M11-Normal Distribution 1 1 Department of ISM, University of Alabama, 1995-2003 Lesson Objective ...

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 randomlyselected item will fall in this region.

Given a probability, find the region for a normally distributed variable that corresponds to this probability.


The The NormalNormal DistributionsDistributions

a.k.a., “a.k.a., “The The Bell Shaped CurveBell Shaped Curve””

Describes the shape for some quantitative, continuous random variables.


= mean determines the location.

= standard deviation determines spread, variation, scatter.

Normal Population Distribution Normal Population Distribution has two has two parametersparameters::


X ~ N( = 66, = 9) or N(66, 9)

Z = the number of standard deviations that an X - value is from the mean.

Notation:Notation:

X - Z =

Z ~ N( = 0, = 1 ) or N(0,1)

Z follows the “Standard Normal Distribution”


-4 -3 -2 -1 0 1 2 3 4

____, ±1

______, ±2

______, ±3

Empirical Rule of the Normal Distribution

Where does this

come from?


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


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

Step 1 for all problems:

DTDPDTDP


0 1.72

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

P(0 < z < ?) =

FindP(0 < z < 1.72) =

Up to the 1st decimal placeUp to the 1st decimal place

2nd decimal place2nd decimal place

.4573.4573

Standard Standard Normal Normal

TableTable


TableTable

M12 Normal Distribution 2 9 Department of ISM, University of Alabama, 1995-2003

P(-1.23 < Z < 2.05) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z2.05

What proportion of Z values are between –1.23 and +2.05?

P(1.23 < Z < 2.05) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z

What proportion of Z values are between +1.23 and +2.05?

= =

=

= 1.23

-1.23

2.05

?


P( 10.0 < X < 15.72) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

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.

10 X15.72

Z = 15.72 – 10.0 4.0

=

Z = 10.0 – 10.0 4.0

=

X = weight of packages.

X ~ N( = 10, = 4.0)

=


P( X > 15.72) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.43

P( Z > 1.43)

Same situation.

What proportion of packages weigh more than 15.72 lbs?

10 X15.72

Z = 15.72 – 10.0 4.0

= 1.43


X ~ N( = 10, = 4.0)

==

?


P( X < 14.2) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.05

.3531

P( Z < 1.05)

Same situation.

What proportion of packages weigh less than 14.2 lbs?

10 X14.2

Z = 14.2 – 10.0 4.0

= 1.05


X ~ N( = 10, = 4.0)

= .5 + .3531= .8531

.5000


-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

Same situation.

What proportion of packages weigh between 5.08 and 18.2 lbs?


X ~ N( = 10, = 4.0)


Same situation.

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


X ~ N( = 10, = 4.0)

Homework


P( X < ?) = .10

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z-1.28

.10

P( Z < )

Same situation.

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

10 X?


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.


0


P(0 < z < ?) =


TableTable


TableTable

FindP( __ < z < 0) =.40

.4000.4000

-1.28

Find the Z value to cut off the top 10%.


0


P(0 < z < ?) =


TableTable


TableTable

.25.25

?

Find the Z values that define the middle 50%.

.25.25

?



P(0 < z < ?) =


TableTable


TableTable

Find the Z values that define the middle 95%.


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 ).


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.


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 followingpractice problemsto pass this course.


1. .02742. .99483. 1.0 – .7881 = .21194. .05. .9934 – .6255 = .06896. .9599 – .0401 = .91987. 1.00008. .00009. .039810. -1.175

Practice problem answers


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.


P( 0 < Z < 1.43) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.43

What proportion of Z values are between 0 and 1.43?

.4236

= .4236

P(-1.43 < Z < 0) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z-1.43

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

.4236

= .4236

ori


P(Z > 1.43) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.43

What proportion of Z values are greater than 1.43?

.4236

P(Z < 1.43) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z

What proportion of Z values are less than 1.43?

.4236

.5000

.0764

= .5 - .4236 = .0764

= .5 + .4236 = .9236 1.43

.5000


P(-1.23 < Z < 2.05) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z2.05

What proportion of Z values are between –1.23 and 2.05?

.3907

P(1.23 < Z < 2.05) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z

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


P( 4.28 < X < 10.0) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z-1.43

.4236

= P( -1.43 < Z < 0)

Same situation.

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

10 X4.28

Z = 4.28 – 10.0 4.0

= -1.43

Z = 10.0 – 10.0 4.0

= 0


X ~ N( = 10, = 4.0)

= .4236


P( 13.0 < X < 17.84) = ?

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.96

.4750

= P( .75 < Z < 1.96)

Same situation.

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

10 X17.84

Z = 17.84 – 10.0 4.0

= 1.96


X ~ N( = 10, = 4.0)

= .4750 - .2734= .2016

.2734

.7513

Z = 13.0 – 10.0 4.0

= .75

?


-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

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 ~ N( = 10, = 4.0)

M11-Normal Distribution 1 1 Department of ISM, University of Alabama, 1995-2003 Lesson Objective ...

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