Normal Distributions

1

1

The Normal Distribution

2

Normal Probability

Density Function

- < x <

Notation: N (, 2) A normal distribution with

mean and standard deviation

2

2

1

2

1)(

-

-

x

exf

The continuous random variable

X has a normal distribution if

its p.d.f. is

3

Normal Distribution

The mean, variance, and m.g.f. of a

continuous random variable X that has a

normal distribution are:

2][

][

XVar

XE

4

Normal Distribution

1. “Bell-Shaped” &

Symmetrical

X

f(X)

Mean

Median

Mode

2. Mean, Median,

Mode Are Equal

3. Random Variable

Has Infinite Range

- < x <

5

Example

N (72, 5) A normal distribution with mean 72

and variance 5.

Possible situations: Test scores, pulse rates, …

6

Effect of Varying Parameters ( & )

x

f(x)

A C

B

Normal Distributions

2

7

Normal Distribution

Probability

dxxfdxcPd

c )()(

c dx

f(x)

Probability is

area under

curve!

8

X

f(X)

Infinite Number

of Tables

Normal distributions differ by

mean & standard deviation.

Each distribution would

require its own table.

That’s an infinite number!

9

Standard Normal Distribution

Standard Normal Distribution:

A normal distribution with

mean = 0 and standard deviation = 1.

Notation:

Z ~ N ( = 0, 2 = 1)

0 Z

= 1

Cap letter Z 10

Area under Standard Normal Curve

How to find the proportion of the are under the standard normal curve below z or say P ( Z < z ) = ?

Use Standard Normal Table!!!

0 z

Z

11 12

Standard Normal Distribution

P(Z < 0.32) = F(0.32) Area below .32 = ?

0 .32

0.6255

Normal Distributions

3

13

Standard Normal Distribution

P(Z > 0.32) = Area above .32

0 .32

Areas in the upper tail of the standard normal distribution

= 1 - .6255 = .3745

14

Standard Normal Distribution

P(0 < Z < 0.32) = Area between 0 and .32 = ?

0 .32

= 0.6255 – 0.5 = 0.1255

15

-1.00 0 1.00

P ( -1.00 < Z < 1.00 ) = __?___

.3413 .3413

.6826

0.8413 - 0.5 16

-2.00 0 2.00

P ( -2.00 < Z < 2.00 ) = _____

.4772 .4772

.9544 = .4772 + .4772

.9544

17

-3.00 0 3.00

P ( -3.00 < Z < 3.00 ) = _____

… .4987 .4987

.9974

18

- 1.40 0 2.33

P ( -1.40 < Z < 2.33 ) = ____

.4901 .4192

.9093

.0099 .0808

Normal Distributions

4

19

Normal Distribution

= 0

= 1

= 50

= 3

= 70

= 9

…

Standard Normal

20

Standardize the

Normal Distribution

X

Normal

Distribution

ZX

-

One table!

= 0

= 1

Z

Standardized

Normal Distribution

21

Theorem

If X is N(, 2), then Z = is N(0,1). (X – )

-F-

-F

-

-

ab

bZ

aPbXaP )(

22

N ( 0 , 1)

0

-

-

bZ

aPbXaP )(

Standardize the

Normal Distribution

a b

N ( , )

Normal

Distribution

X

Standard

Normal

Distribution

Z

-a

-b

23

Example 1

Normal

Distribution

For a normal distribution that has

a mean = 5 and s.d. = 10, what

percentage of the distribution is

between 5 and 6.2?

