Density Curve Continuous Distributions -...

Normal Distribution

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Continuous Distributions

Random Variables of the Continuous Type

2

Density Curve

Percent

A smooth curve that fit

the distribution

10 20 30 40 50 60 70 80 90 100 110

Test scores

Density

function, f (x)

f(x)

3

Density Curve

A smooth curve that fit

the distribution

Percent

Use a mathematical model to describe the variable.

10 20 30 40 50 60 70 80 90 100 110

Test scores Shows the

probability

density for all

values of x.

Probability density is not probability!

Probability

Density

Function, f (x)

f(x)

4

Continuous Distribution

1. f (x) 0, x S,

2. S f(x) dx = 1, (Total area under curve is 1.)

3. For a and b in S,

Probability Density Function (p.d.f.) of a random

variable X of continuous type with a space S is an

integrable function, f(x), that satisfying the following

conditions:

b

adxxfbXaP )()(

f(x)

x a b

5

Meaning of Area Under Curve

Example: What percentage of the

distribution is in between 72 and 86?

f(x)

X (Height) 72 86

P(X = 72) =

P(72 X 86)

0, density is not probability.

6

The Normal Distribution

1 The Normal Distribution

Normal Distribution

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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

8

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

continuous random variable X that has a

normal distribution are:

2/

2

22

)(

,][ ,][

ttetM

XVarXE

9

m.g.f. of Normal Distribution

Example: If the m.g.f. of a random variable X is

M(t) = exp(2t + 16t 2), what is the distribution of

this random variable? What are the mean and

standard deviation of this distribution and what

is the p.d.f. of this random variable?

Distribution ?

Mean ?

Standard deviation ?

p.d.f. ? 10

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 <

11

Example

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

and variance 5.

Possible situations: Test scores, pulse rates, …

N (130, 24) A normal distribution with mean

130 and variance 24.

Possible situations: Weight, Cholesterol levels, …

12

Effect of Varying Parameters ( & )

x

f(x)

A C

B

Normal Distribution

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Normal Distribution

Probability

dxxfdxcPd

c )()(

c dx

f(x)

Probability is

area under

curve!

14

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!

15

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 16

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

17 18


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

0 .32

0.6255

Normal Distribution

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P(Z > 0.32) = Area above .32

0 .32

Areas in the upper tail of the standard normal distribution

= 1 - .6255 = .3745

20


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

0 .32

= 0.6255 – 0.5 = 0.1255

21

-1.00 0 1.00

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

.3413 .3413

.6826

0.8413 - 0.5

22

-2.00 0 2.00

P ( -2.00 < Z < 2.00 ) = _____

.4772 .4772

.9544 = .4772 + .4772

.9544

23

-3.00 0 3.00

P ( -3.00 < Z < 3.00 ) = _____

… .4987 .4987

.9974

24

- 1.40 0 2.33

P ( -1.40 < Z < 2.33 ) = ____

.4901 .4192

.9093

.0099 .0808

Normal Distribution

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Normal Distribution

= 0

= 1

= 50

= 3

= 70

= 9

…

Standard Normal

26

Standardize the

Normal Distribution

X

Normal

Distribution

ZX

-

One table!

= 0

= 1

Z

Standardized

Normal Distribution

27

Theorem

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

-F-

-F

-

-

ab

bZ

aPbXaP )(

28

N ( 0 , 1)

0

-

-

bZ

aPbXaP )(

Standardize the Normal Distribution

a b

N ( , )

Normal

Distribution

X

Standard

Normal

Distribution

Z

-a

-b

29

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

30

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 Distribution

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Z= 0

= 1

.12

Example 1

Standardized Normal

Distribution Table

Area = .5478 - .5 = .0478 32

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

33

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

34

Example

P(2.9 X 7.1)

5

= 10

2.9 7.1 X

Normal

Distribution

21.10

51.7

21.10

59.2

-

-

--

-

XZ

XZ

0

= 1

-.21 Z.21

Standardized

Normal Distribution

.1664

Area = .0832 + .0832 = .1664

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

.1664

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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

36

X = 5

= 10

8

Normal

Distribution

62% 38%

Value 8 is the 62nd percentile

Example

P(X > 8)

Normal Distribution

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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) = ?

38

P(42 X 60) = ?

Normal

Distribution

29

4260

-

-

XZ

Standardized

Normal Distribution

09

4242

-

-

XZ

X

= 200

2400 Z 0

= 1

2.0

9

42 60 2

P(42 Z 60)

P(0 Z 2)

.4772(47.72%) .4772

39

P(X 20) = ?

Normal

Distribution

Standardized

Normal Distribution

44.29

4220-

-

-

XZ

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

40

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

41

Finding X Values


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 42

Finding X Values


Normal Distribution Standardized Normal Distribution

.80 .80

0 7

( )( ) 008.82.184.7 ZX

X = 8.008 z = .84

80th percentile

Normal Distribution

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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% ? %

44

Theorem

If the random variable X is N(, 2), 2 > 0.

then the random variable

V = = Z 2 is c2(1). (X – )2

2

45

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?

46


2 Random Functions Associated

with Normal Distributions

47

Theorem (M.G.F.)

