Continuous probability distributions Examples: The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is __________ The probability that a given part will fail before 1000 hours of use is In general, __________ P(90 < T < 95) P(X < 1000) Probability density function f(x) ETM 620 - 09U

Understanding continuous distributions The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is The probability that a given part will fail before 1000 hours of use is ETM 620 - 09U

Continuous probability distributions The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R The cumulative distribution, F(x) Expectation and variance μ = E(X) = ______________ σ2 = ________________ Example: the uniform distribution (i.e., f(x) = 1, 1 < x < 2) 1. what is the area of the rectangle? (1) The total area under the curve is P(S) and so will always be 1. ETM 620 - 09U

{ An example x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere 1st – what does the function look like? P(X < 1.2) = ___________________ P(0.5 < X < 1) = ___________________ E(X) = -∞∫∞ x f(x) dx = ________________________ { P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) = = ∫01 x dx + ∫11.2(2-x)dx = (x2/2)|01 + (2x - x2/2)| 11.2 =0.5 + [1.68-1.5] =0.68 P(.5 < X < 1) = ∫0.51 x dx = (x2/2)|.51 = 0.375 E(x) = ∫01 x2 dx + ∫12x(2-x)dx = x3/3 |10 + (x2 – x3/3)|12 = 1 ETM 620 - 09U

Continuous probability distributions Many continuous probability distributions, including: Uniform Exponential Gamma Weibull Normal Lognormal Bivariate normal Ch. 6: Uniform Exponential Gamma Weibull - Ch. 7: Normal – Lognormal Bivariate normal ETM 620 - 09U

Uniform distribution Simplest – characterized by the interval endpoints, A and B. A ≤ x ≤ B = 0 elsewhere Mean and variance: and draw distribution ETM 620 - 09U

Example A circuit board failure causes a shutdown of a computing system until a new board is delivered. The delivery time X is uniformly distributed between 1 and 5 days. What is the probability that it will take 2 or more days for the circuit board to be delivered? interval = [1,5] f(x) = 1/(B-A) = 1/(5-1) = ¼, 1 < x < 5 (0 elsewhere) First: show the distribution and demonstrate the “intuitive” answer Then – P(X>2) = ∫25 (1/4)dx = 0.75 ETM 620 - 09U

Gamma & Exponential distributions Recall the Poisson Process Number of occurrences in a given interval or region “Memoryless” process Sometimes we’re interested in the time or area until a certain number of events occur. For example An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the probability that up to a minute will elapse before 2 calls arrive? How long before the next call? ETM 620 - 09U

Gamma distribution The density function of the random variable X with gamma distribution having parameters α (number of occurrences) and β (time or region, 1/λ). x > 0. Describes the time until a specified # of Poisson events occurs. ETM 620 - 09U

Exponential distribution Special case of the gamma distribution with α = 1. x > 0. Describes the time until or time between Poisson events. μ = β σ2 = β2 ETM 620 - 09U

Example β = ________ α = ________ An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the probability that up to a minute will elapse before 2 calls arrive? β = ________ α = ________ P(X ≤ 1) = _________________________________ β = 1/λ = 1/(2.7) = 0.3704 α = 2 P(X < 1) = 0∫1 (1/ β2) x e-x/ β dx = 2.72 0∫1 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x]01 = 1 – e-2.7 (1 + 2.7) = 0.7513 ETM 620 - 09U

Example (cont.) β = ________ α = ________ What is the expected time before the next call arrives? β = ________ α = ________ μ = _________________________________ Exponential distribution μ = β = 1/2.7 = 0.3407 min. ETM 620 - 09U

Another example Example 6-6, page 136. Note: in Excel, use =GAMMADIST(x,alpha,beta,cumulative) to find the probability associated with a specific value of x. Use =GAMMAINV(probability,alpha,beta) to find the value of x given a probability. ETM 620 - 09U

Wiebull Distribution Used for many of the same applications as the gamma and exponential distributions, but does not require memoryless property of the exponential Uses – ETM 620 - 09U

Example Designers of wind turbines for power generation are interested in accurately describing variations in wind speed, which in a certain location can be described using the Weibull distribution with α = 0.02 and β = 2. A designer is interested in determining the probability that the wind speed in that location is between 3 and 7 mph. P(3 < X < 7) = ___________________________ In Excel, =WEIBULL(x,alpha,beta,cumulative) P(3 < X < 7) = F(7) – F(3) = [1 – e – (0.02)7^2] – [1 – e – (0.02)3^2] = 0.45996 = 0.46 ETM 620 - 09U

Normal distribution The “bell-shaped curve” Also called the Gaussian distribution The most widely used distribution in statistical analysis forms the basis for most of the parametric tests we’ll perform later in this course. describes or approximates most phenomena in nature, industry, or research Random variables (X) following this distribution are called normal random variables. the parameters of the normal distribution are μ and σ (sometimes μ and σ2.) note: nonparametric tests are distribution-free, assume no underlying distribution (see ch 16) ETM 620 - 09U

