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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 ath 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 , x0 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 / x0 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 ew/ 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 xm 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 bm am P (a X b) P Z s s m= 5 6.2 X 39 Example X m 40 Example 55 0 10 X m 6.25 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? am 0 bm 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 55 0 10 6.25 .12 10 Standardized Normal P(5 X 6.2) P(0 Z .12) 0.0478 4.78% Distribution s =1 Z X m s 3.85 .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) 85 .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 4242 0 9 6042 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 2042 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(1p) = 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(1p) = 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