FIT Normal Curve Concepts and Exercises 2015

FIT Normal Curve Concepts and Exercises 2015 - FIT...

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FIT Statistics Class Normal Curve Concepts and Exercises Created by Kenneth W. Lewis, Ph.D., ALU 1 Statistics Handout Finding Areas Under the Normal Curve. A normal distribution is nothing more than a histogram with very small (infinitesimally small) intervals and their respective relative cumulative frequencies. When you talk about a variable that is normally distributed you usually say something like, “ The time it takes to complete a certain personnel action (X) is normally distributed with a mean of = 80 days and a standard deviation of = 15 days .” You can rewrite the previous statement in the following statistical shorthand X ~ n (80, 15). Now since X is normally distributed with a mean of 80 and a standard deviation of 15, maybe I want to know what proportion of personnel actions are completed in less than 62 days. How do I figure this out? Here is the procedure: Find the P(X < 62) Change 62 to a Z score. What is a Z score? A Z score is a standardized number or value that tells you how far your score is located from the mean in terms of standard deviation units. How do you compute a Z score? Z = (X ) / Z = (The actual score the population mean) / the population standard deviation. Z = (62 80) / 15 = -18 / 15 = -1.20 So the value of X = 62 corresponds to a Z score value of -1.20. That means that the score of 62 is 1.20 standard deviations below the mean. How do I use the Z score and the Z table to find the appropriate proportion? Now that you have a Z score, go to the cumulative normal curve table and do the following: 1. Since the Z score you computed is a negative number, look at the first page of Z score values which has the negative Z scores. 2. Look down the left column of page one until you find -1.2. 3. Then move your finger over to the column heading 0.00, which is the second decimal of the Z-score.
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FIT Statistics Class Normal Curve Concepts and Exercises Created by Kenneth W. Lewis, Ph.D., ALU 2 4. You should see the value .1151 5. So, therefore, 11.51% of all personnel actions were completed in less than 62 days. NOW LET’S TRY A FEW MORE OF THESE. How do I figure this out? Find the P(X < 100) Change 100 to a Z score. Z = (100 80) / 15 = 20 / 15 = 1.33 So the value of X = 100 corresponds to a Z score value of 1.33. That means that the value of 100 is 1.33 standard deviations above the mean. 1. Since the Z score you computed is a positive number, look at the second page of Z score values which has the positive Z scores. 2. Look down the left column of page one until you see the value, 1.3. 3. Then move your finger over to the column heading 0.03, which is the second decimal of the Z-score. 4. You should see the value 0.9082. 5. So, therefore, 90.82% of all personnel actions were completed in less than 100 days. NOW LET’S TRY A FEW MORE OF THESE. How do I figure this out? Find the P(X > 72) First we have to rewrite this problem in terms of X < something. WHY? Because the normal curve table is a cumulative table and it only tells us what proportion of the curve is less than something. So we have to do the following: P(X > 72) = 1 P(X < 72).
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