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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

Individual

observations

Means for

n = 5

Means for

n = 20

149 153.0 151.6

146.

.

.

146.4.

.

.

151.3.

.

.

µ = 150 lbs µ = 150 lbs µ = 150 lbs

2 = 100lbs

= 10 lbs

22 220 lbsx n

2

2 25 lbsx n

4.47 lbsxn

2.23 lbsxn

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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.

21

21( )

2

x

f x e

Normal Distribution Density Function

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%

%

Pro

bab

ilit

y / %

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)

0

1

N

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Using the Normal Tables

Z

2

(0,1)

0

1

N

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

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Using the Normal Tables

Z

2

(0,1)

0

1

N

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

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Using the Normal Tables

Z

2

(0,1)

0

1

N

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

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Using the Normal Tables

2

(0,1)

0

1

N

(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)

0

1

N

(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)

0

1

N

(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)

0

1

N

(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)

0

1

N

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

Z

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Standard Normal Distribution

2

(0,1)

0

1

N

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

Z

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Calculating z-values

and ~ (0,1)Z N

If ~ ( , )x xX N

then the corresponding z value for x is given as

x

x

xz

i.e. µz = 0 and z2 = 1

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

150 150when = 150; 0

10170 150 20

when = 170; 210 10

xx x

x

xz X i e

x z

x z

Calculating z-values

2

~ (0,1)

0

1

z

z

Z N

Z

~ ( , )

150

10

x x

x

x

X N

110 120 130 140 150 160 170 180 190 2x x 2x x 1x x 3x x 1x x x 3x x

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The following questions reference a normal distribution with a mean = 150 lbs, a variance 2 = 100 lbs2, and a standard deviation = 10 lbs. Such a distribution is often indicated by the symbols N(,) = N(150, 10).

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

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

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

Some Questions

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Solution to Question 1

0.38292

X

155 1500.5

10Upper x

Upperx

xz

Area between z upper and z lower = 0.38292

145 150 0.5

10Lower x

Lowerx

xz

, Area up to z upper = 0.69146

, Area up to z lower = 0.30854

~ (150,10)

1

150

10

x

x

X N

n

lbs

lbs

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Solution to Question 2

Area between z upper and z lower = 0.73728

0.73728

155 1501.12

4.47Upper x

Upperx

xz

145 1501.12

4.47Lower x

Lowerx

xz

, Area up to z upper = 0.86864

, Area up to z lower = 0.13136

~ ( , )

5

150

104.47

5

x x

x

x

X N

n

n

X

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Solution to Question 3

0.97490

155 1502.24

2.23Upper x

Upperx

xz

145 1502.24

2.23Lower x

Lowerx

xz

, 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 N

n

lbs

lbsn

X

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1150 10

nlbslbs

0.38292

5150

= 4.47 x

nlbs

lbsn

20150

= 2.23 x

nlbs

lbsn

0.73728

0.97490

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• When centered about = 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 - (??) standard deviations to + (??) standard deviations covers (??) % of the distribution?

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

Some More Questions