1 Chapter 5 The Standard Deviation and the Normal Model

1

Chapter 5The Standard Deviation and

the Normal Model

2

68-95-99.7 rule

Mean andStandard Deviation

(numerical)

Histogram(graphical)

68-95-99.7 rule

3

The 68-95-99.7 rule; applies only to mound-shaped data

approximately 68% of the measurements

are within 1 standard deviation of the mean,

that is, in ( , )

approx. 95% of the measurements are within

2 stand. dev. of the mean, i.e., in ( 2 , 2 )

almos

y s y s

y s y s

t all the measurements are within 3 stan.

dev of the mean, i.e., in ( 3 , 3 )y s y s

4

68-95-99.7 rule: 68% within 1 stan. dev. of the mean

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

-5

-4.5 -4

-3.5 -3

-2.5 -2

-1.5 -1

-0.5 0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

68%

34%34%

y-s y y+s

5

68-95-99.7 rule: 95% within 2 stan. dev. of the mean

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

-5

-4.5 -4

-3.5 -3

-2.5 -2

-1.5 -1

-0.5 0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

95%

47.5% 47.5%

y-2s y y+2s

6

Example: textbook costs

375.48

42.72

50

y

s

n

286 291 307 308 315 316 327 328340 342 346 347 348 348 349 354355 355 360 361 364 367 369 371373 377 380 381 382 385 385 387390 390 397 398 409 409 410 418422 424 425 426 428 433 434 437440 480

7

Example: textbook costs (cont.)286 291 307 308 315 316 327 328340 342 346 347 348 348 349 354355 355 360 361 364 367 369 371373 377 380 381 382 385 385 387390 390 397 398 409 409 410 418422 424 425 426 428 433 434 437440 480

375.48 42.72

( , ) (332.76, 418.20)

32percentage of data values in this interval 64%;

5068-95-99.7 rule: 68%

y s

y s y s

1 standard deviation interval about the mean

8


375.48 42.72

( 2 , 2 ) (290.04, 460.92)


5068-95-99.7 rule: 95%

y s

y s y s


9


375.48 42.72

( 3 , 3 ) (247.32, 503.64)


5068-95-99.7 rule: 99.7%

y s

y s y s


10

The best estimate of the standard deviation of the men’s weights

displayed in this dotplot is

1 2 3 4

4%9%

71%

16%

1. 10

2. 15

3. 20

4. 40

Changing Units of Measurement

Shifting data and rescaling data, and how shifting and rescaling data affect graphical and numerical summaries of data.

Shifting and rescaling: linear transformations

Original data x1, x2, . . . xn

Linear transformation:x* = a + bx, (intercept a, slope b)

x

x*

0

aShifts data by a

Changes scale

Linear Transformationsx* = a+ b x

Examples: Changing1. from feet (x) to inches (x*): x*=12x2. from dollars (x) to cents (x*):

x*=100x3. from degrees celsius (x) to degrees

fahrenheit (x*): x* = 32 + (9/5)x 4. from ACT (x) to SAT (x*): x*=150+40x5. from inches (x) to centimeters (x*):

x* = 2.54x

0 120 10032 9/5150 400 2.54

Shifting data only: b = 1x* = a + x

Adding the same value a to each value in the data set: changes the mean, median, Q1 and Q3

by a The standard deviation, IQR and

variance are NOT CHANGED. Everything shifts together. Spread of the items does not change.

Shifting data only: b = 1x* = a + x (cont.)

weights of 80 men age 19 to 24 of average height (5'8" to 5'10") x = 82.36 kg

NIH recommends maximum healthy weight of 74 kg. To compare their weights to the recommended maximum, subtract 74 kg from each weight; x* = x – 74 (a=-74, b=1)

x* = x – 74 = 8.36 kg

1. No change in shape

2. No change in spread

3. Shift by 74

Shifting and Rescaling data: x* = a + bx, b > 0

Original x data:x1, x2, x3, . . ., xn

Summary statistics:mean xmedian m1st quartile Q1

3rd quartile Q3

stand dev svariance s2

IQR

x* data: x* = a + bxx1*, x2*, x3*, . . ., xn*

Summary statistics:new mean x* = a + bxnew median m* = a+bmnew 1st quart Q1*= a+bQ1

new 3rd quart Q3* = a+bQ3

new stand dev s* = b snew variance s*2 = b2 s2

new IQR* = b IQR

Rescaling data: x* = a + bx, b > 0 (cont.)

