VCU STAT 210 - Lecture20 (71 pages)

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Lecture20



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Lecture20

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Pages:
71
School:
Virginia Commonwealth University
Course:
Stat 210 - Basic Practice of Statistics
Basic Practice of Statistics Documents

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STAT 210 Lecture 20 Normal Distributions October 11 2017 Practice Problems Pages 162 through 165 Relevant problems VI 4 Recommended problems VI 4 Additional Reading and Examples Read pages 158 through 160 TOP HAT 2 Properties The normal curve is bell shaped The peak of the curve is the population mean m The normal curve is symmetric about m The center and spread are completely specified by specifying the values of the population mean m and the population standard deviation s The total area under the normal curve is 1 or 100 Notation X N m s X is distributed normal with mean m and standard deviation s Standard Normal Distribution Denoted by Z Has population mean m 0 center Has population standard deviation s 1 spread Shape is normal symmetric bell curve No unusual features Z N 0 1 Probabilities are tabled on pages 338 339 Normal Table Gives the probability that the standard normal variable Z falls below some specified value z less than problems Read the value of z down the left most column and across the top row and read the probability from the body of the table Normal Variables A standard normal Z distribution requires that the mean is 0 and that the standard deviation is 1 How many real variables weight height grades on a test etc do you think have a mean of 0 and a standard deviation of 1 Answer I cannot think of any Z Score Transformation Skip to page 147 Suppose X is distributed normal with some mean m not equal to 0 and or some standard deviation s not equal to 1 X N m s Z Score Transformation We convert to a standard normal variable Z N 0 1 Z X m value mean s standard deviation Z Score Transformation Subtraction of m converts the mean to 0 Division by s converts the standard deviation to 1 Z X m s s m s 1 m 0 Z Score Transformation P a X b P a m X m b m s P a m s s s Z b m s Once converted from X to Z the standard normal table on pages 338 and 339 is used to find the probability just as in Examples 34 through 41 Example 47 X N 10 5 P 12 X 20 s 5 12 20 Example 47 X N 10 5 P 12 X 20 P 12 10 Z 20 10 5 5 s 5 12 20 Example 47 X N 10 5 P 12 X 20 P 12 10 Z 20 10 5 5 s 5 P 0 40 Z 2 00 12 20 s 1 0 4 2 Example 47 X N 10 5 P 12 X 20 P 12 10 Z 20 10 5 5 s 5 P 0 40 Z 2 00 12 20 P Z 2 00 P Z 0 40 s 1 0 4 2 00 Example 47 X N 10 5 P



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