X = 5

= 10

6.2 24

Example 1

X= 5

= 10

6.2

Normal

Distribution

12.10

52.6

-

-

XZ

Z= 0

= 1

.12

Standardized

Normal Distribution P(5 X 6.2)

P(0 Z .12)

010

55

-

-

XZ

Normal Distributions

5

25

Z= 0

= 1

.12

Example 1

Standardized Normal

Distribution Table

Area = .5478 - .5 = .0478 26

Example 1

X= 5

= 10

6.2

Normal

Distribution

12.10

52.6

-

-

XZ

Z= 0

= 1

.12

Standardized

Normal Distribution P(5 X 6.2)

P(0 Z .12)

0.0478 4.78%

010

55

-

-

XZ

27

Example

P(3.8 X 5)

X = 5

= 10

3.8

Normal

Distribution

12.10

58.3-

-

-

XZ

Z = 0

= 1

-.12

Standardized

Normal Distribution

.0478

Area = .0478

P(3.8 X 5) =P(-.12 Z 0)

.0478

28

Example

P(2.9 X 7.1)

5

= 10

2.9 7.1 X

Normal

Distribution

0

= 1

-.21 Z.21

Standardized

Normal Distribution

.1664

Area = .0832 + .0832 = .1664

P(2.9 X 7.1) =P(-.21 Z .21)

.1664

29

Example

P(X > 8)

X = 5

= 10

8

Normal

Distribution

Standardized

Normal Distribution

30.10

58

-

-

XZ

Z = 0

= 1

.30

.3821

Area = .3821

P(X > 8) =P(Z > .30)

=.3821

30

X = 5

= 10

8

Normal

Distribution

62% 38%

Value 8 is the 62nd percentile

Example

P(X > 8)

Normal Distributions

6

31

More on Normal Distribution

The work hours per week for residents in Ohio has a

normal distribution with = 42 hours & = 9 hours.

Find the percentage of Ohio residents whose work hours

are

A. between 42 & 60 hours.

P(42 X 60) =?

B. less than 20 hours.

P(X 20) = ?

32

P(42 X 60) = ?

Normal

Distribution

Standardized

Normal Distribution

X

= 200

2400 Z 0

= 1

2.0

9

42 60 2

P(42 Z 60)

P(0 Z 2)

.4772(47.72%) .4772

33

P(X 20) = ?

Normal

Distribution

Standardized

Normal Distribution

X

= 200

2400 Z 0

= 1

2.0

9

20 42 -2.44

P(X 20) = P(Z -2.44)

= 0.0073 = 0.73%

0.0073

34

Finding Z Values

for Known Probabilities

Standardized Normal

Distribution Table

What is z given

P(Z < z) = .80 ?

0

.80 .20

z = .84

z.20 = .84

Def. za : P(Z ≥ za) = a ; P(Z < za) = 1 – a

35

Finding X Values

for Known Probabilities

Example: The weight of new born

infants is normally distributed with a

mean 7 lb and standard deviation of

1.2 lb. Find the 80th percentile.

Area to the left of 80th percentile is 0.800.

In the table, there is a area value 0.7995

(close to 0.800) corresponding to a z-score

of .84.

80th percentile = 7 + .84 x 1.2 = 8.008 lb 36

Finding X Values

for Known Probabilities

Normal Distribution Standardized Normal Distribution

.80 .80

0 7

( )( ) 008.82.184.7 ZX

X = 8.008 z = .84

80th percentile

Normal Distributions

7

37

Stanine Score

1 2 3 4 5 6 7 8 9

-1.75 -1.25 -.75 -.25 0 .25 .75 1.25 1.75

4% 4% ? %

38

More Examples

• The pulse rates for a certain population follow a

normal distribution with a mean of 70 per minute

and s.d. 5. What percent of this distribution that is

in between 60 to 80 per minute?

• The weights of a population follow a normal

distribution with a mean 130 and s.d. 10. What

percent of this population that is in between 110

and 150 lbs?

39

The Normal Distribution

Approximation for Discrete Distributions

(Normal approximation to Binomial)

40

Normal Approximation for

Binomial Probability

0 1 2 3 4 5 6 7

= np = 3.5

2= np(1-p) = 1.75

?)3( bXP

41

)5.2()3( > Nb XPXP

0 1 2 3 4 5 6 7

= np = 3.5

2 = np(1-p) = 1.75

7764.

)76.(

)75.1

5.35.2(

->

->

ZP

ZP

Binomial Table:

P(Xb ≥ 3) = .7734

Normal Approximation for

Binomial Probability