Uniform and Normal Distributions Chp. 6 Uniform and Normal EGR 252

Continuous Probability Distributions Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull Weibull PDF Source: www.itl.nist.gov Uniform Normal – Gamma Exponential Chi-Squared Lognormal Weibull - Chp. 6 Uniform and Normal EGR 252

Uniform Distribution Simplest – characterized by the interval endpoints, A and B. A ≤ x ≤ B = 0 elsewhere Mean and variance: and draw distribution Chp. 6 Uniform and Normal EGR 252

Example: Uniform Distribution 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) Chp. 6 Uniform and Normal EGR 252

In Class Examples 6.3 Chp. 6 Uniform and Normal EGR 252

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) Chp. 6 Uniform and Normal EGR 252

Normal Distribution The density function of the normal random variable X, with mean μ and variance σ2, is all x. (μ = 5, σ = 1.5) properties of the curve: peak is both the mean and the mode and the median 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. Chp. 6 Uniform and Normal EGR 252

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 We cannot integrate this expression simply, so we will use the Z transformation. Chp. 6 Uniform and Normal EGR 252

Standard Normal Distribution Table A.3 Pages 735-736: “Areas under the Normal Curve” Body of table – area under curve Chp. 6 Uniform and Normal EGR 252

Examples P(Z ≤ 1) = P(Z ≥ -1) = P(-0.45 ≤ Z ≤ 0.36) = draw the area on the picture … 2012 1. P(Z < 1) = 0.8413 2. P(Z ≥ -1) = 0.1587 3. P(-0.45 ≤ Z ≤ 0.36) = P(Z < 0.36) – P(Z< -0.45) = 0.6406 – 0.3264 = 0.3142 Chp. 6 Uniform and Normal EGR 252

Name:________________________ Use Table A.3 to determine (draw the picture!) 1. P(Z ≤ 0.8) = 2. P(Z ≥ 1.96) = 3. P(-0.25 ≤ Z ≤ 0.15) = 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 Chp. 6 Uniform and Normal EGR 252

Applications of the Normal Distribution A certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms. What percentage of the resistors will have a resistance less than 44 ohms? Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44 P(X<44) = P(Z< +2.0) = 0.9772 Therefore, we conclude that 97.72% will have a resistance less than 44 ohms. What percentage will have a resistance greater than 44 ohms? Chp. 6 Uniform and Normal EGR 252

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. Step 1 If P(Z < z) = 0.45, z = _______ (from Table A.3) Why is z negative? Step 2 X = _________ 45% Will X be greater than 40? k = -0.125 Z = -0.125 = (X – 40)/6 X = (-0.125*6)+40 = 39.25 Chp. 6 Uniform and Normal EGR 252

In-Class Exercise 6.14 The finished inside diameter of a piston ring is normally distributed with a mean of 10 centimeters and a standard deviation of 0.03 centimeter. What proportion of rings will have inside diameters exceeding 10.075 centimeters? (b) What is the probability that a piston ring will have an inside diameter between 9.97 and 10.03 centimeters? (c) Below what value of inside diameter will 15% of the piston rings fall? of 0.03 centimeter. Chp. 6 Uniform and Normal EGR 252

In-Class Exercise 6.14 Solution What proportion of rings will have inside diameters exceeding 10.075 centimeters? (b) What is the probability that a piston ring will have an inside diameter between 9.97 and 10.03 centimeters? (c) Below what value of inside diameter will 15% of the piston rings fall? of 0.03 centimeter. Chp. 6 Uniform and Normal EGR 252

In-Class Exercise 6.17 The average life of a certain type of small motor is 10 years with a standard deviation of 2 years. The manufacturer replaces free all motors that fail while under guarantee. If she is willing to replace only 3% of the motors that fail, how long a guarantee should be offered? Assume that the lifetime of a motor follows a normal distribution. Chp. 6 Uniform and Normal EGR 252

In-Class Exercise 6.17 The average life of a certain type of small motor is 10 years with a standard deviation of 2 years. The manufacturer replaces free all motors that fail while under guarantee. If she is willing to replace only 3% of the motors that fail, how long a guarantee should be offered? Assume that the lifetime of a motor follows a normal distribution. 3% Solve for X X = (2 * -1.88) + 10 = 6.24 A z-value of -1.88 corresponds to 3% of area under the curve Chp. 6 Uniform and Normal EGR 252