bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-25_Lec-19_Statistics-1.ppt1

Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engr/Math/Physics 25

Chp7Statistics-

1



Learning Goals Use MATLAB to solve Problems in

• Statistics• Probability

Use Monte Carlo (random) Methods to Simulate Random processes

Properly Apply Interpolation or Extrapolation to Estimate values between or outside of know data points



Histogram Histograms are

COLUMN Plots that show the Distribution of Data• Height Represents

Data Frequency Some General

Characteristics• Used to represent

continuous grouped, or BINNED, data– BIN SubRange

within the Data

• Usually Does not have any gaps between bars

• Areas represent %-of-Total Data



HistoGram ≡ Frequency Chart A HistoGram shows how OFTEN some

event Occurs• Histograms are

often constructedusing FrequencyTables



Histograms In MATLAB MATLAB has 6

Forms of the Histogram Cmd

The Simplest Hist(y)

The Plot Statement

• Generates a Histogram with 10 bins

Example: Max Temp at Oakland AirPort in Jul-Aug08

TmaxOAK = [70, 75, 63, 64, 65, 66, 65, 65, 67, 78, 75, 73, 79, 71, 72, 67, 69, 69, 70, 74, 71, 72, 71, 74, 77, 77, 86, 90, 90, 70, 71, 66, 66, 72, 68, 73, 72, 82, 91, 82, 76, 75, 72, 72, 69, 70, 68, 65, 67, 65, 63, 64, 72, 70, 68, 71, 77, 65, 63, 69, 69, 67]

hist(TmaxOAK), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title('Oakland Airport - Jul-Aug08')



hist Result for Oakland

It was COLD in Summer 08

Bin Width = (91-63)/10 = 2.8 °F

60 65 70 75 80 85 90 950

5

10

15

No.

Day

s

Max. Temp (°F)

Oakland Airport - Jul-Aug08



Histograms In MATLAB Next Example:

Max Temp at Stockton AirPort in Jul-Aug08Hist(y)

The Plot Statement

• Generates a Histogram with 10 bins

TmaxSTK = [94, 98, 93, 94, 91, 96, 93, 87, 89, 94, 100, 99, 103, 103, 103, 97, 91, 83, 84, 90, 89, 95, 94, 99, 97, 94, 102, 103, 107, 98, 86, 89, 95, 91, 84, 93, 98, 104, 105, 107, 103, 91, 90, 96, 93, 86, 92, 93, 95, 95, 86, 81, 93, 97, 96, 97, 101, 92, 89, 92, 93, 94]

hist(TmaxSTK), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title(‘Stockton Airport - Jul-Aug08')



hist Result for Stockton

It was HOT in Summer 08

Bin Width = (107-81)/10 = 2.6 °F

80 85 90 95 100 105 1100

2

4

6

8

10

12

14

16Stockton Airport - Jul-Aug08

No.

Day

s

Max. Temp (°F)



hist Command Refinements Adjust The

number and width of the bins using hist(y,N)hist(y,x)• Where

– N an integer specifying the NUMBER of Bins

– x A vector that Specs CENTERs of the Bins

Consider Summer 08 Max-Temp Data from Oakland and Stockton

Make 2 Histograms• 17 bins• 60F→110F by 2.5’s



hist Plots 17 Bins>> hist(TmaxSTK,17), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title('Stockton, CA - Jul-Aug08')>>

hist(TmaxOAK,17), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title('Oakland, CA - Jul-Aug08')

80 85 90 95 100 105 1100

1

2

3

4

5

6

7

8

9

10Stockton, CA - Jul-Aug08

No.

Day

s

Max. Temp (°F)60 65 70 75 80 85 90 950

1

2

3

4

5

6

7

8

9

10Oakland, CA - Jul-Aug08

No.

Day

s

Max. Temp (°F)



hist Plots Same Scale>> x = [60:2.5:110];>> hist(TmaxSTK,x), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title('Stockton, CA - Jul-Aug08')

>> x = [60:2.5:110];hist(TmaxOAK,x), ylabel('No. Days'), xlabel('Max. Temp (°F)'), title('Oakland, CA - Jul-Aug08')

60 65 70 75 80 85 90 95 100 105 1100

2

4

6

8

10

12

14

16Oakland, CA - Jul-Aug08

No.

