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Examples: Continuous Probability Distributions Examples: Normal Distribution 1/156) Given a standard normal distribution find the area under the curve which lies a) to the left of z = 1.43; b) to the right of z = -0.89; c) between z = -2.16 and z = -0.65; d) to the left of z = -1.39; e) to the right of z = 1.96; f) between z = -0.48 and z = 1.74. Solution: a. P(Z < 1.43), from Table A.3, is equal to 0.9236. b. P(Z > -0.89) = 1 - P(Z < -0.89) = 1 - 0.1867 (from Table A.3) = 0.8133 c. P(-2.16 < Z < -0.65) = P(Z < -0.65) - P(Z < -2.16) From Table A.3 = 0.2578 - 0.0154 = 0.2424 d. P(Z < -1.39), From Table A.3, is equal to 0.0823. e. P(Z > 1.96) = 1 - P(Z< 1.96) = 1 - 0.975 = 0.0250 f. P(-0.48 < Z < 1.74) = P(Z < 1.74) - P(Z < -0.48) From Table A.3 = 0.9591 - 0.3156 = 0.6435 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 2/156) Find the value of z if the area under a standard normal curve a) to the right of z is 0.3622; b) to the left of z is 0.1131; c) between 0 and z, with z > 0, is 0.4838; d) between -z and z, with z > 0, is 0.9500. Solution: a. P(Z > z) = 0.3622 P(Z < z) = 1 - P(Z > z) = 1 - 0.3622 = 0.6378 P(Z < 0.35) = 0.6378, therefore z = 0.35. b. P(Z < z) = 0.1131, From Table A.3, z = -1.21 c. P(0 < Z < z) = 0.4838 = P(Z < z) - P(Z < 0) 0.4838 = P(Z < z) - 0.5000 P(Z < z) = 0.9838 Therefore, From Table A.3, z = 2.14 d. P(-z < Z < z) = 0.9500 = P(Z < z) - P(Z < -z) = 0.9750 - 0.025 Therefore, From Table A.3, z = 1.96. ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 4/156) Given the normal distribution with µ = 30 and σ = 6, find a) the normal-curve area to the right of x = 17; b) the normal-curve area to the left of x = 22; c) the normal-curve area between x = 32 and x = 41; d) the value of x that has 80% of the normal curve area to the left; e) the two values of x that contain the middle 75% of the normalcurve area. Solution: a. z = ( x - µ)/σ = (17 - 30) / 6 = -2.17 P(Z > z ) = P(Z > 2.17) = 1 - P(Z - 2.17) = 1 - 0.015 (From Table A.3) = 0.9850 b. z = (22 - 30)/6 = -1.33 P(Z < - 1.33) = 0.0918 c. x1 = (32 - 30)/6 = 0.33 x2 = (41 - 30)/6 = 1.83 P(0.33 < Z < 1.83) = 0.9664 - 0.6293 (From Table A.3) = 0.3371 d. From Table A.3, P(Z < z ) = 0.80, z = 0.84 x = zσ + µ = 0.84(6) + 30 x = 35.04 e. P(z < Z < z) = 0.75 = P(Z < z) - P( Z < -z) From Table A.3, z = 1.15 Therefore, x1 = -1.15(6) + 30 = 23.1 x2 = 1.15(6) + 30 = 36.9 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 9/157) A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, a) what fraction of the cups will contain more than 224 milliliters? b) what is the probability that a cup contains between 191 and 209 milliliters? c) how many cups will probably overflow if 230 milliliter cups are used for the next 1000 drinks? d) below what value do we get the smallest 25% of the drinks? Solution: Given: µ = 200 ml/cup, σ = 15 ml a. P(X > 224) = P(Z > 1.6) z = (224 - 200)/15 = 1.6 = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0548 b. (From Table A.3) P(191 < X < 209) = P( -0.6 < Z < 0.6) z1 = (191 - 200)/15 = -0.6 z2 = (209 - 200)/15 = 0.6 = P(Z < 0.6) - P(Z < -0.6) = 0.7257 - 0.2743 (From Table A.3) = 0.4514 c. P(X > 230) = P(Z > 2) z = (230 - 200)/15 = 2 = 1 - P(Z < 2) = 1 - 0.9772 (From Table A.3) = 0.0228 Hence, no. of cups that will overflow = 1000(0.0228) = 22.8 ≈ 23 cups d) P(Z < z) = 0.25 From Table A.3, z = -0.675 x = zσ + µ = -0.670(15) + 200 = 189.95 ml ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 21/158) In a mathematics examination the average grade was 82 and the standard deviation was 5. All students with grades from 88 to 94 received a grade of B. If the grades are approximately normally distributed and 8 students received a B grade, how many students took the examination? Solution: µ = 82 Given: σ=5 Let S = the no. of students z1 = (87.5 - 82)/5 = 1.1 z2 = (94.5 - 82)/5 = 2.5 P(87.5 < X < 94.5) = P(1.1 < Z < 2.5) = 0.9938 - 0.8643 = 0.1295 Therefore, (0.1295)S = 8 S = 62 ENGSTAT Notes of AM Fillone (From Table A.3) Examples: Continuous Probability Distributions 20/158) (Similar) The average rainfall, recorded to the nearest hundredth of a centimeter, in Tagaytay, for the month of March is 9.22 centimeters. Assuming a normal distribution with a standard deviation of 2.83 centimeters, find the probability that next March Tagaytay receives a) less than 1.84 centimeters of rain; b) more than 5 centimeters but not over 7 centimeters of rain; c) more than 13.8 centimeters of rain. Solution: µ = 9.22 centimeters Given: σ = 2.83 centimeters a. P(X < 1.84) = P(Z < -2.61) = 0.0045 (From Table A.3) z = (1.84 - 9.22)/2.83 = -2.61 b. P(5.01 < X < 7.00) = P(-1.487 < Z < -0.784) = 0.2177 - 0.0681 (From Table A.3) = 0.1496 z1 = (5.01 - 9.22)/2.83 = -1.488 z2 = (7.00 - 9.22)/2.83 = -0.784 c. P(X > 13.8) = P(Z > 1.62) = 1 - P(Z < 1.62) = 1 - 0.9474 (From Table A.3) = 0.0526 z = (13.8 - 9.22)/2.83 = 1.62 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions Example: Gamma and Exponential Distribution 1/175) If a random variable X has a gamma distribution with α = 2 and β = 1, find P(1.8 < X < 2.4). 2.4 P(1.8 < X < 2.4) = ∫ (1/β)e -x/β 1.8 =∫ 2.4 dx = ∫ (1/1) e-x/1 dx 1.8 2.4 1.8 x e-x dx Using integration by parts: Let u = x, hence du = dx dv = e-xdx, hence v = -e-x Since, ∫ u dv = uv + ∫ v du 2.4 -x -x = ∫ x e dx = x (-e ) therefore, 1.8 -x = -xe + e 2.4 -x 1.8 2.4 1.8 2.4 + ∫ -e-x dx 1.8 = 0.00523 7/175) The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. What is the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days? Sol’n: β = 4 minutes 3 P(T < 3) = ∫ (1/4) e -t/4 0 = -[e-t/4] 3 dt = - ∫ e-t/4 (-1/4 dt) 0 3 0 = 0.52763 6 P(X ≥ 4) = Σ b(x; 6, 0.52763) = 0.39688 x=4 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 5.31/183) (M/S) The lifetime y (in hours) of the central processing unit of a certain type of microcomputer is an exponential random variable with parameter β = 1,000. a. Find the mean and variance of the lifetime of the central processing unit. b. What is the probability that a central processing unit will have a lifetime of at