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M11-Normal Distribution 1 1  Department of ISM, University of Alabama, 1995-2003 Lesson Objective  Learn the mechanics of using the table for the Normal.

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Presentation on theme: "M11-Normal Distribution 1 1  Department of ISM, University of Alabama, 1995-2003 Lesson Objective  Learn the mechanics of using the table for the Normal."— Presentation transcript:

1 M11-Normal Distribution 1 1  Department of ISM, University of Alabama, 1995-2003 Lesson Objective  Learn the mechanics of using the table for the Normal Distribution.  Given a region for a variable that follows the Normal Distribution, find the probability that a randomly selected item will fall in this region.  Given a probability, find the region for a normally distributed variable that corresponds to this probability.

2 M12-Normal Distribution 2 2  Department of ISM, University of Alabama, 1995-2003 The Normal Distributions a.k.a., “The Bell Shaped Curve” Describes the shape for some quantitative, continuous random variables.

3 M12-Normal Distribution 2 3  Department of ISM, University of Alabama, 1995-2003  = mean determines the location.  = standard deviation determines spread, variation, scatter. Normal Population Distribution has two parameters:

4 M12-Normal Distribution 2 4  Department of ISM, University of Alabama, 1995-2003 X ~ N(  = 66,  = 9) or N(66, 9) Z = the number of standard deviations that an X - value is from the mean. Notation: X -   Z = Z ~ N(  = 0,  = 1 ) or N(0,1) Z follows the “Standard Normal Distribution”

5 M12-Normal Distribution 2 5  Department of ISM, University of Alabama, 1995-2003 ____, ±1  ______, ±2  ______, ±3  Empirical Rule of the Normal Distribution Where does this come from?

6 M12-Normal Distribution 2 6  Department of ISM, University of Alabama, 1995-2003 Recall The “area” under the curve within a range of X values is equal to proportion of the population within that range of X values. Question: How do we compute “areas”? Geometry formulas Calculus (integration) Tables Excel Minitab

7 M12-Normal Distribution 2 7  Department of ISM, University of Alabama, 1995-2003 Reading the Standard Normal Table (finding areas under the normal curve) Step 1 for all problems: DTDP

8 M12-Normal Distribution 2 8  Department of ISM, University of Alabama, 1995-2003 0 1.72 Table gives P(0 < z < ?) = Find P(0 < z < 1.72) = Up to the 1 st decimal place 2 nd decimal place.4573 Standard Normal Table

9 M12 Normal Distribution 2 9  Department of ISM, University of Alabama, 1995-2003 P(-1.23 < Z < 2.05) = ? 0Z 2.05 What proportion of Z values are between –1.23 and +2.05? P(1.23 < Z < 2.05) = ? 0Z What proportion of Z values are between +1.23 and +2.05? = = = = 1.23 -1.23 2.05 ?

10 M12 Normal Distribution 2 10  Department of ISM, University of Alabama, 1995-2003 P( 10.0 < X < 15.72) = ? Z P( 0 < Z < 1.43) Weights of packages are normally distributed with mean of 10 lbs. and standard deviation of 4.0 lbs. Find the proportion of packages that weigh between 10 and 15.72 lbs. 10X 15.72 Z = 15.72 – 10.0 4.0 = Z = 10.0 – 10.0 4.0 = X = weight of packages. X ~ N(  = 10,  = 4.0) =

11 M12 Normal Distribution 2 11  Department of ISM, University of Alabama, 1995-2003 P( X > 15.72) = ? 0Z 1.43 P( Z > 1.43) Same situation. What proportion of packages weigh more than 15.72 lbs? 10X 15.72 Z = 15.72 – 10.0 4.0 = 1.43 X = weight of packages. X ~ N(  = 10,  = 4.0) = = ?

12 M12 Normal Distribution 2 12  Department of ISM, University of Alabama, 1995-2003 P( X < 14.2) = ? 0Z 1.05.3531 P( Z < 1.05) Same situation. What proportion of packages weigh less than 14.2 lbs? 10X 14.2 Z = 14.2 – 10.0 4.0 = 1.05 X = weight of packages. X ~ N(  = 10,  = 4.0) =.5 +.3531 =.8531.5000

13 M12-Normal Distribution 2 13  Department of ISM, University of Alabama, 1995-2003 Same situation. What proportion of packages weigh between 5.08 and 18.2 lbs? X = weight of packages. X ~ N(  = 10,  = 4.0)

14 M12-Normal Distribution 2 14  Department of ISM, University of Alabama, 1995-2003 Same situation. What proportion of packages weigh either less than 2.4 lbs or greater than 11.0 lbs? X = weight of packages. X ~ N(  = 10,  = 4.0) Homework

15 M12 Normal Distribution 2 15  Department of ISM, University of Alabama, 1995-2003 P( X < ?) =.10 0Z -1.28.10 P( Z < ) Same situation. Find the weight such that 10% of all packages weigh less than this weight. 10X ? X = weight of packages. X ~ N(  = 10,  = 4.0) =.10.40 – 1.28 = ? – 10 4 ? = 10 – 1.28 4 = 10 – 5.12 = 4.88 pounds -1.28 This is a backwards problem! We are given the probability; we need to find the boundary. 10% weigh less than 4.88 pounds; 90% weigh more than 4.88 pounds. 10% weigh less than 4.88 pounds; 90% weigh more than 4.88 pounds.

