Second graded Homework Assignment: 1.80; 1.84; 1.92; 1.110; 1.115; 1.141 (optional: 1.127) Due in Labs on Sept 14-15

Second graded Second graded Homework Assignment:Homework Assignment:

1.80; 1.84; 1.92; 1.110; 1.115; 1.141

(optional: 1.127)

Due in Labs on Sept 14-15

Last Time:Last Time:

Two Formulae for Variance!

22

1

1XX

ns iDefinitional Formula:

2

1

1

1

2

1

1 n

iin

n

ii XX

nComputational

Formula:

Comment about “Variance”• The concept and definition of variance that

we have so far talked about in class is often referred to as

Sample Variance• We will later discuss a different concept,

definition (and different formula) for what is commonly referred to as

Population VarianceThe latter is a theoretical variance, not a data variance.

a

b

InterceptSlope

bx+a

x

Reminder: (Simple) Linear Function y=a+bx

1 2 3 4 5 6 7 8 9 10

X

1 2 3 4 5 6 7 8 9 10

X+5

If we add the constant a=5to each observation,then we obtain newdata that are shifted by 5

A shift affects the meanbut not the variance

New mean = Old mean + 5

New variance = Old variance

1 2 3 4 5 6 7 8 9 10

X

1 2 3 4 5 6 7 8 9 10

2X

If we multiply each observation by the constant b=2, then we obtain a new data set that looks rather different

A multiplicative constant affects both the mean

and the variance

New mean = Old mean x 2

New variance = Old variance x 4

The effect of linear transformations:

Suppose that you have original data

nXXX ,...,, 21

Suppose that you need to recode the data as follows:

0 with say,

,,...,, 2211

b

bXaYbXaYbXaY nn


0 with say,, bbXaY ii

XbaY

Median of Y’s = a + b (Median of X’s)

Same with other quartiles and percentiles

IQR of Y’s = b (IQR of X’s)


0 with say,, bbXaY ii

222XY sbs

XY bss

Example:

Original Data: 1, 2, 4, 6, 7

Transformed Data: 4, 7, 13, 19, 22

Mean: 4

Mean: 13 = 1 + 3 (4)

Median: 4

Median: 13 = 1 + 3 (4)

IQR: 6.5 – 1.5 = 5

IQR: 20.5 – 5.5 = 15= 3 (5)

Example:Original Data: 1, 2, 4, 6, 7

Transformed Data: 4, 7, 13, 19, 22

Variance:

Variance:

22222

41 4746444241

5.694049 426

41

22222

41 132213191313137134

5.58813603681 4234

41 5.63 2

Y = 1 + 3 X

0Z 12 Zs 1Zs

Transformation to Z-scores:

X

iiss

Xi s

XXXZ

XX

1

Z-scores have Mean equal to zero and (Variance) Standard Deviation equal to one

Today:Today:

The Normal DistributionThe Normal Distribution

Histogram of the Frequency Counts

Type A

14

19 1713

3 30 00

5

10

15

20

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

Sample Size: 60

Type A

00.050.1

0.150.2

0.250.3

0.35

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

Histogram of the Proportions

Type A

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

Ca

libra

te a

pp

rop

riat

ely

Choose vertical unit such that the total orange area is equal to 1

Type A

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

Ca

libra

te a

pp

rop

riat

ely

Choose Vertical Unit such that the total orange area is equal to 1

3/6019/60

Type A

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

(4+19+17)/60

Type A

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

2/3

2/3 of all Type A respondents had measurements between 55 and 69

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

DataWorld

TheoryWorld

50-54

55-59

60-64

65-69

70-74

75-79

80-84

85-89

90-94

Density Function

It can be mathematically easier to work with such a curve instead of a histogram

Density Function

The marked area is the relative frequencywith which observations between a and b occur

ba

Examples of Density Functions

Median

75th percentile25th percentile

Area p below pth percentile

Symmetric

= Mean

IQR


Median Mean

Positively Skewed (Skewed to the right)


MedianMean

Negatively Skewed (Skewed to the left)

THE NORMAL DISTRIBUTIONTHE NORMAL DISTRIBUTION(“Gaussian”, “bell-shaped”)

Probably the single most important distribution in Statistics

Symmetric

THE NORMAL DISTRIBUTIONTHE NORMAL DISTRIBUTION

Notation: X ~ N( , )

A normal distribution is fully specified with just two ‘parameters’, the mean of the distribution () and the standard deviation of the distribution ().

