V6: Normal Dist. 1. Edd Sum `95 V06_nor

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.

 Department of ISM, University of Alabama, 1995-2003

M11-Normal Distribution 1

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The Normal Distributions

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

M12-Normal Distribution 2

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Normal Population Distribution has two parameters:

 

= mean determines the location.

= standard deviation determines spread, variation, scatter.

 Department of ISM, University of Alabama, 1995-2003

M12-Normal Distribution 2

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Notation: X ~ N(  = 66,  = 9) or N(66, 9) Z = the number of standard deviations that an X - value is from the mean.

Z =

 

X-

Z ~ N(  = 0,  = 1 ) or N(0,1) Z follows the “Standard Normal Distribution”  Department of ISM, University of Alabama, 1995-2003

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Empirical Rule of the Normal Distribution ______, ±3 ______, ±2 ____, ±1

-4

-3

-2

-1

0

1

 Department of ISM, University of Alabama, 1995-2003

2

3

4

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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  Department of ISM, University of Alabama, 1995-2003

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Reading the Standard Normal Table (finding areas under the normal curve)

Step 1 for all problems:

 Department of ISM, University of Alabama, 1995-2003

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Standard Normal Table 2nd decimal place

Table gives P(0 < z < ?) = Find P(0 < z < 1.72) =

.4573

Up to the 1st decimal place -4.0

 Department of ISM, University of Alabama, 1995-2003

-3.0

-2.0

-1.0

0

0.0

1.72

1.0

M12-Normal Distribution 2

2.0

8

3.0

What proportion of Z values are between –1.23 and +2.05? P(-1.23 < Z < 2.05) = ? = = -4.0

-3.0

What proportion of Z values are between +1.23 and +2.05? P(1.23 < Z < 2.05) = ? = = -4.0

-3.0

 Department of ISM, University of Alabama, 1995-2003

-1.23

-2.0

-1.0

0

0.0

1.0

2.05 2.0

3.0

Z

4.0

3.0

Z

4.0

?

-2.0

-1.0

0 1.23 2.05 0.0

1.0

2.0

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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.

X = weight of packages. X ~ N( = 10,  = 4.0)

P( 10.0 < X < 15.72) = ? P( 0 < Z < 1.43) =

-4.0

-3.0

-2.0

-1.0

10 15.72 0.0

1.0

2.0

3.0

X Z

4.0

15.72 – 10.0 Z= = 4.0 10.0 – 10.0 Z= = 4.0

 Department of ISM, University of Alabama, 1995-2003

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Same situation. What proportion of packages weigh more than 15.72 lbs?

X = weight of packages. X ~ N( = 10,  = 4.0)

P( X > 15.72) = ? P( Z > 1.43) = =

?

-4.0

-3.0

-2.0

-1.0

10 0 0.0

15.72 1.43 1.0

2.0

3.0

X Z

4.0

15.72 – 10.0 Z= = 1.43 4.0

 Department of ISM, University of Alabama, 1995-2003

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.5000

Same situation. What proportion of packages weigh less than 14.2 lbs?

X = weight of packages. X ~ N( = 10,  = 4.0)

P( X < 14.2) = ? P( Z < 1.05) = .5 + .3531 = .8531

.3531

-4.0

-3.0

-2.0

-1.0

10 14.2 0 1.05 0.0

1.0

2.0

3.0

X Z

4.0

14.2 – 10.0 Z= = 1.05 4.0

 Department of ISM, University of Alabama, 1995-2003

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Same situation. What proportion of packages weigh between 5.08 and 18.2 lbs?

X = weight of packages. X ~ N( = 10,  = 4.0)

-4.0

-3.0

 Department of ISM, University of Alabama, 1995-2003

-2.0

-1.0

0.0

1.0

2.0

3.0

M12-Normal Distribution 2

4.0

13

Same situation. What proportion of packages weigh either less than 2.4 lbs or greater than 11.0 lbs?

Homework

X = weight of packages. X ~ N( = 10,  = 4.0)

 Department of ISM, University of Alabama, 1995-2003

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Same situation. Find the weight such that 10% of all packages weigh less than this weight.

.40

.10

This is a backwards problem! We are given the probability; we need to find the boundary.

X = weight of packages. X ~ N( = 10,  = 4.0) P( X < ?) = .10 P( Z < -1.28 ) = .10

-4.0

-3.0

? -1.28 -2.0

-1.0

10 0 0.0

1.0

2.0

3.0

X Z

4.0

? – 10 –1.28 = 4 ? = 10 – 1.28 • 4 = 10 – 5.12 = 4.88 pounds

10% weigh less than 4.88 pounds; 90% weigh moreM12 than 4.88 pounds. Normal Distribution 2  Department of ISM, University of Alabama, 1995-2003

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Standard Normal Table Table gives P(0 < z < ?) = Find P( __ < z < 0) =.40

Find the Z value to cut off the top

.4000

10%. -4.0

 Department of ISM, University of Alabama, 1995-2003

-3.0

-1.28

-2.0

-1.0

0

0.0

1.0

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2.0

16

3.0

Standard Normal Table Table gives P(0 < z < ?) =

.25 .25

Find the Z values that define the middle 50%. -4.0

 Department of ISM, University of Alabama, 1995-2003

-3.0

-2.0

?

