is distribution are:

  (a  b ) a 1 b 1  x (1  x) , for 0  x  1 a ab f ( x)    (a ) (b ) m  s 2  2  0 , elsewhere. a  b (a  b ) (a  b  1) for a > 0 and b > 0.

Γ(훼)Γ(훽) 1 퐵푒푡푎 function : = 푥훼−1(1 − 푥)훽−1푑푥 Γ(훼+훽) 0

17 18

CD - 3 Continuous Distributions

Normal Probability Normal Distribution Density Function s The mean, variance, and m.g.f. of a Definition 6.6: The continuous continuous random variable X that has a random variable X has a normal normal distribution are: distribution if its p.d.f. is m

2 1  x  m     2 1 2  s  E[ X ]  m , Var [ X ]  s , f ( x)  e  < x <  s 2p m t s 2 t 2 / 2 M (t )  e

Notation: N (m, s 2)  A normal distribution with mean m and standard deviation s 19 20

m.g.f. of Normal Distribution Normal Distribution

Example: If the m.g.f. of a random variable X is 1. “Bell-Shaped” & f(X) M(t) = exp(2t + 16t 2), what is the distribution of Symmetrical this random variable? What are the mean and standard deviation of this distribution and what 2. Mean, Median, is the p.d.f. of this random variable? Mode Are Equal X 3. Random Variable Distribution  ? Has Infinite Range Mean Median Mean  ?   < x <  Mode Standard deviation  ? p.d.f.  ? 21 22

Example Effect of Varying Parameters (m & s)

N (72, 5)  A normal distribution with mean 72 and variance 5. f(x) Possible situations: Test scores, pulse B rates, …

N (130, 24)  A normal distribution with mean A C 130 and variance 24. Possible situations: Weight, Cholesterol levels, … x

23 24

CD - 4 Continuous Distributions

Infinite Number Standard Normal Distribution of Tables Standard Normal Distribution: Normal distributions differ by mean & Each distribution would require standard deviation. its own table. Definition 6.7: A normal distribution with mean = 0 and standard deviation = 1, is referred to as the f(X) Standard Normal Distribution.

s = 1 Notation: Z ~ N (m = 0, s2 = 1) X Z That’s an infinite number! 0 Cap letter Z 25 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 28

Standard Normal Distribution 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 ( -1.00 < Z < 1.00 ) = __.6826?___ P(0 < Z < 0.32) = Area between 0 and .32 = ? = 0.6255 – 0.5 = 0.1255 .3413 .3413

0 .32 -1.00 0 1.00

0.8413  0.5 31 32

P ( -2.00 < Z < 2.00 ) = _____.9544 P ( 3.00 < Z < 3.00 ) = _____.9974

… .4772 .4772 .4987 .4987

-2.00 0 2.00 -3.00 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

.0808 .0099 s = 3 s = 9

- 1.40 0 2.33 … m = 50 m = 70

35 36

CD - 6 Continuous Distributions

Standardize the Theorem 6.7: If X has a normal distribution Normal Distribution 푋 − 휇 with mean m and variance s2, then Z = 휎 Xm Z has a standard normal distribution. Normal s Standardized Normal 푥2 1 푥−휇 2 Distribution Distribution 1 − 푃 푥1 < 푋 < 푥2 = 푒 2 휎 푑푥 2휋휎 푥1 s s = 1 푧 푥 − 휇 1 2 1 2 1 −2 푧 푧1 = = 푒 푑푥 휎 2휋 푧1 푥 − 휇 푧2 1 푧 = 2 1 − 푧 2 2 휎 = 푒 2 푑푥 m m= 0 푧 2휋 X Z 1

One table! 37 = 푃 푧1 < 푍 < 푧2 38

Standardize the Normal Distribution N ( 0 , 1) Example

N ( m, s) Standard Normal Normal Distribution Distribution Z X For a normal distribution that has Normal a mean = 5 and s.d. = 10, what Distribution percentage of the distribution is s = 10 between 5 and 6.2? a m b a  m 0 b  m

s s

 a  m b  m  P (a  X  b)  P  Z   m = 5 6.2 X  s s  39 40

Example Example Xm 55 Z  0 s 10 Xm 6.25 Standardized Normal Z  .12 Distribution Table Normal s 10 Standardized Normal Distribution P(5  X  6.2) Distribution s = 1  P(0  Z  .12) s = 10 s = 1

m= 0 .12 Z

m= 5 6.2 X m= 0 .12 Z Area = .5478 - .5 = .0478 41 42

CD - 7 Continuous Distributions

Example Example P(3.8  X  5) Xm 55 Z  0 s 10 Xm 3.85 Xm 6.25 Z  .12 Z  .12 s 10 Normal s 10 Standardized Normal Normal Standardized Normal Distribution P(5  X  6.2) Distribution Distribution P(3.8  X  5) Distribution  P(0  Z  .12) =P(.12  Z  0) s = 10 s = 1 s = 10 s = 1  0.0478  4.78% .0478 .0478

m= 5 6.2 X m= 0 .12 Z 3.8 m = 5 X -.12 m = 0 Z 43 44 Area = .0478

Example Example P(X > 8) P(X > 8)

