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Chapter 5 Normal Probability Distributions. Chapter 5 Normal Probability Distributions Section 5-3 – Normal Distributions: Finding Values A.We have learned.

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Presentation on theme: "Chapter 5 Normal Probability Distributions. Chapter 5 Normal Probability Distributions Section 5-3 – Normal Distributions: Finding Values A.We have learned."— Presentation transcript:

1 Chapter 5 Normal Probability Distributions

2 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 z- score; 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)2 nd VARS invNorm(probability) = z-score. b)2 nd VARS invNorm(probability, mean, standard deviation) = x-value

3 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 x- value. 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)

4 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)0.758013)P 1 2)0.209014)P 15 3)0.633116)P 55 InvNorm(.7580,0,1) =.700 InvNorm(.2090,0,1) = -.810 InvNorm(.6331,0,1) =.340 InvNorm(.01,0,1) = -2.326 InvNorm(.15,0,1) = -1.036 InvNorm(.55,0,1) = 1.126

5 Your Assignments are: Classwork: Pages 266-267 #25-38 All Homework: Pages 267-268 #39-46 All


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