weights of 80 men age 19 to 24, of average height (5'8" to 5'10")

x = 82.36 kg min=54.30 kg max=161.50 kg range=107.20 kg s = 18.35 kg

Change from kilograms to pounds:x* = 2.2x (a = 0, b = 2.2)

x* = 2.2(82.36)=181.19 pounds min* = 2.2(54.30)=119.46 pounds max* = 2.2(161.50)=355.3 pounds range*= 2.2(107.20)=235.84 pounds s* = 18.35 * 2.2 = 40.37 pounds

Example of x* = a + bx

4 student heights in inches

(x data)62, 64, 74, 72x = 68 inchess = 5.89 inches

Suppose we wantcentimeters instead:x* = 2.54x(a = 0, b = 2.54)

4 student heights in centimeters:

157.48 = 2.54(62)162.56 = 2.54(64)187.96 = 2.54(74)182.88 = 2.54(72)x* = 172.72 centimeterss* = 14.9606 centimeters

Note thatx* = 2.54x = 2.54(68)=172.2s* = 2.54s =

2.54(5.89)=14.9606

not necessary!UNC method

Go directly to this. NCSU method

Example of x* = a + bxx data:Percent returns from 4investments during2003:5%, 4%, 3%, 6%x = 4.5%s = 1.29%Inflation during 2003:2%x* data:Inflation-adjusted returns.x* = x – 2%(a=-2, b=1)

x* data:

3% = 5% - 2%2% = 4% - 2%1% = 3% - 2%4% = 6% - 2%x* = 10%/4 = 2.5%s* = s = 1.29%

x* = x – 2% = 4.5% –2%s* = s = 1.29% (note!

thats* ≠ s – 2%) !!

not necessary!

Go directly to this

Example Original data x: Jim Bob’s jumbo watermelons from

his garden have the following weights (lbs):

23, 34, 38, 44, 48, 55, 55, 68, 72, 75s = 17.12; Q1=37, Q3 =69; IQR = 69 – 37 = 32

Melons over 50 lbs are priced differently; the amount each melon is over (or under) 50 lbs is:

x* = x 50 (x* = a + bx, a=-50, b=1)-27, -16, -12, -6, -2, 5, 5, 18, 22, 25

s* = 17.12; Q*1 = 37 - 50 =-13, Q*3 = 69 - 50 = 19

IQR* = 19 – (-13) = 32 NOTE: s* = s, IQR*= IQR

SUMMARY: Linear Transformations x* = a + bx

Assembly Time (seconds)

0

5

10

15

20

25

30

Fre

qu

en

cy

Linear transformations do not affect the shape of the distribution of the data-for example, if the original data is right-skewed, the transformed data is right-skewed

Assembly Time (minutes)

0

10

20

30

Fre

qu

en

cy

SUMMARY: Shifting and Rescaling data, x* = a + bx, b > 0

* * *1 2 3 1 2 3

*

*

*1 1 1

*3 3 3

original data , , ,... transformed data , , ,...

summary statistics summary statistics

mean new mean

median new median

1st new

3rd new

st dev

x x x x x x

x x a bx

m m a bm

Q Q a bQ

Q Q a bQ

*

2 2 2 2

new st dev

var. new var. *

new *

s s bs

s s b s

IQR IQR bIQR

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Z-scores: Standardized Data Values

Measures the distance of a number from the mean in units of

the standard deviation

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z-score corresponding to y

where

original data value

the sample mean

s the sample standard deviation

the z-score corresponding to

y yz

s

y

y

z y

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Exam 1: y1 = 88, s1 = 6; exam 1 score: 91

Exam 2: y2 = 88, s2 = 10; exam 2 score: 92

Which score is better?

1

2

91 88 3z .5

6 692 88 4

z .410 10

91 on exam 1 is better than 92 on exam 2

If data has mean and standard deviation ,

then standardizing a particular value of

indicates how many standard deviations

is above or below the mean .

y s

y

y

y

27

Comparing SAT and ACT Scores

SAT Math: Eleanor’s score 680

SAT mean =500 sd=100 ACT Math: Gerald’s score 27

ACT mean=18 sd=6 Eleanor’s z-score: z=(680-500)/100=1.8 Gerald’s z-score: z=(27-18)/6=1.5 Eleanor’s score is better.