least 2,000 hours? c. What is the probability that a central processing unit will have a lifetime of at most 1,500 hours? Solution: a. µ = β = 1,000 σ2 = β2 = 1,000,000 b. f(y ≥ 2,000) = 1 – f(y < 2,000) 2,000 2,000 -y/1,000 = 1 - ∫ (1/1,000) e dy = 1 + ∫ e-y/1,000 (–dy/1000) 0 0 2,000 -y/1,000 = 1 + [e-2 – e0] =1+e 0 -2 = e = 0.1353 1,500 1,500 -y/1,000 c. f(y ≤ 1,500) = ∫ (1/1,000) e dy = - ∫ e-y/1,000 (-dy/1,000) 0 0 1,500 -y/1,000 = -[e-1.5 – e0] = 0.777 = -e 0 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions Weibull Distribution 12/175) Suppose that the service life, in years, of a hearing aid battery is a random variable having a Weibull distribution with α = 1/2 and β = 2. a. How long can such a battery be expected to last? b. What is the probability that such a battery will still be operating after 2 years? Soln: a. µ = α-1/β Γ[(β+1)/β] = 0.5-1/2 Γ(1.5) but Γ(1.5) = 0.88623 from Table = 1.4142*0.88623 = 1.2533 b. F(x > 2) = 1 – F(x ≤ 2) 2 =1- ∫ 0 αβxβ-1 e-αxβ dx 2 = 1 - ∫ αβxβ-1 e-αxβ dx 0 β let z = x , then dz = βxβ-1 dx 2 = 1 + ∫ e-αz (-αdz) 0 2 -αz =1+ e 0 2 = 1 + e-αxβ = 1 + [e-0.5(2)2 – e0] = 0.135335 0 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions Examples: Normal Distribution 1/156) Given a standard normal distribution find the area under the curve which lies g) to the left of z = 1.43; h) to the right of z = -0.89; i) between z = -2.16 and z = -0.65; j) to the left of z = -1.39; k) to the right of z = 1.96; l) between z = -0.48 and z = 1.74. Solution: a) P(Z < 1.43), from Table A.3, is equal to 0.9236. b) P(Z > -0.89) = 1 - P(Z < -0.89) = 1 - 0.1867 (from Table A.3) = 0.8133 c) P(-2.16 < Z < -0.65) = P(Z < -0.65) - P(Z < -2.16) From Table A.3 = 0.2578 - 0.0154 = 0.2424 d) P(Z < -1.39), From Table A.3, is equal to 0.0823. e) P(Z > 1.96) = 1 - P(Z< 1.96) = 1 - 0.975 = 0.0250 f) P(-0.48 < Z < 1.74) = P(Z < 1.74) - P(Z < -0.48) From Table A.3 = 0.9591 - 0.3156 = 0.6435 2/156) Find the value of z if the area under a standard normal curve e) to the right of z is 0.3622; f) to the left of z is 0.1131; g) between 0 and z, with z > 0, is 0.4838; h) between -z and z, with z > 0, is 0.9500. Solution: a) P(Z > z) = 0.3622 P(Z < z) = 1 - P(Z > z) = 1 - 0.3622 = 0.6378 P(Z < 0.35) = 0.6378, therefore z = 0.35. d. P(Z < z) = 0.1131, From Table A.3, z = -1.21 e. P(0 < Z < z) = 0.4838 = P(Z < z) - P(Z < 0) 0.4838 = P(Z < z) - 0.5000 P(Z < z) = 0.9838 Therefore, From Table A.3, z = 2.14 f. P(-z < Z < z) = 0.9500 = P(Z < z) - P(Z < -z) = 0.9750 - 0.025 Therefore, From Table A.3, z = 1.96. ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 3/156) Given the standard normal distribution, find the value of k such that a) P(Z < k) = 0.0427; b) P(Z > k) = 0.2946; c) P(-0.93 < Z < k) = 0.7235. Solution: a) P(Z < k) = 0.0427, From Table A.3, k = -1.72. b) P(Z > k) = 0.2946 P(Z > k) = 1 - P(Z < k) Therefore, P(Z < k) = 1 - 0.2946 = 0.7054 From Table A.3, k = 0.540 c) P(-0.93 < Z < k) = 0.7235 = P(Z < k) - P(Z < -0.93) From Table A.3, P(Z < -0.93) = 0.1762 P(Z < k) = 0.7235 + 0.1762 = 0.8997 Therefore, from Table A.3, k = 1.28. 