-1.0

0

0.0

1.0

M12 Normal Distribution 2

?

2.0

17

3.0

Standard Normal Table Table gives P(0 < z < ?) =

Find the Z values that define the middle 95%. -4.0

 Department of ISM, University of Alabama, 1995-2003

-3.0

-2.0

-1.0

0.0

1.0

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2.0

18

3.0

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 ).  Department of ISM, University of Alabama, 1995-2003

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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.  Department of ISM, University of Alabama, 1995-2003

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Practice problems. You MUST know 1. P( Z < –1.92) = how to work ALL 2. P( Z < 2.56) = of these problems and the following 3. P( Z > 0.80) = practice problems 4. P( Z = 1.42) = to pass this course. 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.  Department of ISM, University of Alabama, 1995-2003

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Practice problem answers

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  Department of ISM, University of Alabama, 1995-2003

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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”? Use the Universal Translator

Z=

X–



Example: Suppose X ~ N(120, 10). 11. Find P ( X > 150 ). 12. Find the quartiles of this distribution.  Department of ISM, University of Alabama, 1995-2003

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What proportion of Z values are between 0 and 1.43?

ori

.4236

P( 0 < Z < 1.43) = ? = .4236 -4.0

-3.0

-2.0

What proportion of Z values are between -1.43 and 0?

-1.0

0

0.0

1.43 1.0

2.0

3.0

Z

4.0

0

1.0

2.0

3.0

Z

4.0

.4236

P(-1.43 < Z < 0) = ?

= .4236 -4.0

-3.0

 Department of ISM, University of Alabama, 1995-2003

-1.43 -2.0

-1.0

0.0

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What proportion of Z values are greater than 1.43? P(Z > 1.43) = ? = .5 - .4236 = .0764 -4.0

.5000 .4236

-3.0

-2.0

-1.0

0

0.0

1.43 1.0

.0764

2.0

3.0

Z

4.0

2.0

3.0

Z

4.0

.5000 .4236

What proportion of Z values are less than 1.43? P(Z < 1.43) = ? = .5 + .4236 = .9236

-4.0

-3.0

 Department of ISM, University of Alabama, 1995-2003

-2.0

-1.0

0

0.0

1.43 1.0

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What proportion of Z values are between –1.23 and 2.05? P(-1.23 < Z < 2.05) = ? = .4798 + .3907 = .8705 -4.0

.4798 .3907

-3.0

-1.23

-2.0

-1.0

0

0.0

-4.0

2.05 2.0

3.0

Z

4.0

3.0

Z

4.0

.4798 .4236

What proportion of Z values are between 1.23 and 2.05? P(1.23 < Z < 2.05) = ? = .4798 - .4236 = .0562

1.0

.0562

-3.0

 Department of ISM, University of Alabama, 1995-2003

-2.0

-1.0

0 1.23 2.05 0.0

1.0

2.0

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Same situation. Find the proportion of packages that weigh between 4.28 and 10.0 lbs.

X = weight of packages. X ~ N( = 10,  = 4.0)

P( 4.28 < X < 10.0) = ? = P( -1.43 < Z < 0) = .4236

-4.0

.4236

-3.0

4.28 10 -1.43 0

-2.0

-1.0

0.0

1.0

2.0

3.0

X Z

4.0

4.28 – 10.0 Z= = -1.43 4.0 10.0 – 10.0 Z= =0 4.0

 Department of ISM, University of Alabama, 1995-2003

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.4750 .2734 ?

Same situation. What proportion of the packages weigh between 13.0 and 17.84 lbs?

X = weight of packages. X ~ N( = 10,  = 4.0)

P( 13.0 < X < 17.84) = ? = P( .75 < Z < 1.96) = .4750 - .2734 = .2016

-4.0

-3.0

-2.0

-1.0

10 13 17.84 0 .75 1.96 0.0

1.0

2.0

3.0

X Z

4.0

17.84 – 10.0 Z= = 1.96 4.0 13.0 – 10.0 Z= = .75 4.0

 Department of ISM, University of Alabama, 1995-2003

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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.

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

X Z

4.0

X = weight of packages. X ~ N( = 10,  = 4.0)

 Department of ISM, University of Alabama, 1995-2003

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