1

This is the most important and most

widely used probability distribution

http://www.youtube.com/watch?v=NYd6wz

YkQIM&feature=relmfu

2

The Normal Distribution

• Properties

– Bell shaped

– Area under curve equals 1

– Symmetric around the mean

– Mean = median = Mode

– Two tails approach the horizontal axis – never

touch axis

– Empirical rule applies

– Two parameters – μ and σ

1

μ

3

How does the standard deviation affect the shape of the distribution

s = 2

s =3

s =4

m = 10 m = 11 m = 12

How does the mean affect the location of the distribution m = 11

s = 2

4

Mathematical model expressed as:

21

21( ) ,

2

3.14159 2.71828

x

f x e x

where and e

m

s

s

The Normal Distribution

- notation N(μ ; σ2)

STANDARDISING THE

RANDOM VARIABLE • Seen how different means and std dev’s

generate different normal distributions

• This means a very large number of

probability tables would be needed to

provide all possible probabilities

• We therefore standardise the random

variable x so that only one set of tables is

needed

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STANDARDISING THE

RANDOM VARIABLE • A normal random variable x can be

converted to a standard normal variable

(denoted Z) by using the following

standardisation formula:-

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, any value of the random variable x

z x Xm

s

7

• Different values of μ and σ generate

different normal distributions

• The random variable X can be

standardised

– mean = μ = 0

– standard deviation = σ = 1

The Standard Normal Distribution

, any value of the random variable x

z x Xm

s

8

z-values on the horizontal axis

• distance between the mean and the point

represented by z in terms of standard deviation

-3 -2 -1 0 1 2 3

z

μ = 0

s = 1

The Standard Normal Distribution

9

• Example

– Marks for a semester test is normally distributed, with a mean of 60 and a standard deviation of 8

– X ~ N(60 , 82)

Finding Normal Probabilities

50 60 65 x

– If we need to determine the

probability that the mark will

be between 50 and 65,

we need to determine the size of the shaded area

– Before calculating the probabilities the x-values need to be transformed to z-values

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The tabulated probabilities

correspond to the area between

Z = -∞ and some z > 0 0 z

Standard normal probabilities have been

calculated and are provided in a table

z 0.00 0.01 → 0.05 0.06 → 0.09

0.0 0.5000 0.5040 0.5199 0.5239 0.5359

0.1 0.5398 0.5438 0.5596 0.5636 0.5753

↓

1.0 0.8413 0.8438 0.8531 0.8554 0.8621

1.1 0.8643 0.8665 0.8749 0.8770 0.8830

1.2 0.8849 0.8869 0.8944 0.8962 0.9015

↓

P(-∞ < Z < z)

11

• Example continue

– If X denotes the test mark, we seek the

probability

– P(50 < X < 65)

– Transform the X to the standard normal

variable Z

XZ

m

s

E(Z)

μ = 0

V(Z)

σ2 = 1

Every normal variable

with some m and s,

can be transformed

into this Z

Therefore, once

probabilities for Z are

calculated, probabilities

of any normal variable

can be found

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P(50 < X < 65) = P( < Z < ) X - m

s

X - m

s 8

- 60 65 - 60

• Example continue

50

8

= P(-1.25 < Z < 0.63)

To complete the calculation we need to compute

the probability under the standard normal distribution

Mean = μ = 60 minutes

Standard deviation = σ = 8 minutes

X - Z =

m

s

13

• How to use the z-table to calculate

probabilities Example

Determine the following probability: P(Z > 1.05) = ?

1.05

1 - P(Z < 1.05)

= P(Z > 1.05)

0

P(Z > 1.05) = 1 – 0.8531 = 0.1469

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-2.12 1.32 0 -2.12

P(-2.12 < Z < 1.32) = (0.9066 + 0.9830) - 1 = 0.8896

P(-2.12 < Z < 1.32) = ?

P(-∞ < Z < 1.32) = 0.9066

1.32

P(-2.12 < Z < +∞) = 0.9830

0,5

0,5

+

0.9066

0.9830

• How to use the z-table to calculate probabilities

Example

Determine the following probability:

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0.73 1.40 0

P(0.73 < Z < 1.40) =

• Example

– Determine the following probabilities:

P(0.73 < Z < 1.40) = ?

P(-∞ < Z < 0,65) = 0,7422

0.73

1.40

P(-∞ < Z < 0.73) = 0.7673

P(-∞ < Z < 1.40) = 0.9192

0.9192

0.7673

0.9192 – 0.7673 = 0.1519

16 P(-z < Z < +∞) P(-∞ < Z < z)

The symmetry of the normal distribution

makes it possible to calculate probabilities

for negative values of Z using the table as

follows:

=

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• Example student marks - continued

z 0.00 0.01 → 0.05 0.06 → 0.09

0.0 0.5000 0.5040 0.5199 0.5239 0.5359

0.1 0.5398 0.5438 0.5596 0.5636 0.5753

↓

1.0 0.8413 0.8438 0.8531 0.8554 0.8621

1.1 0.8643 0.8665 0.8749 0.8770 0.8830

1.2 0.8849 0.8869 0.8944 0.8962 0.9015

↓

In this example z = -1.25 , because of symmetry read 1.25

0.8944 0.8944 0.8944 0.8944 0.8944

= 0.8944 + 0.7357 – 1

= 0.6301

0.8944 P(50 < X < 65) = P(-1.25 < Z < 0.63)

z = 0.63

0.7357