4/156) Given the normal distribution with µ = 30 and σ = 6, find a) the normal-curve area to the right of x = 17; b) the normal-curve area to the left of x = 22; c) the normal-curve area between x = 32 and x = 41; d) the value of x that has 80% of the normal curve area to the left; e) the two values of x that contain the middle 75% of the normal-curve area. Solution: a) z = ( x - µ)/σ = (17 - 30) / 6 = -2.17 P(Z > z ) = P(Z > 2.17) = 1 - P(Z - 2.17) = 1 - 0.015 (From Table A.3) = 0.9850 g. z = (22 - 30)/6 = -1.33 P(Z < - 1.33) = 0.0918 h. x1 = (32 - 30)/6 = 0.33 x2 = (41 - 30)/6 = 1.83 P(0.33 < Z < 1.83) = 0.9664 - 0.6293 (From Table A.3) = 0.3371 d) From Table A.3, P(Z < z ) = 0.80, z = 0.84 x = zσ + µ = 0.84(6) + 30 x = 35.04 i. P(z < Z < z) = 0.75 = P(Z < z) - P( Z < -z) From Table A.3, z = 1.15 Therefore, x1 = -1.15(6) + 30 = 23.1 x2 = 1.15(6) + 30 = 36.9 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 9/157) A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, e) what fraction of the cups will contain more than 224 milliliters? f) what is the probability that a cup contains between 191 and 209 milliliters? g) how many cups will probably overflow if 230 milliliter cups are used for the next 1000 drinks? h) below what value do we get the smallest 25% of the drinks? Solution: Given: µ = 200 ml/cup σ = 15 ml a) P(X > 224) = P(Z > 1.6) z = (224 - 200)/15 = 1.6 = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0548 (From Table A.3) j. P(191 < X < 209) = P( -0.6 < Z < 0.6) z1 = (191 - 200)/15 = -0.6 z2 = (209 - 200)/15 = 0.6 = P(Z < 0.6) - P(Z < -0.6) = 0.7257 - 0.2743 (From Table A.3) = 0.4514 k. P(X > 230) = P(Z > 2) z = (230 - 200)/15 = 2 = 1 - P(Z < 2) = 1 - 0.9772 (From Table A.3) = 0.0228 Hence, no. of cups that will overflow = 1000(0.0228) = 22.8 ≈ 23 cups e) P(Z < z) = 0.25 From Table A.3, z = -0.675 x = zσ + µ = -0.670(15) + 200 = 189.95 ml ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 11/157) (Similar) A engineer commutes daily from his suburban home to his midtown office. On the average the trip one way takes 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times to be normally distributed. a) What is the probability that a trip will take at least ½ hour? b) If the office opens at 9:00 AM and he leaves his house at 8:45 AM daily, what percentage if the time is he late for work? c) If he leaves the house at 8:35 AM and coffee is sered at the office from 8:50 AM until 9:00 AM, what is the probability that he misses coffee? d) Find the length of time above which we find the slowest 15% of the trips. e) Find the probability that 2 of the next 3 trips will take at least ½ hour. Solution: Given: µ = 24 min. σ = 3.8 min. a) P(X > 30 min) = P(Z > 1.58) z = (30 - 24)/3.8 = 1.58 = 1 - P(Z < 1.58) = 1 - 0.9429 (From Table A.3) = 0.0571 b) P(X < 15 min) = P(Z < -2.37) = 0.0089 z = (15 - 34)/3.8) = -2.368 ≈ -2.37 Therefore, percent late = 0.0089(100) = 0.89% c) P(X > 25 min) = P(Z > 0.26) = 1 - P(Z < 0.26) z = (25 - 34)/3.8 = 0.263 ≈ 0.26 = 1 - 0.6026 (From Table A.3) = 0.3974 d) P(Z > z) = 0.15 = 1 - P(Z < z) P(Z < z) = 0.85 From Table A.3, z = 1.04 x = zσ + µ = 1.04(3.8) + 24 = 27.952 min. e) P(X > 30 min) = 0.0571 3 2 (0.0571)2(0.9429)1 = 0.0092 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 21/158) In a mathematics examination the average grade was 82 and the standard deviation was 5. All students with grades from 88 to 94 received a grade of B. If the grades are approximately normally distributed and 8 students received a B grade, how many students took the examination? Solution: Given: µ = 82 σ=5 Let S = the no. of students z1 = (87.5 - 82)/5 = 1.1 z2 = (94.5 - 82)/5 = 2.5 P(87.5 < X < 94.5) = P(1.1 < Z < 2.5) = 0.9938 - 0.8643 (From Table A.3) = 0.1295 Therefore, (0.1295)S = 8 S = 62 17/158) The tensile strength of a certain metal component is normally distributed with a mean 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter. a) What proportion of these components exceed 10,150 kilogram per square centimeter in tensile strength? b) If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap? Solution: Given: µ = 10,000 kg/sq.cm. σ = 100 kg/sq.cm. Measurement recorded to the nearest 50 kg/sq.cm. a) P(X > 10,175) = P(Z > 1.75) = 1 - P(Z < 1.75) = 1 - 0.9599 (From Table A.3) = 0.0401 Z = (10,175 - 10,000)/100 = 1.75 b) P(9775 < X < 10225) = P(-2.25 < Z < 2.25) = P(Z < 2.25) - P(Z < -2.25) = 0.9878 - 0.0122 (From Table A.3) = 0.9756 z1 = (9775 - 10000)/100 = - 2.25 z2 = (10225 - 10000)/100 = 2.25 Proportion of pieces scrapped = 1 - 0.9756 = 0.0244 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 20/158) (Similar) The average rainfall, recorded to the nearest hundredth of a centimeter, in Tagaytay, for the month of March is 9.22 centimeters. Assuming a normal distribution with a standard deviation of 2.83 centimeters, find the probability that next March Tagaytay receives d) less than 1.84 centimeters of rain; e) more than 5 centimeters but not over 7 centimeters of rain; f) more than 13.8 centimeters of rain. Solution: Given: µ = 9.22 centimeters σ = 2.83 centimeters a) P(X < 1.84) = P(Z < -2.61) = 0.0045 (From Table A.3) z = (1.84 - 9.22)/2.83 = -2.61 l. P(5.01 < X < 7.00) = P(-1.487 < Z < -0.784) = 0.2177 - 0.0681 (From Table A.3) = 0.1496 z1 = (5.01 - 9.22)/2.83 = -1.488 z2 = (7.00 - 9.22)/2.83 = -0.784 m. P(X > 13.8) = P(Z > 1.62) = 1 - P(Z < 1.62) = 1 - 0.9474 (From Table A.3) = 0.0526 z = (13.8 - 9.22)/2.83 = 1.62 Example: Gamma and Exponential Distribution 1/175) If a random variable X has a gamma distribution with α = 2 and β = 1, find P(1.8 < X < 2.4). 2.4 2.4 P(1.8 < X < 2.4) = ∫ (1/β)e-x/βdx = ∫ (1/1) e-x/1 dx 1.8 1.8 2.4 = ∫ x e-x dx 1.8 Using integration by parts: Let u = x, hence du = dx dv = e-xdx, hence v = -e-x Since, ∫ u dv = uv + ∫ v du 2.4 2.4 2.4 therefore, = ∫ x e-x dx = x (-e-x) + ∫ -e-x dx 1.8 1.8 1.8 = -xe-x + e-x 2.4 = 0.00523 1.8 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 7/175) The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. What is the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days? Sol’n: β = 4 minutes 3 3 -t/4 P(T < 3) = ∫ (1/4) e dt = - ∫ e-t/4 (-1/4 dt) 0 0 3 = -[e-t/4] = 0.52763 0 6 P(X ≥ 4) = Σ b(x; 6, 0.52763) = 0.39688 x=4 5.31/183) (M/S) The lifetime y (in hours) of the central processing unit of a certain type of microcomputer is an exponential random variable with parameter β = 1,000. d. Find the mean and variance of the lifetime of the central processing unit. e. What is the probability that a central processing unit will have a lifetime of at least 2,000 hours? f. What is the probability that a central processing unit will have a lifetime of at most 1,500 hours? Solution: c. µ = β = 1,000 σ2 = β2 = 1,000,000 d. f(y ≥ 2,000) = 1 – f(y < 2,000) 2,000 2,000 = 1 - ∫ (1/1,000) e-y/1,000 dy = 1 + ∫ e-y/1,000 (–dy/1000) 0 0 2,000 = 1 + e-y/1,000 = 1 + [e-2 – e0] 0 = e-2 = 0.1353 1,500 1,500 c. f(y ≤ 1,500) = ∫ (1/1,000) e-y/1,000 dy = - ∫ e-y/1,000 (-dy/1,000) 0 0 1,500 = - e-y/1,000 = -[e-1.5 – e0] = 0.777 0 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions 5.43/186) (M/S) Suppose the random variable x has a Weibull density function with β = 4 and α = 100. a. Find F(5). b. Find P(x ≥ 3). c. Find µ and σ. Soln: x a. F(x) = ∫ f(x) dx = 0 x ∫ F(x) = (β/α)xβ-1exβ/αdx 0 by letting z = xβ , dz = βxβ-1 dx x ∫ F(x) = 5 (1/α) e 0 -z/α dz = - ∫ e -z/α (-1/α) dz 0 5 5 -x /α] = - [ e-z/α ] 0 = -[e -5 β 0 = - [e (4) /100 - e0 ] = 1 – e-6.25 = 0.9981 b. P(x ≥ 3) = 1 – P(x < 3) = 1 – F(3) 3 =1- ∫ (β/α)xβ-1exβ/αdx 0 by letting z = x , dz = βxβ-1 dx β 3 =1- ∫ 0 3 (1/α) e -z/α dz = 1 - ∫ e -z/α 0 3 = 1 + [ e-z/β ] (-1/α) dz 3 -x /α] = 1+[e ENGSTAT Notes of AM Fillone β -3 = 1 + [e (4) /100 - e0 ] Examples: Continuous Probability Distributions 0 0 = 1 + e-.81 – 1 = 0.44486 c. µ = α1/β Γ[(β+1)/β] = 1001/4 Γ(1.25) From Table, Γ(1.25) = 0.90640 µ = 3.16228 (0.90640) = 2.8663 = α2/β{Γ(1+2/β) – [Γ(1+1/β)2]}. σ2 = 1002/4 {Γ(1+2/4) – [Γ(1+1/4)2]}. = 0.8042 12/175) Suppose that the service life, in years, of a hearing aid battery is a random variable having a Weibull distribution with α = 1/2 and β = 2. b. How long can such a battery be expected to last? b. What is the probability that such a battery will still be operating after 2 years? Soln: c. µ = α-1/β Γ[(β+1)/β] = 0.5-1/2 Γ(1.5) but Γ(1.5) = 0.88623 from Table = 1.4142*0.88623 = 1.2533 d. F(x > 2) = 1 – F(x ≤ 2) 2 =1- ∫ αβx β-1 0 e-αx β dx e-αx β dx 2 =1- ∫ αβx β-1 0 let z = xβ, then dz = βxβ-1 dx 2 =1+ ∫ 0 e-αz (-αdz) = 1 + e-αz 2 0 2 ENGSTAT Notes of AM Fillone Examples: Continuous Probability Distributions = 1 + e-αxβ = 1 + [e-0.5(2)2 – e0] = 0.135335 0 The Beta Probability Distribution The probability density function for a beta-type random variable is given by yα-1(1-y)β-1 ------------------f(y) = βαβ 0 if 0 ≤ y ≤ 1; α > 0; β > 0 elsewhere where β(α,β) = ∫ yα-1(1-y)β-1dy = Γ(α)Γ(β) / Γ(α+β) The mean and variance of a beta random variable are, respectively, µ = α/(α+β) and σ2 = αβ / (α+β)2(α+β+1) [Recall that ∞ Γ(a) = ∫ ya-1 e-y dy 0 and Γ(α) = (α-1)! when α is a positive integer.] ENGSTAT Notes of AM Fillone