Day

s

Max. Temp (°F)60 65 70 75 80 85 90 95 100 105 1100

2

4

6

8

10

12

14

16Stockton, CA - Jul-Aug08

No.

Day

s

Max. Temp (°F)



hist Numerical Output Hist can also

provide numerical Data about the Histogramn = hist(y)• Gives the number of

values in each of the (default) 10 Bins

For the Stockton data

k =2 5 1 10 16 7 9 2 7 3 We can also spec

the number and/or Width of Bins

>> k13 = hist(TmaxSTK,13)k13 =2 2 4 4 6 10 10 7 5 2 6 2 2

>> k2_5s = hist(TmaxOAK,x)



hist Numerical Output Bin-Count and Bin-Locations

(Frequency Table) for the Oakland Data>> [u, v] = hist(TmaxOAK,x)u =0 3 11 7 159 6 4 1 2 1 0 3 0 0 0 0 0 0 0 0v = 60.0000 62.5000 65.0000 67.5000 70.0000 72.5000 75.0000 77.5000 80.0000 82.5000 85.0000 87.5000 90.0000 92.5000 95.0000 97.5000 100.0000 102.5000 105.0000 107.5000 110.0000



Histogram Commands - 1Command Descriptionbar(x,y) Creates a bar chart of y versus x.

hist(y)Aggregates the data in the vector y into 10 bins evenly spaced between the minimum and maximum values in y.

hist(y,n)Aggregates the data in the vector y into n bins evenly spaced between the minimum and maximum values in y.

hist(y,x)Aggregates the data in the vector y into bins whose center locations are specified by the vector x. The bin widths are the distances between the centers.



Histogram Commands - 2Command Description

[z,x] = hist(y)Same as hist(y) but returns two vectors z and x that contain the frequency count and the 10 bin locations.

[z,x] = hist(y,n) Same as hist(y,n) but returns two vectors z and x that contain the frequency cnt and the n bin locations.

[z,x] = hist(y,x)

Same as hist(y,x) but returns two vectors z and x that contain the frequency count and the bin locations. The returned vector x is the same as the user-supplied vector x.



Data Statistics Tool - 1 Make Line-

Plot of Temp Data for Stockton, CA

Use the Tools Menu to find the Data Statistics Tool



Data Statistics Tool - 2 Use the

Tool to Add Plot Lines for• The

Mean• ±StdDev



Data Statistics Tool - 3 Quite a

Nice Tool, Actually

The Result The Avg

Max Temp Was 96.97 °F



Probability Probability The LIKELYHOOD that a

Specified OutCome Will be Realized• The “Odds” Run from 0% to 100%

Class Question: What are the Odds of winning the California MEGA-MILLIONS Lottery?



175 711 536 ... EXACTLY???!!! To Win the MegaMillions Lottery

• Pick five numbers from 1 to 56 • Pick a MEGA number from 1 to 46

The Odds for the 1st ping-pong Ball = 5 out of 56

The Odds for the 2nd ping-pong Ball = 4 out of 55, and so On

The Odds for the MEGA are 1 out of 46



175 711 536 ... Calculated Calc the OverAll Odds as the

PRODUCT of each of the Individual OutComes

536,711,1751

320,384,085,21120

461

!56!51!5

461

521

532

543

554

565

Odds

• This is Technically a COMBINATION



175 711 536 ... is a DEAL! The ORDER in Which the Ping-Pong

Balls are Drawn Does NOT affect the Winning Odds

If we Had to Match the Pull-Order:

Current theX120320,384,085,21

1!5646

!51461

521

531

541

551

561

Odds

• This is a PERMUTATION



Normal Distribution - 1 Consider Data on the Height of a

sample group of 20 year old Men

Ht (in) No.64 164.5 065 065.5 066 266.5 467 567.5 468 868.5 1169 1269.5 1070 970.5 871 771.5 572 472.5 473 373.5 174 174.5 075 1