Normal distribution The density function of the normal random variable X, with mean μ and variance σ2, is , all x. In Excel, =NORMDIST(x,mean,standard_dev,cumulative) and =NORMINV(probability,mean,standard_dev) (μ = 5, σ = 1.5) properties of the curve: peak is both the mean and the mode and occurs at x = μ curve is symmetrical about a vertical axis through the mean total area under the curve and above the horizontal axis = 1. points of inflection are at x = μ + σ ETM 620 - 09U

Standard normal RV … Note: the probability of X taking on any value between x1 and x2 is given by: To ease calculations, we define a normal random variable where Z is normally distributed with μ = 0 and σ2 = 1 and, ETM 620 - 09U

Standard normal distribution Table II, pg. 601-602: “Cumulative Standard Normal Distribution” In Excel, =NORMSDIST(z) and =NORMSINV(probability) ETM 620 - 09U

Examples P(Z ≤ 1) = Φ(1) = P(Z ≥ -1) = P(-0.45 ≤ Z ≤ 0.36) = draw the area on the picture … 1. P(Z < 1) = 0.8413 2. P(Z ≥ -1) = 1 – Φ(-1) = 0.1587 3. P(-0.45 ≤ Z ≤ 0.36) = Φ(0.36) – Φ(-0.45) = 0.6406 – 0.3246 = 0.316 ETM 620 - 09U

Your turn … Use Excel to determine (draw the picture!) 1. P(Z ≤ 0.8) = 4. P(Z ≤ -2.0 or Z ≥ 2.0) = 1. P(Z ≤ 0.8) = 0.7881 2. P(Z ≥ 1.96) = 1 – 0.975 = 0.025 (=P(Z < -1.96)) Note symmetry!! 3. P(-0.25 ≤ Z ≤ 0.15) = 0.5596 – 0.4013 = 0.1583 4. P(Z ≤ -2.0 or Z ≥ 2.0) = 2 * 0.0228 = 0.0456 ETM 620 - 09U

The normal distribution “in reverse” Example: Given a normal distribution with μ = 40 and σ = 6, find the value of X for which 45% of the area under the normal curve is to the left of X. If Φ(k) = 0.45, k = ___________ Z = _______ X = _________ Will X be greater than 40? k = -0.125 Z = -0.125 = (X – 40)/6 Z = (-0.125*6)+40 = 39.25 ETM 620 - 09U

Normal approximation to the binomial If n is large and p is not close to 0 or 1, or if n is smaller but p is close to 0.5, then the binomial distribution can be approximated by the normal distribution using the transformation: NOTE: add or subtract 0.5 from X to be sure the value of interest is included (draw a picture to know which) Look at example 7-10, pg. 156 ETM 620 - 09U

Another example The probability of a patient making a full recovery from a rare heart ailment is 0.4. If 100 people are diagnosed with this ailment, what is the probability that fewer than 30 of them recover? p = 0.4 n = 100 μ = ____________ σ = ______________ if x = 30, then z = _____________________ and, P(X < 30) = P (Z < _________) = _________ μ= np = 100*0.4 = 40 σ = sqrt(npq) = sqrt(100*0.4*0.6) = 4.899 draw the picture! z = ((30-0.5) – 40)/4.899 = -2.14 P(Z < -2.14) = 0.0162 ETM 620 - 09U

Your Turn Refer to the previous example, DRAW THE PICTURE!! What is the probability that more than 50 survive? What is the probability that exactly 45 survive? x > 50, z = ((50 + 0.5) – 40)/4.899 = 2.14 P(Z > 2.14) = P(Z < -2.14) = 0.0162 (by symmetry) 2. x = 45, z1 = (45.5-40)/4.899 = 1.12 z2 = (44.5 – 40)/4.899 = 0.9816 P(X = 45) = P(z < 1.12) – P(z < 0.98) = 0.8686 – 0.8365 = 0.0321 NOTE: b(45;100,0.4) = (100 choose 45)*0.4450.655 = 0.0478 ETM 620 - 09U

Lognormal Distribution When the random variable Y = ln(X) is normally distributed with mean μ and standard deviation σ, then X has a lognormal distribution with the density function, Uses – reliability and maintainability; environmental engineers – concentration of pollutants, particle size in emissions; long term rate of return on stock investments 26 ETM 620 - 09U

Example The average rate of water usage (in thousands of gallons per hour) by a certain community is know to involve the lognormal distribution with parameters μ = 5 and σ =2. What is the probability that, for any given hour, more than 50,000 gallons are used? P(X > 50,000) = __________________________ P(X > 50,000) = 1 – P(X < 50,000) = 1 – P (Z < (ln(50,000) – 5)/ 2) = 1 – P (Z < 2.91) = 1 - 0.9982 = 0.0018 27 ETM 620 - 09U

Bivariate Normal distribution Random variables [X, Y] that follow a bivariate normal distribution have a joint density function, and a joint probability, This would be pretty nasty to calculate, except … X ∼ N (μX, σ2X ) and Y ∼ N (μY , σ2Y ) ETM 620 - 09U

So… It can be shown (see pp. 161-162) that, This allows us to use the standard normal approach to solve several important problems. ETM 620 - 09U

Example Example7-14, pg. 163 Given the information regarding shear strength and weld diameter in the example, what is the probability that the strength specification (1080 lbs.) can be met if the weld diameter is 0.18 in.? ETM 620 - 09U