DataWorld

TheoryWorld

ssX ,, 2

,, 2 No formulae

yet

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( -2 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( 0 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( 2 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( 2 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( 0 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

n

N( -2 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

va

lue

of

de

ns

ity

fu

nc

tio

n

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

nN( 0 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

va

lue

of

de

ns

ity

fu

nc

tio

n

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

va

lue

of

de

ns

ity

fu

nc

tio

n

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

val

ue

of

de

nsi

ty f

un

ctio

n

N( 0 , 1 )

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

value of r.v.

va

lue

of

de

ns

ity

fu

nc

tio

n

Equation of the Density Function:

2

21

2

1

X

e

According to the Germans:“That’s right on the money”

2

21

2

1

X

e

According to the Germans:“That’s right on the money”

2

21

2

1

X

e

HerrGauss

The (theoretical) proportion of people in the population

between a and b: P(a X b) = area between a and b under the curve defined by the density function.

P(a < X < b) is equal to this area

a b

THE NORMAL DISTRIBUTIONTHE NORMAL DISTRIBUTIONProperties of X ~ N( , )

The proportion of a normally distributed X within:

•one standard deviation from its mean is .6826 P( - < X < + ) = .6826

•two standard deviations from its mean is .9544 P( - 2 < X < + 2 ) = .9544

•three standard deviations from its mean is .9974 P( - 3 < X < + 3 ) = .9974

True for any value of and

Example:The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of .08: Thus, X ~ N(1.1 , .08).

a) P(1.1 < X) = ?

b) P(1.02 < X < 1.18) = ?

c) How to calculate P(1.1 < X < 1.25) ?

d) How to calculate P(X > 1) ?

e) How to find x such that P(X <x) = .75 ?

0064.,08.,1.1 2

STANDARD NORMAL DISTRIBUTIONSTANDARD NORMAL DISTRIBUTION

Z ~ N( 0, 1)

-4 -3 -2 -1 0 1 2 3 4

Know everything about Z ~ N(0,1):

Table in your book (inside cover) tabulates values P(Z<z)

z

STANDARD NORMAL DISTRIBUTIONSTANDARD NORMAL DISTRIBUTION

Z ~ N( 0, 1)

-4 -3 -2 -1 0 1 2 3 4

Know everything about Z ~ N(0,1):

Table in your book (inside cover) tabulates values P(Z<z)

(note the table goes over two pages)

Note: you can think of values z of Z ~ N(0,1) as

“z many standard deviations from the mean”

z

AMAZING PROPERTY OF AMAZING PROPERTY OF

NORMAL DISTRIBUTIONSNORMAL DISTRIBUTIONS

If X is normally distributedthen a+bX (b>0) is also normally distributed.

More precisely: X ~ N( , ) (a+bX) ~ N(a+b , b)

Note:

This type of relationship is not necessarily true

for other distributions

)1,0(~then ),(~ If NZX

NX

AMAZING PROPERTY OF AMAZING PROPERTY OF

NORMAL DISTRIBUTIONSNORMAL DISTRIBUTIONS

Consequence:Any normally distributed quantity can be standardized:

iprelationsh following obtain the to,1

let Simply

ab

This enables us to use Z~N(0,1) (which we know everything about) whenever information about X ~ N( , ) is needed.


a) P(1.1 < X) =

b) P(1.02 < X < 1.18) =



e) How to find x such that P(X <x) = .75 ?

0064.,08.,1.1 2


a) P(1.1 < X) =

0064.,08.,1.1 2

Mean is 1.1

X~N(1.1,.08)


b) P(1.02 < X < 1.18)

0064.,08.,1.1 2

Mean is 1.1

X~N(1.1,.08)


1. Draw a picture!2. Standardize:

3. Draw a picture again!4. Find the probability (using the table)

P(0<Z<1.875) =?

Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of .08: Thus, X ~ N(1.1 , .08).

Z~N(0,1)

X~N(1.1,.08)

1.1

0

c) How to calculate P(0 < Z < 1.875) ?


Z~N(0,1)

0

0

Z~N(0,1)



3. Draw a picture again!4. Find the probability (using the table)

P(Z > -1.25) =?


X~N(1.1,.08)

1.1

Z~N(0,1)

0

d) How to calculate P(Z > -1.25) ?


Z~N(0,1)

0

Z~N(0,1)

0

e) How to find x such that P(X < x) =.75?


3. Draw a picture again!4. Find z (using the table)

P(Z <z)5. Solve for x using


1.1

X~N(1.1,.08)

0

Z~N(0,1)

Documents

Second graded Homework Assignment: 1.80; 1.84; 1.92; 1.110; 1.115; 1.141 (optional: 1.127) Due in Labs on Sept 14-15