Z-scores: a special linear transformation a + bx

1 1where ,

x x x xz x a bx a b

s s s s s

Example. At a community college, if a student takes x credit hours the tuition is x* = $250 + $35x. The credit hours taken by students in an Intro Stats class have mean x = 15.7 hrs and standard deviation s = 2.7 hrs.

Question 1. A student’s tuition charge is $941.25. What is the z-score of this tuition?

x* = $250+$35(15.7) = $799.50; s* = $35(2.7) = $94.50

941.25 799.50 141.75 1.594.50 94.50

z

Z-scores: a special linear transformation a + bx (cont.)Example. At a community college, if a student takes x credit hours the tuition is x* = $250 + $35x. The credit hours taken by students in an Intro Stats class have mean x = 15.7 hrs and standard deviation s = 2.7 hrs.

Question 2. Roger is a student in the Intro Stats class who has a course load of x = 13 credit hours. The z-score isz = (13 – 15.7)/2.7 = -2.7/2.7 = -1.What is the z-score of Roger’s tuition?

Roger’s tuition is x* = $250 + $35(13) = $705

Since x* = $250+$35(15.7) = $799.50; s* = $35(2.7) = $94.50

705-799.50 -94.50z= = =-194.50 94.50

This is why z-scores are so useful!!

The linear transformation did not change the z-score!

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Z-scores add to zero

Student/Institutional Support to Athletic Depts For the 8 Public ACC Schools: 2008 ($ millions)

School Support y - ybar Z-score

Clemson 4.5 -3.713 -0.8806

FSU 7.5 -0.7125 -0.1690

GaTech 6.0 -2.213 -0.5248

Maryland 17.1 8.8875 2.1082

NCSU 5.5 -2.713 -0.6434

UNC 6.4 -1.813 -0.4299

UVA 11.9 3.6875 0.8747

VaTech 6.8 -1.413 -0.3351

Mean=8.2125, s=4.216

Sum = 0 Sum = 0

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Average IQ by BrowserNationally:Mean IQ=100sd = 15

81 100 19z 1.27

15 15

127 100 27

z 1.8015 15

Story was exposed as a hoax

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NORMAL PROBABILITY MODELS

The Most Important Model for Data in Statistics

33

X83 6 9 120

A family of bell-shaped curves that differ only in their means and standard deviations.

µ = the mean

= the standard deviation

µ = 3 and = 1

34

Normal Probability Models

The mean, denoted ,can be any number

The standard deviation can be any nonnegative number

The total area under every normal model curve is 1

There are infinitely many normal distributions

35

Total area =1; symmetric around µ

36

The effects of The effects of and and

How does the standard deviation affect the shape of f(x)?

= 2

=3 =4

= 10 = 11 = 12How does the expected value affect the location of f(x)?

37X3 6 9 120

X3 6 9 120

µ = 3 and = 1

µ = 6 and = 1

38X83 6 9 120

X83 6 9 120

µ = 6 and = 2

µ = 6 and = 1

39

area under the density curve between 6 and 8.

3 6 9 12

µ = 6 and = 2

0 X

40

area under the density curve between 6 and 8

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Standardizing

Suppose X~N( Form a new random variable by

subtracting the mean from X and dividing by the standard deviation :

(X This process is called standardizing the

random variable X.

42

Standardizing (cont.)

(X is also a normal random variable; we will denote it by Z:

Z = (X has mean 0 and standard deviation

1:E(Z) = = 0; SD(Z) =

The probability distribution of Z is called

the standard normal distribution.

43

Standardizing (cont.) If X has mean and stand. dev. , standardizing

a particular value of x tells how many standard deviations x is above or below the mean .

Exam 1: =80, =10; exam 1 score: 92

Exam 2: =80, =8; exam 2 score: 90

Which score is better?