If X1, X2, …, Xn are independent random

variables with m.g.f.’s , i = 1, 2,…, n,

then the m.g.f. of is

)(tMiX

n

i

iXY taMtMi

1

)()(

n

i ii XaY1

48

Theorem (Linear Function)

If X1, X2, …, Xn are mutually independent

random variables from normal distributions

with means 1, 2, 3, …, n, and variances

12, 2

2, 32, …, n

2, then the linear function

has the normal distribution N(Scii , Sci2i

2).

n

i

ii XcY1

.1

if n

cXY i

Normal Distribution

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Proof:

2

1 1

2/

2

1

22

1

222

)()(

tctc

n

i

n

i

tctc

iXY

n

iii

n

i

ii

iiii

i

e

etcMtM

m.g.f. of

n

i

n

i

ii iiccN

1

22

1

,

50

Example: An elevator can take 1000 lb maximum, and

it is posted in the elevator that the maximum capacity is

5 adults. The weight distribution of adults that use this

elevator is normally distributed with mean =160 lb and

standard deviation = 20 lb.

1. If an adult person walk into this elevator, how likely

that his or her weight is over 250 lb?

2. If 5 adult persons walk into this elevator, how likely

that the total weight will be greater than 1000 lb? (The

same as “the average is over 200.”)

51

Theorem (Distribution of )

If X1, X2, …, Xn are observations of a random

sample of size n from the normal distribution

N(, 2), then the distribution of the sample

mean is N(, 2/n)

n

i

iXn

X1

1

X

nX

X

Standard Error

52

Theorem (Chi-square distribution)

If Z1, Z2, …, Zn have the standard normal

distribution N(0, 1). If these random variables

are independent then W =Z12 + Z2

2 + …+ Zn

2

has a Chi-square distribution with n degrees

of freedom, c2(n).

53

Theorem (Distribution of and S 2) X

)1( is )1( 2

2

2

--

nSn

c

X



N(, 2). The statistics, sample mean, , and

sample variance, S 2, are independent and

n

i iXn

X1

1

-

-

n

i i XXn

S1

22 )(1

154

Student’s t Distribution

Let Z be a random variable that is N(0, 1),

and U be a random variable that is c2(r), and

Z and U are independent. Then, the random

variable

has a t-distribution with degrees of freedom r.

rU

ZT

Normal Distribution

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Student’s t Distribution



N(, 2). The statistics, sample mean, , and

sample variance, 2, are independent and

has a t-distribution with d.f. n – 1.

nS

XT

-

X

t-statistic

nS

X

nSn

n

X

T

-

--

-

)1()1(

2

2

Z

U d.f. of U

57


3 The Central Limit Theorem

58

Theorem (Distribution of )


sample of size n from a distribution that has a

mean and a finite variance 2, then the

distribution of is N(0, 1), as n ,

X

n

XZ

/

-

n

nXZ

n

i i -

1

Xand the distribution of the sample mean

is N(, 2/n), as n .

59

A Random Sample from Population

Population mean = 19.9, standard deviation = 12.6

Random Sample of Size 400 from Population

110.0

100.090.0

80.070.0

60.050.0

40.030.0

20.010.0

0.0

120

100

80

60

40

20

0

Std. Dev = 12.92

Mean = 20.7

N = 400.00

60

Simulated Sampling Distribution of Means

SIZE2

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

70

60

50

40

30

20

10

0

Std. Dev = 8.88

Mean = 20.3

N = 400.00

n=2 SIZE4

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

70

60

50

40

30

20

10

0

Std. Dev = 5.40

Mean = 19.4

N = 400.00

n=4 SIZE10

77.073.0

69.065.0

61.057.0

53.049.0

45.041.0

37.033.0

29.025.0

21.017.0

13.09.0

5.01.0

100

80

60

40

20

0

Std. Dev = 4.32

Mean = 19.9

N = 400.00

n=10

SIZE25

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

200

100

0

Std. Dev = 2.23

Mean = 19.84

N = 400.00

n=25 SIZE50

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

200

100

0

Std. Dev = 1.64

Mean = 19.75

N = 400.00

n=50 SIZE100

77.00

73.00

69.00

65.00

61.00

57.00

53.00

49.00

45.00

41.00

37.00

33.00

29.00

25.00

21.00

17.00

13.009.00

5.001.00

300

200

100

0

Std. Dev = 1.20

Mean = 19.81

N = 400.00

n=100

Normal Distribution

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Probability Related to Mean

Example: Consider the distribution of serum

cholesterol levels for 40- to 70-year-old

males living in community A has a mean of

211 mg/100 ml, and the standard deviation

of 46 mg/100 ml. If a random sample of

100 individuals is taken from this population,

what is the probability that the average

serum cholesterol level of these 100

individuals is higher than 225?

62

P(X > 225) = ?

225

3.04

225 - 211

4.6 = 3.04

.001

Cholesterol Level has a mean 211, s.d. 46.

001.0

)04.3()225(

>> ZPXP

211

n = 100

x

)6.4,211( xxNX

0

z

63


4 Approximation for Discrete Distributions

(Normal approximation to Binomial)

64

Normal Approximation for

Binomial Probability

0 1 2 3 4 5 6 7

= np = 3.5

2= np(1-p) = 1.75

?)3( bXP

65

)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

Density Curve Continuous Distributions -...

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