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Math/Stat 370: Engineering Statistics, Washington State University Haijun Li [email protected] Department of Mathematics Washington State University Week 3 Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 1 / 15 Normal (or Gaussian) Distribution Normal density function N(µ, σ 2 ): −(x−µ)2 1 ϕµ,σ2 (x) = √ e 2σ2 , −∞ < x < ∞. 2πσ Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 2 / 15 Normal (or Gaussian) Distribution Normal density function N(µ, σ 2 ): −(x−µ)2 1 ϕµ,σ2 (x) = √ e 2σ2 , −∞ < x < ∞. 2πσ Normal random variable: X has a normal density function N(µ, σ 2 ). µ = location parameter, σ = scale parameter. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 2 / 15 Properties of Normal Distributions 1 2 3 4 5 Normal density curve is symmetric about µ. The curve reaches the maximum at µ. E(X ) = µ. V (X ) = σ 2 . P(µ − σ < X < µ + σ) ≈ 0.68. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 3 / 15 Normal Approximation Many histograms of large samples have similar shapes as that of normal distributions. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 4 / 15 Standardizing Normal Random Variables Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 5 / 15 Standardizing Normal Random Variables Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1. If X has N(µ, σ 2 ), then Z = X −µ , (called the standardization!) σ has the standard normal N(0, 1). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 5 / 15 Standardizing Normal Random Variables Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1. If X has N(µ, σ 2 ), then Z = X −µ , (called the standardization!) σ has the standard normal N(0, 1). CDF of N(0, 1): Φ(z) = P(Z ≤ z). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 5 / 15 Standardizing Normal Random Variables Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1. If X has N(µ, σ 2 ), then Z = X −µ , (called the standardization!) σ has the standard normal N(0, 1). CDF of N(0, 1): Φ(z) = P(Z ≤ z). Appendix A Table I lists all the values of Φ(z) from z = −3.9 to z = 3.9. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 5 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) = 0.8849 − 0.0107 = 0.8742. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) = 0.8849 − 0.0107 = 0.8742. 4 P(|X | > 3.1). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) = 0.8849 − 0.0107 = 0.8742. 4 P(|X | > 3.1). P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) = 0.8849 − 0.0107 = 0.8742. 4 P(|X | > 3.1). P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194. 5 Find z such that P(−z < Z < z) = 0.95. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Standard Normal Probabilities Example: Find the following. 1 P(Z ≤ −2.3). P(Z ≤ −2.3) = Φ(−2.3) = 0.0107. 2 P(Z > 1.2) P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 3 P(−2.3 < Z < 1.2). P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) = 0.8849 − 0.0107 = 0.8742. 4 P(|X | > 3.1). P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194. 5 Find z such that P(−z < Z < z) = 0.95. First, we must have that P(Z < z) = 0.975. Second, we find from the normal table that z = 1.96. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 6 / 15 Normal Probabilities If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization leads to Z z }| { a − µ X −µ b − µ P(a < X < b) = P < < σ σ σ Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 7 / 15 Normal Probabilities If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization leads to Z z }| { a − µ X −µ b − µ P(a < X < b) = P < < σ σ σ = P Haijun Li a−µ b−µ <Z < σ σ Math/Stat 370: Engineering Statistics, Washington State University Week 3 7 / 15 Normal Probabilities If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization leads to Z z }| { a − µ X −µ b − µ P(a < X < b) = P < < σ σ σ a−µ b−µ = P <Z < σ σ b−µ a−µ = Φ −Φ . σ σ Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 7 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 What is the probability that the diameter of a dot exceeds 0.0026 inch? Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 2 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. What is the probability that the diameter of a dot between 0.0014 and 0.0026 inch? Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 2 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. What is the probability that the diameter of a dot between 0.0014 and 0.0026 inch? P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) = 0.933 − 0.067 = 0.866. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 2 3 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. What is the probability that the diameter of a dot between 0.0014 and 0.0026 inch? P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) = 0.933 − 0.067 = 0.866. What is the diameter that is exceeded 90% of time? Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 2 3 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. What is the probability that the diameter of a dot between 0.0014 and 0.0026 inch? P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) = 0.933 − 0.067 = 0.866. What is the diameter that is exceeded 90% of time? Solution: Find P(X > t) = 0.9. That is, t such that t−0.002 t−0.002 1 − Φ 0.0004 = 0.9 or Φ 0.0004 = 0.1. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Example: The diameter X of the dot produced produced by a printer is normally distributed with a mean diameter of µ = 0.002 inch and a standard deviation of σ = 0.0004 inch. 1 2 3 What is the probability that the diameter of a dot exceeds 0.0026 inch? P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 = 1 − Φ(1.5) = 0.067. What is the probability that the diameter of a dot between 0.0014 and 0.0026 inch? P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) = 0.933 − 0.067 = 0.866. What is the diameter that is exceeded 90% of time? Solution: Find P(X > t) = 0.9. That is, t such that t−0.002 t−0.002 1 − Φ 0.0004 = 0.9 or Φ 0.0004 = 0.1. From the normal table, t−0.002 = −1.28 and t = 0.001482 0.0004 inch. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 8 / 15 Log-normal Distribution X ≥ 0 has a log-normal distribution if ln(X ) has a normal distribution with mean θ and variance ω 2 , N(θ, ω 2 ). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 9 / 15 Log-normal Distribution X ≥ 0 has a log-normal distribution if ln(X ) has a normal 2 2 distribution with mean θ and variance ω ,N(θ, ω ). P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ . ω Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 9 / 15 Log-normal Distribution 0/02 5:20 PM Page 136 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:MONTGOMERY:REVISES UPLO D C 136 X ≥ 0 has a log-normal distribution if ln(X ) has a normal 2 2 distribution with mean θ and variance ω ,N(θ, ω ). P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ . ω Compound returns of stock prices and component lifetimes CHAPTER 4 CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS can be modeled using log-normal distributions. 1 ω 2 = 0.25 ω2 = 1 ω 2 = 2.25 0.9 0.8 0.7 0.6 f (x) 0.5 0.4 0.3 0.2 0.1 0 –0.1 Haijun Li 0 1 2 3 x 4 5 Math/Stat 370: Engineering Statistics, Washington State University 6 Week 3 9 / 15 Example The lifetime X of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 10 / 15 Example The lifetime X of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. 1 What is the probability that lifetime exceeds 10,000 hours? Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 10 / 15 Example The lifetime X of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. 1 What is the probability that lifetime exceeds 10,000 hours? P(X > 10, 000) = 1 − P(X ≤ 10, 000) = = 1 − Φ(−0.53) = 0.702. 1 − Φ ln 10,000−10 1.5 Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 10 / 15 Example The lifetime X of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. 1 What is the probability that lifetime exceeds 10,000 hours? P(X > 10, 000) = 1 − P(X ≤ 10, 000) = = 1 − Φ(−0.53) = 0.702. 1 − Φ ln 10,000−10 1.5 2 What lifetime is exceeded by 99% of lasers? Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 10 / 15 Example The lifetime X of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. 1 What is the probability that lifetime exceeds 10,000 hours? P(X > 10, 000) = 1 − P(X ≤ 10, 000) = = 1 − Φ(−0.53) = 0.702. 1 − Φ ln 10,000−10 1.5 2 What lifetime is exceeded by 99% of lasers? Solution: Find t such that P(X > t) = 0.99. That is, ln t − 10 P(X ≤ t) = 0.01, and Φ = 0.01. 