Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 5 Normal Probability Distributions Chapter 5 Normal Probability Distributions Section 5-3 – Normal Distributions: Finding Values A. We have learned how to calculate the probability given an x-value or a z-score. In this lesson, we will explore how to find an x-value or z-score when given the probability (cumulative area under the curve). 1. The cumulative area under the curve is a direct variation of the zscore; as the z-score goes up, so does the cumulative area. a. Because this is a one to one function, it also has an inverse function. 1) Lucky for us, the calculator has an operation to find the inverse of the cumulative area. a) 2nd VARS invNorm(probability) = z-score. b) 2nd VARS invNorm(probability, mean, standard deviation) = x-value Chapter 5 Normal Probability Distributions Section 5-3 – Normal Distributions: Finding Values 2. Once you have the z-score, you can also find the matching xvalue. a. If we take the formula for finding the z-score and solve it for x, we get that x = μ + zσ. 1) In other words, x is equal to the mean plus the z-score times the standard deviation. B. The key here is going to be using the correct area under the curve to find the z or x value that we are looking for. 1. Practice will make this a LOT easier. a. As a general rule, if we want the area below a given percentile, we use the given percentile. b. If we want the area above a given percentile, we subtract the given percentile from 1 and use the answer. (Complement Rule) Chapter 5 Normal Probability Distributions Section 5-3 – Examples Page 266, #1-3 and 13, 14, and 16 Find the z-score that corresponds to the given cumulative area or percentile. 1) 0.7580 InvNorm(.7580,0,1) = .700 2) 0.2090 InvNorm(.2090,0,1) = -.810 3) 0.6331 InvNorm(.6331,0,1) = .340 13) P1 InvNorm(.01,0,1) = -2.326 14) P15 InvNorm(.15,0,1) = -1.036 16) P55 InvNorm(.55,0,1) = 1.126 Your Assignments are: Classwork: Pages 266-267 #25-38 All Homework: Pages 267-268 #39-46 All