Xm 85 Z  .30 s 10 Normal Standardized Normal Normal Distribution P(X > 8) Distribution Distribution =P(Z > .30) s = 10 s = 1 s = 10 =.3821 62% 38% .3821 Value 8 is the 62nd percentile m = 5 8 X m = 0 .30 Z m = 5 8 X 45 46 Area = .3821

More on Normal Distribution P(42  X  60) = ? Xm 4242 The work hours per week for residents in Ohio has a Z  0 s 9 normal distribution with m = 42 hours & s = 9 hours. Xm 6042 Find the percentage of Ohio residents whose work Z  2 Normal s 9 Standardized Normal hours are Distribution P(42  Z  60) Distribution A. between 42 & 60 hours.  P(0  Z  2) P(42  X  60) =? s = 2009 s = 1  .4772(47.72%) B. less than 20 hours. .4772 P(X  20) = ?

42 602400 X 0 2 2.0 Z 47 48

CD - 8 Continuous Distributions

Finding Z Values P(X  20) = ? for Known Probabilities

Xm 2042 What is z given Z  2.44 Standardized Normal Distribution s 9 P(Z < z) = .80 ? Table Normal Standardized Normal Distribution P(X  20) Distribution = P(Z  2.44) .80 .20 s = 2009 = 0.0073 = 0.73% s = 1

0 z = .84 0.0073 z.20 = .84 20 42 2400 X -2.44 0 2.0 Z 49 Def. za : P(Z ≥ za) = a ; P(Z < za) = 1 – a 50

Finding X Values Finding X Values for Known Probabilities for Known Probabilities

Example: The weight of new born Normal Distribution Standardized Normal Distribution infants is normally distributed with a mean 7 lb and standard deviation of 1.2 lb. Find the 80th percentile. .80 .80 Area to the left of 80th percentile is 0.800. In the table, there is a area value 0.7995 7 X = 8.008 0 z = .84 (close to 0.800) corresponding to a z-score 80th percentile of .84. 80th percentile = 7 + .84 x 1.2 = 8.008 lb XmZs 7(.84)(1.2)8.008 51 52

6.6 Normal Approximation for More Examples Binomial Probability • The pulse rates for a certain population follow a m = np = 3.5 normal distribution with a mean of 70 per minute P ( X b  3)  ? s2= np(1p) = 1.75 and s.d. 5. What percent of this distribution that is in between 60 to 80 per minute?

• 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 1 2 3 4 5 6 7 53 54

CD - 9 Continuous Distributions

Normal Approximation for Binomial Probability 6.7 The Bivariate Normal Distribution Definition 6.8: The pair of random variables X and Y P ( X b  3)  P ( X N > 2.5) m = np = 3.5 have a bivariate normal distribution if and only if 2.5  3.5 s2 = np(1p) = 1.75  P (Z > ) their joint probability density function is given by 1.75  P (Z > .76 ) 풇 풙, 풚 ퟐ ퟐ ퟏ 풙−흁ퟏ 풙−흁ퟏ 풚−흁ퟐ 풚−흁ퟐ − ퟐ 흈 −ퟐ흆 흈 흈 + 흈  .7764 풆 ퟐ ퟏ−흆 ퟏ ퟏ ퟐ ퟐ = for  < x <  and ퟐ흅흈 < 흈y < ퟏ, −where흆ퟐ s1 > 0, s2 > 0, Binomial Table: ퟏ ퟐ and 1  r < 1. P(Xb ≥ 3) = .7734 for  < x <  and  < y < , where s1 > 0, s2 > 0, 0 1 2 3 4 5 6 7 and 1  r < 1. 55 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 Theorem 6.10: If X and Y have a bivariate normal distribution, the conditional density of Y given X = x is distribution, they are independent if and only if r = 0. a normal distribution with

휎2 휇푌|푥=휇2 + 휌 (푥 − 휇1) 휎1 2 2 2 휎푌|푥 = 휎2 (1 − 휌 ) 휎1 휇푋|푦=휇1 + 휌 (푦 − 휇2) 휎2 2 2 2 휎푋|푦 = 휎1 (1 − 휌 ) 57 58

CD - 10