1.5 Solve it, we have hours. Haijun Li ln t−10 1.5 = −2.33 and t = e6.505 = 668.48 Math/Stat 370: Engineering Statistics, Washington State University Week 3 10 / 15 Gamma Distribution A random variable X is said to have a gamma distribution if its pdf Z ∞ λk x k −1 e−λx f (x) = , x > 0, where Γ(k ) = x k −1 e−x dx, Γ(k ) 0 Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 11 / 15 Gamma Distribution A random variable X is said to have a gamma distribution if its pdf Z ∞ λk x k −1 e−λx f (x) = , x > 0, where Γ(k ) = x k −1 e−x dx, Γ(k ) 0 If k is a positive integer, then the gamma function Γ(k ) = (k − 1)!. Widely used in reliability engineering and telecommunication. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 11 / 15 Gamma Distribution A random variable X is said to have a gamma distribution if its pdf Z ∞ λk x k −1 e−λx f (x) = , x > 0, where Γ(k ) = x k −1 e−x dx, Γ(k ) 0 If k is a positive integer, then the gamma function Γ(k ) = (k − 1)!. Widely used in reliability engineering and telecommunication. E(X ) = λk , V (X ) = λk2 . k = shape parameter, λ = rate parameter. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 11 / 15 k = shape parameter, λ = rate parameter, θ = λ−1 = scale parameter. A special case: When k = n (integer), the distribution is called an Erlang distribution (widely used in telecommunication). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 12 / 15 k = 1: Exponential distribution. The pdf is given by f (x) = λe−λx , x ≥ 0. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 13 / 15 k = 1: Exponential distribution. The pdf is given by f (x) = λe−λx , x ≥ 0. The CDF of the exponential is given by F (x) = 1 − e−λx . Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 13 / 15 k = 1: Exponential distribution. The pdf is given by f (x) = λe−λx , x ≥ 0. The CDF of the exponential is given by F (x) = 1 − e−λx . E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 13 / 15 k = 1: Exponential distribution. The pdf is given by f (x) = λe−λx , x ≥ 0. The CDF of the exponential is given by F (x) = 1 − e−λx . E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale. f (x) Failure rate h(x) = 1−F = λ, a constant. (x) Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 13 / 15 Weibull Distribution A random variable X with pdf β x β−1 −( x )β e δ , x > 0, f (x) = δ δ is said to have a Weibull distribution with parameters β (shape), and δ (scale). Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 14 / 15 Weibull Distribution A random variable X with pdf β x β−1 −( x )β e δ , x > 0, f (x) = δ δ is said to have a Weibull distribution with parameters β (shape), and δ (scale). If β = 1, the Weibull becomes an exponential. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 14 / 15 Weibull Distribution A random variable X with pdf β x β−1 −( x )β e δ , x > 0, f (x) = δ δ is said to have a Weibull distribution with parameters β (shape), and δ (scale). If β = 1, the Weibull becomes an exponential. x β The CDF F (x) = 1 − e−( δ ) . Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 14 / 15 Weibull Distribution A random variable X with pdf β x β−1 −( x )β e δ , x > 0, f (x) = δ δ is said to have a Weibull distribution with parameters β (shape), and δ (scale). If β = 1, the Weibull becomes an exponential. x β The CDF F (x) = 1 − e−( δ ) . The failure rate f (x) β x β−1 h(x) = . = 1 − F (x) δ δ h(x) is increasing if β > 1; decreasing if β < 1; and a constant if β = 1. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 14 / 15 Weibull Distribution A random variable X with pdf β x β−1 −( x )β e δ , x > 0, f (x) = δ δ is said to have a Weibull distribution with parameters β (shape), and δ (scale). If β = 1, the Weibull becomes an exponential. x β The CDF F (x) = 1 − e−( δ ) . The failure rate f (x) β x β−1 h(x) = . = 1 − F (x) δ δ h(x) is increasing if β > 1; decreasing if β < 1; and a constant if β = 1. E(X ) = δΓ(1 + β1 ), V (X ) = δ 2 Γ(1 + β2 ) − δ 2 [Γ(1 + β1 )]2 . Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 14 / 15 Weibull PDF k = β = shape parameter, λ = δ = scale parameter. Haijun Li Math/Stat 370: Engineering Statistics, Washington State University Week 3 15 / 15