16 M12 Normal Distribution 2 16  Department of ISM, University of Alabama, 1995-2003 0 Table gives P(0 < z < ?) = Standard Normal Table Find P( __ < z < 0) =.40.4000 -1.28 Find the Z value to cut off the top 10%.

17 M12 Normal Distribution 2 17  Department of ISM, University of Alabama, 1995-2003 0 Table gives P(0 < z < ?) = Standard Normal Table.25 ? Find the Z values that define the middle 50%..25 ?

18 M12 Normal Distribution 2 18  Department of ISM, University of Alabama, 1995-2003 Table gives P(0 < z < ?) = Standard Normal Table Find the Z values that define the middle 95%.

19 M12-Normal Distribution 2 19  Department of ISM, University of Alabama, 1995-2003 Normal Functions in Excel NORMDIST – Used to compute areas under any normal curve. Can also compute height of curve (not useful except for drawing normal curves). NORMSDIST - Used to compute areas under a standard normal ( N(0,1) or Z curve ).

20 M12-Normal Distribution 2 20  Department of ISM, University of Alabama, 1995-2003 Normal Functions in Excel NORMINV - Used to find the X value corresponding to a given cumulative probability for any normal distribution. NORMSINV - Used to find the Z value corresponding to a given cumulative probability for a standard normal distribution.

21 M12-Normal Distribution 2 21  Department of ISM, University of Alabama, 1995-2003 1. P( Z < –1.92) = 2. P( Z < 2.56) = 3. P( Z > 0.80) = 4. P( Z = 1.42) = 5. P(.32 < Z < 2.48) = 6. P( -1.75 < Z < 1.75) = 7. P( Z < 4.25) = 8. P( Z > 4.25) = 9. P(-.05 < Z <.05) = 10. Find Z such that only 12% are smaller. Practice problems. You MUST know how to work ALL of these problems and the following practice problems to pass this course.

22 M12-Normal Distribution 2 22  Department of ISM, University of Alabama, 1995-2003 1..0274 2..9948 3. 1.0 –.7881 =.2119 4..0 5..9934 –.6255 =.0689 6..9599 –.0401 =.9198 7. 1.0000 8..0000 9..0398 10. -1.175 Practice problem answers

23 M12-Normal Distribution 2 23  Department of ISM, University of Alabama, 1995-2003 Question: What do we do when we have a normal population distribution, but the mean is not “0” and/or the standard deviation is not “1”? Z = X –   Use the Universal Translator Example: Suppose X ~ N(120, 10). 11. Find P ( X > 150 ). 12. Find the quartiles of this distribution.

24 M12-Normal Distribution 2 24  Department of ISM, University of Alabama, 1995-2003 P( 0 < Z < 1.43) = ? 0Z 1.43 What proportion of Z values are between 0 and 1.43?.4236 =.4236 P(-1.43 < Z < 0) = ? 0Z -1.43 What proportion of Z values are between -1.43 and 0?.4236 =.4236 ori

25 M12-Normal Distribution 2 25  Department of ISM, University of Alabama, 1995-2003 P(Z > 1.43) = ? 0Z 1.43 What proportion of Z values are greater than 1.43?.4236 P(Z < 1.43) = ? 0Z What proportion of Z values are less than 1.43?.4236.5000.0764 =.5 -.4236 =.0764 =.5 +.4236 =.9236 1.43.5000

26 M12-Normal Distribution 2 26  Department of ISM, University of Alabama, 1995-2003 P(-1.23 < Z < 2.05) = ? 0Z 2.05 What proportion of Z values are between –1.23 and 2.05?.3907 P(1.23 < Z < 2.05) = ? 0Z What proportion of Z values are between 1.23 and 2.05?.4236.4798 =.4798 +.3907 =.8705 =.4798 -.4236 =.0562 1.23.4798 -1.23 2.05.0562

27 M12-Normal Distribution 2 27  Department of ISM, University of Alabama, 1995-2003 P( 4.28 < X < 10.0) = ? 0Z -1.43.4236 = P( -1.43 < Z < 0) Same situation. Find the proportion of packages that weigh between 4.28 and 10.0 lbs. 10X 4.28 Z = 4.28 – 10.0 4.0 = -1.43 Z = 10.0 – 10.0 4.0 = 0 X = weight of packages. X ~ N(  = 10,  = 4.0) =.4236

28 M12-Normal Distribution 2 28  Department of ISM, University of Alabama, 1995-2003 P( 13.0 < X < 17.84) = ? 0Z 1.96.4750 = P(.75 < Z < 1.96) Same situation. What proportion of the packages weigh between 13.0 and 17.84 lbs? 10X 17.84 Z = 17.84 – 10.0 4.0 = 1.96 X = weight of packages. X ~ N(  = 10,  = 4.0) =.4750 -.2734 =.2016.2734.75 13 Z = 13.0 – 10.0 4.0 =.75 ?

29 M12-Normal Distribution 2 29  Department of ISM, University of Alabama, 1995-2003 Z Same situation. Find the weight such that a. 16% weigh more less than this value. b. You have the boundaries of the middle 80%. c. The top 25% weigh more. d. You have the quartiles. X X = weight of packages. X ~ N(  = 10,  = 4.0)

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