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Module 13: Normal Distributions

This module focuses on the normal distribution and how to use it.

Reviewed 05 May 05/ MODULE 13

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Sampling Distributions

σ = 10 lbs

σ2 = 100lbs

µ = 150 lbsµ = 150 lbsµ = 150 lbs

151.3...

146.4...

146...

151.6153.0149

Means for n = 20

Means forn = 5

Individual observations

22 220 lbsx n

σσ = =2

2 25 lb sx nσσ = =

4.47 lbsx nσσ = = 2.23 lbsx n

σσ = =

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The normal distribution is defined by the density function:

This function happens to be Symmetrical, Bell-shaped,

and easy to use tables are available.

2121( )

2

x

f x eμ

σ

σ π

−⎛ ⎞− ⎜ ⎟⎝ ⎠=

Normal Distribution Density Function

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%

%

Prob

abili

ty /

%

Normal Distribution

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Population Distributions

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Population Distributions

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We can use the normal tables to obtain probabilities for measurements for which this frequency distribution is appropriate. For a reasonably complete set of probabilities, see TABLE MODULE 1: NORMAL TABLE.

This module provides most of the z-values and associated probabilities you are likely to use; however, it also provides instructions demonstrating how to calculate those not included directly in the table.

Using the Normal Tables

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Normal Tables (contd.)

The table is a series of columns containing numbers for z and for P(z). The z represents the z-value for a normal distribution and P(z) represents the area under the normal curve to the left of that z-value for a normal distribution with mean µ = 0 and standard deviation σ = 1.

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Using the Normal Tables

Z

(1) Area Below z = -2; P(z < -2) = 0.0228

2

(0,1)01

Nμ

σ

=

=

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Using the Normal Tables

Z

2

(0,1)01

Nμ

σ

=

=

(2) Area Below z = -1; P(z < -1) = 0.1587

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Using the Normal Tables

Z

2

(0,1)01

Nμ

σ

=

=

(1) Area Below z = +2; P(z > +2) = 0.0228

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Using the Normal Tables

Z

2

(0,1)01

Nμ

σ

=

=

(2) Area Below z = +1; P(z > +1) = 0.1587

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Using the Normal Tables

2

(0,1)01

Nμ

σ

=

=

(3) Area Below z = 0; P(z > 0) = 0.5000

Z

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Calculating the Area Under the Normal Curve

Z

2

(0,1)01

Nμ

σ

=

=

(1) Area between -1, +1; P( -1 < z < +1)up to z = +1: .8413up to z = -1 : .1587

.6826

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Calculating the Area Under the Normal Curve

Z

2

(0,1)01

Nμ

σ

=

=

(2) Area between -2, +2; P( -2 < z < +2)up to z = +2: .9772up to z = -2 : .0228

.9544

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Calculating the Area Under the Normal Curve

2

(0,1)01

Nμ

σ

=

=

(3) Area between -2, +1; P( -2 < z < +1)up to z = +1: .8413up to z = -2 : .0228

.8185

Z

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Standard Normal Distribution

2

(0,1)01

Nμ

σ

=

=

(1) Values of z that bracket middle 95% -1.96 to +1.96

Z

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Standard Normal Distribution

2

(0,1)01

Nμ

σ

=

=

(1) Values of z that bracket middle 99% -2.576 to +2.576

Z

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Calculating z-values

a n d ~ ( 0 , 1 )Z NIf ~ ( , )x xX N μ σ

then the corresponding z value for x is given as

x

x

xz μσ−

=

i.e. µz = 0 and σz2 = 1

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; if ~N( 150,10) . . 150, 10

150 150when = 150; 010

170 150 20when = 170; 210 10

xx x

x

xz X i e

x z

x z

μ μ σσ−

= = =

−= =

−= = =

Calculating z-values

2

~ (0,1) 0

1z

z

Z Nμσ

==

Z

~ ( , ) 150

10

x x

x

x

X N μ σμσ

==

110 120 130 140 150 160 170 180 190 2x xμ σ− 2x xμ σ+ 1x xμ σ− 3x xμ σ− 1x xμ σ+xμ 3x xμ σ+

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The following questions reference a normal distribution with a mean μ = 150 lbs, a variance σ2 = 100 lbs2, and a standard deviation σ = 10 lbs. Such a distribution is often indicated by the symbols N(μ,σ) = N(150, 10).

1. What is the likelihood that a randomly selected individual observation is within 5 lbs of the population mean μ = 150lbs?

2. What is the likelihood that a mean from a random sample of size n = 5 is within 5 lbs of μ = 150 lbs?

3. What is the likelihood that a mean from a random sample of size n = 20 is within 5 lbs of μ = 150 lbs?

Some Questions

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Solution to Question 1

0.38292

X

155 150 0.510

Upper xUpper

x

xz

μσ− −

= = =

Area between z upper and z lower = 0.38292

145 150 0.510

Lower xLower

x

xz μσ− −

= = =−

, Area up to z upper = 0.69146

, Area up to z lower = 0.30854

~ (150,10)1

150 10

x

x

X Nn

lbslbs

μσ

===

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Solution to Question 2

Area between z upper and z lower = 0.73728

0.73728

1 5 5 1 5 0 1 .1 24 .4 7

U p p e r xU p p e r

x

xz

μσ

− −= = =

1 4 5 1 5 0 1 .1 24 .4 7

L ow er xL ow er

x

xz μσ− −

= = = −

, Area up to z upper = 0.86864

, Area up to z lower = 0.13136

~ ( , )5

15010 4.47

5

x x

x

x

X Nn

n

μ σ

μσσ

==

= = =

X

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Solution to Question 3

0.97490

155 150 2 .242 .23

U pper xU pper

x

xz

μσ− −

= = =

1 4 5 1 5 0 2 .2 42 .2 3

L o w er xL o w er

x

xz μσ

− −= = = −

, Area up to z upper = 0.98745

, Area up to z lower = 0.01255

Area between z upper and z lower = 0.97490

~ ( , )20

150

2.23

x x

x

x

X Nn

lbs

lbsn

μ σ

μσσ

==

= =

X

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1150 10

nlbs

lbsμσ

===

0.38292

5150

= 4.47 x

nlbs

lbsn

μσσ

==

=

20150

= 2.23 x

nlbs

lbsn

μσσ

==

=

0.73728

0.97490

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• When centered about μ = 150 lbs, what proportion of the total distribution does an interval of length 10 lbs cover?

• How many standard deviations long must an interval be to cover the middle 95% of the distribution?

• From μ - (??) standard deviations to μ + (??) standard deviations covers (??) % of the distribution?

All these questions require that the value for μ be known and that it be placed in the center of these “intervals”.

Some More Questions