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MM207 Unit 5: Normal Distribution, Probabilities, and Sampling Distributions

Examines normal distribution and sampling techniques.

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    MM207 Unit 5 Chapter 5 Project ROBYN GALINDO 1 MM207 Unit 5: Normal Distribution, Probabilities, and Sampling Distributions Total points are 4 0 . Points are included with each problem. Included with each section or problem are reference examples and end of section exercises that can be used as a guide . Be sure to show your work in case partial credit is awarded . To receive full credit, work must be shown if applicable. Section 5.1: Introduction to Normal Distribution and the Standard Normal Distribution 1. Use the Standard Normal Distribution table to find the indicated area under the standard normal curve. (References: example 3 - 6 pages 244 - 247, end of section exercises 21 – 40 page 249) ( 2 points per each part ) a. Between z = 0 and z = 1.42 The area to the left of z = 0 is 0.5000. The area to the left of z = 1.42 is 0.9222. The area between z = 0 and z = 1.42 is 0.4222. b. To the left of z = 1.86 The area to the left of z = 1.86 is 0.9686. c. Between z = - 1.52 and z = - 0.46 The area to the left of z = - 1.52 is 0.0643. The area to the left of z = - 0.46 is 0.3228. The area between z = - 1.52 and z = - 0.46 is 0.2585. d. To t he right of z = - 1.45 The area to the right of z = - 1.45 is 1 - .0735 = 0.9265 . Section 5.2: Normal Distributions: Find Probabilities (References: example 1 and 2 page 253, end of section exercises 13 - 30 pages 257 - 259 2 . The diameters of a wooden dowel produced by a new machine are normally distributed with a mean of 0.55 inches and a standard deviation of 0.01 inches. What percent of the dowels will have a diameter less than 0.5 6 ? (6 points) Z = (0.56 – 0.55) / .01 = 1 = 0.8413 84.13% of the dowels will have a diameter of less than 0.56.

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