VIEWS: 2 PAGES: 16 POSTED ON: 1/14/2014
Review • Continuous Random Variables – Density Curves • Uniform Distributions • Normal Distributions – Probabilities correspond to areas under the curve. – the total area under the curve is 1, – Only the probability of an event occurring in some interval can be evaluated. – The probability that a continuous random variable takes on any particular value is zero. Normal Distributions Shape of this curve is determined by µ and σ – µ it’s centered, σ is how far it’s spread out. The Normal Distribution Probabilities associated with values or ranges of a random variable correspond to areas under the normal curve Calculating probabilities can be simplified by working with a Standard Normal Distribution A Standard Normal Distribution is a Normal distribution with m =0 and s =1 The standard normal random variable is denoted by the symbol z The Normal Distribution Table for Standard Normal Distribution contains probability for the area between 0 and z Partial table shows components of table The Normal Distribution What is P(-1.33 < z < 1.33)? P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z <1.33) = .4082 + .4082 = .8164 The Normal Distribution What is P(z < .67)? Table gives us area A1 Symmetry about the mean tell us that A2 = .5 P(z < .67) = A1 + A2 = .2486 + .5 = .7486 The Normal Distribution What is P(|z| > 1.96)? P(|z| > 1.96) = A1 + A2 = .0250 + .0250 =.05 The Normal Distribution What if values of interest were not normalized? We want to know P (8<x<12), with μ=10 and σ=1.5 Convert to standard normal using P(8<x<12) = P(-1.33<z<1.33) = 2(.4082) = .8164 The Normal Distribution Steps for Finding a Probability Corresponding to a Normal Random Variable 1) Sketch the distribution, locate mean, shade area of interest 2) Convert to standard z values using 3) Add z values to the sketch 4) Use tables to calculate probabilities, making use of symmetry property where necessary The Normal Distribution Making an Inference How likely is an observation in area A, given an assumed normal distribution with mean of 27 and standard deviation of 3? Z value for x=20 is -2.33 P(x<20) = P(z<-2.33) = .5 - .4901 = .0099 You could reasonably conclude that this is a rare event The Normal Distribution You can also use the table in reverse to find a z-value that corresponds to a particular probability What is the value of z that will be exceeded only 10% of the time? Look in the body of the table for the value closest to .4, and read the corresponding z value z = 1.28 The Normal Distribution Which values of z enclose the middle 95% of the standard normal z values? Using the symmetry property, z0 must correspond with a probability of .475 From the table, we find that z0 and –z0 are 1.96 and -1.96 respectively. The Normal Distribution Given a normally distributed variable x with mean 550 and standard deviation of 100, what value of x identifies the top 10% of the distribution? The z value corresponding with .40 is 1.28. Solving for x0 x0 = 550 +1.28(100) = 550 +128 = 678 Problems Problems 5.28, 5.26, 5.30, 5.36, 5.40, 5.48 Problem 5.54 Keys to success Learn the standard normal table and how to use it. This includes using the table in reverse. We will be using these tables through out the course. Homework • Review Chapter 5.1-5.4 • Read Chapters 6.1-6.3 for next week 16