1 exam on 92 than better is 2 exam on 90

1.258

10

8

8090z

1.210

12

10

8092z

2

1

44

X83 6 9 120

µ = 6 and = 2

Z0 1 2 3-1-2-3

.5.5

µ = 0 and = 1

(X-6)/2

45

Z = standard normal random variable

= 0 and = 1

Z0 1 2 3-1-2-3

.5.5

Standard Normal Model

.5.5

46

Important Properties of Z

#1. The standard normal curve is symmetric around the mean 0

#2. The total area under the curve is 1;

so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2

47

Finding Normal Percentiles by Hand (cont.)

Table Z is the standard Normal table. We have to convert our data to z-scores before using the table.

The figure shows us how to find the area to the left when we have a z-score of 1.80:

48

Areas Under the Z Curve: Using the Table

Proportion of area above the interval from 0 to 1 = .8413 - .5 = .3413

0 1Z

.1587.3413

.50

49

Standard normal areas have been calculated and are provided in table Z.

The tabulated area correspondto the area between Z= - and some z0

Z = z0

Area between - and z0

50

Example – begin with a normal model with mean 60 and stand dev 8

In this example z0 = 1.25

0.89440.8944

0.89440.8944

Proportion of the area to the left of 70

under the original curve is the proportion

70 60of the area to the left of 1.25

8under the standard normal Z curve

= 0.8944

51

Example

Area between 0 and 1.27) =

1.270 z

Area=.3980

.8980-.5=.3980

52

Example

Area to the right of .55 = A1

= 1 - A2

= 1 - .7088 = .2912

0 .55

A2

53

Example

Area between -2.24 and 0 =

Area=.4875

.5 - .0125 = .4875

z-2.24 0Area=.0125

54

Example

Area to the left of -1.85 = .0322

55

Example

A1 A2

0 2.73z

-1.18

Area between -1.18 and 2.73 = A - A1

= .9968 - .1190

= .8778

.1190

.9968

A1

A

56Area between -1 and +1 = .8413 - .1587 =.6826

.8413.1587

.6826

Example

57

Example

Is k positive or negative?

Direction of inequality; magnitude of probability

Look up .2514 in body of table; corresponding entry is -.67

-.67

58

Example

Area to the right of 250

under original curve

area to the right of

250 275 25.58

43 43under the standard normal

curve = 1 .2810 .7190

Z

59

Example

area between 225 and 375 area under

standard normal curve between z = (225 275) 43

= -1.16 and z = (375 275) 43 = 2.33;

the area is .9901 .1230 .8671

.9901.1230

.8671

60

N(275, 43); find k so that areato the left is .9846

.9846 area to the left of k under N(275,43)

curve area to left of z = ( 275) 43 under

275N(0,1) curve 2.1643

(from standard normal table)

2.16(43) 275 367.88

k

k

k

61

Area to the left of z = 2.16 = .9846

0 2.16 Z.1587

.4846

Area=.5

.9846

62

Example

Regulate blue dye for mixing paint; machine can be set to discharge an average of ml./can of paint.

Amount discharged: N(, .4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable.

Determine the setting so that only 1% of the cans of paint will be unacceptable

63

Solution

=amount of dye discharged into can

~N( , .4); determine so that

area to the right of 6 is .01

X

X

64

Solution (cont.)

6.4

(6 ) .4

=amount of dye discharged into can

~N( , .4); determine so that

the area to the right of x= 6 is .01.

.01 area to the right of 6

area to the right of z =

2.33(from standard normal tab

X

X

x

le)

= 6-2.33(.4) = 5.068

65

A random variable X with mean and standard deviation is normally distributed if its probability density function is given by

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

Normal Distributions

66

The Shape of Normal Distributions

Normal distributions are bell shaped, and symmetrical around

Why symmetrical? Let = 100. Suppose x = 110.

2210

)2/1(100110

)2/1(e

21

e2

1)110(f

Now suppose x = 9022

10)2/1(

10090)2/1(

e2

1e

21

)90(f

11090

67

Are You Normal? Normal Probability Plots

Checking your data to determine if a normal model is appropriate

68

Are You Normal? Normal Probability Plots

When you actually have your own data, you must check to see whether a Normal model is reasonable.

Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric.

69

A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot.

If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal.

Are You Normal? Normal Probability Plots (cont)

70

Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example:


71

A skewed distribution might have a histogram and Normal probability plot like this:


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1 Chapter 5 The Standard Deviation and the Normal Model