We can Plot this Frequency Data using bar

>> y_abs=[1,0,0,0,2,4,5,4,8,11,12,10,9,8,7,5,4,4,3,1,1,0,1];>> xbins = [64:0.5:75];>> bar(xbins, y_abs), ylabel('No.'), xlabel('Height (Inches'), title('Height of 20 Yr-Old Men')



Normal Distribution - 2 We can also SCALE the

Bar/Hist such that the AREA UNDER the CURVE equals 1.00, exactly The Game Plan for Scaling• Calc the Height of Each Bar To Get

the Total Area = [Bin Width] x [Σ(individual counts)]

• The individual Bar Area =[Bin Width] x [individual count]

• %-Area any one bar → [Bar Areas]/[Total Area]

Ht (in) No. Area (BW*No.) No./TotArea64 1 0.5 0.0200

64.5 0 0 0.000065 0 0 0.0000

65.5 0 0 0.000066 2 1 0.0400

66.5 4 2 0.080067 5 2.5 0.1000

67.5 4 2 0.080068 8 4 0.1600

68.5 11 5.5 0.220069 12 6 0.2400

69.5 10 5 0.200070 9 4.5 0.1800

70.5 8 4 0.160071 7 3.5 0.1400

71.5 5 2.5 0.100072 4 2 0.0800

72.5 4 2 0.080073 3 1.5 0.0600

73.5 1 0.5 0.020074 1 0.5 0.0200

74.5 0 0 0.000075 1 0.5 0.0200

50.0



Normal Distribution - 3 We can Use bar to Plot

the Scaled-Area Hist. >>y_abs=[1,0,0,0,2,4,5,4,8,11,12,10,9,8,7,5,4,4,3,1,1,0,1];>> xbins = [64:0.5:75];>> TotalArea = sum(0.5*y_abs)

>> y_scale = 100*y_abs/TotalArea;>> bar(xbins, y_scale), ylabel('Fraction (%/inch)'), xlabel('Height (inches)'), title('Height of 20 Yr-Old Men')



Normal Distribution - 4 This is a Good

Time for a UNITS Check• Remember, our

GOAL → the Area Under the Curve = 1

Recall From the Plot the UNITS for the y-axis → %/inch (?)

The Units come from these MATLAB Statements

So TotalArea is in inches•No.

Now y_scale

TotalArea = sum(0.5*y_abs)

Bin Width in INCHES

y_scale = 100*y_abs/TotalArea;

• Cont. on Next Slide



Normal Distribution - 5 The Units

Analysis for y-scale

Recall From MTH1 that for y = f(x) displayed in BAR Form the Area Under the Curve

y_scale = 100*y_abs/TotalArea;

inch%y_scale

No.*inchesNo.*

1%100y_scale

hi

lo

x

xlo

crv

xxxy

xBinWidthxyHgt

A AreasIndividual



Normal Distribution - 6 In this Case

• y(x) → y_scale in %/inch

• Δx → Bin Width = 0.5 in inches

Then The Units Analysis for Our “integration”

Check the integration

inch5.0inch%

y

xxxyAhi

lo

x

xlocrv

Ht (in) No. Area (BW*No.) No./TotArea BW*(No./TotArea)64 1 0.5 0.0200 1.00%

64.5 0 0 0.0000 0.00%65 0 0 0.0000 0.00%

65.5 0 0 0.0000 0.00%66 2 1 0.0400 2.00%

66.5 4 2 0.0800 4.00%67 5 2.5 0.1000 5.00%

67.5 4 2 0.0800 4.00%68 8 4 0.1600 8.00%

68.5 11 5.5 0.2200 11.00%69 12 6 0.2400 12.00%

69.5 10 5 0.2000 10.00%70 9 4.5 0.1800 9.00%

70.5 8 4 0.1600 8.00%71 7 3.5 0.1400 7.00%

71.5 5 2.5 0.1000 5.00%72 4 2 0.0800 4.00%

72.5 4 2 0.0800 4.00%73 3 1.5 0.0600 3.00%

73.5 1 0.5 0.0200 1.00%74 1 0.5 0.0200 1.00%

74.5 0 0 0.0000 0.00%75 1 0.5 0.0200 1.00%

50.0 100.00%

Example



Normal Distribution - 7 Example 71” The 71” Bar Area =

Hgt•Width:

area) total the(of %7

inches 5.0inch%14,71

sclA

Alternatively from the Absolute values

inchNo. 5.3

inches 5.0No.by 7,71

absA

%7inNo. 50inNo. 5.3

,

,71

absall

abs

AA

• The Total Abs Area = 50 No.•inch



Probability Distribution Fcn (PDF) Because the Area

Under the Scaled Plot is 1.00, exactly, The FRACTIONAL Area under any bar, or set-of-bars gives the probability that any randomly Selected 20 yr-old man will be that height

e.g., from the Plot we Find • 67.5 in → 8 %/in• 68 in → 16 %/in• 68.5 in → 22%/in

Summing → 46 %/in

Multiply the Uniform BinWidth of 0.5 in → 23% of 20 yr-old men are 67.25-68.75 inches tall



Random Variable A random variable x takes on a defined set of

values with different probabilities; e.g.. • If you roll a die, the outcome is random (not fixed)

and there are 6 possible outcomes, each of which occur with equal probability of one-sixth.

• If you poll people about their voting preferences, the percentage of the sample that responds “Yes on Proposition 101” is a also a random variable – the %-age will be slightly differently every time you poll.

Roughly, probability is how frequently we expect different outcomes to occur if we repeat the experiment over and over (“frequentist” view)



Random variables can be Discrete or Continuous Discrete random variables have a

countable number of outcomes• Examples: Dead/Alive, Red/Black,

Heads/Tales, dice, counts, etc. Continuous random variables have an

infinite continuum of possible values. • Examples: blood pressure, weight, Air

Temperature, the speed of a car, the real numbers from 1 to 6.



Probability Distribution Functions A Probability Distribution Function

(PDF) maps the possible values of x against their respective probabilities of occurrence, p(x)

p(x) is a number from 0 to 1.0, or alternatively, from 0% to 100%.

The area under a probability distribution function curve is always 1 (or 100%).



Discrete Example: Roll The Die

1/6

1 4 5 62 3

xall

1 xp

x

x p(x)1 p(x=1)=1

/62 p(x=2)=1

/63 p(x=3)=1

/64 p(x=4)=1

/65 p(x=5)=1

/66 p(x=6)=1

/6

xp



Continuous Case The probability function that accompanies a

continuous random variable is a continuous mathematical function that integrates to 1.

The Probabilities associated with continuous functions are just areas under a Region of the curve (→ Definite Integrals)

Probabilities are given for a range of values, rather than a particular value • e.g., the probability of getting a math SAT

score between 700 and 800 is 2%).



Continuous Case PDF Example Recall the negative exponential function

(in probability, this is called an “exponential distribution”):

xexf )(

This Function Integrates to 1 zero to infinity as required for all PDF’s

11000

xx ee



Continuous Case PDF Example

x

p(x)=e-x

1

For example, the probability of x falling within 1 to 2:

The probability that x is any exact value (e.g.: 1.9976) is 0 • we can ONLY assign

Probabilities to possible RANGES of x

x

1

1 2

p(x)=e-x

NO Area Under a

LINE

23% 23.368.135.

2)(1

12

21

2

1

ee

eexp xx



Gaussian Curve The Man-Height HistroGram had some

Limited, and thus DISCRETE, Data If we were to Measure 10,000 (or more)

young men we would obtain a HistoGram like this As We increase the

number and fineness of the measurements The PDF approaches a CONTINUOUS Curve



Gaussian Distribution A Distribution that

Describes Many Physical Processes is called the GAUSSIAN or NORMAL Distribution

Gaussian (Normal) distribution• Gaussian → famous “bell-shaped curve”

– Describes IQ scores, how fast horses can run, the no. of Bees in a hive, wear profile on old stone stairs...

• All these are cases where:– deviation from mean is equally probable in either

direction– Variable is continuous (or large enough integer

to look continuous)



Normal Distribution Real-valued PDF: f(x) → −∞ < x < +∞ 2 independent fitting parameters:

µ , σ (central location and width) Properties:

• Symmetrical about Mode at µ ,• Median = Mean = Mode,• Inflection points at ±σ

Area (probability of observing event) within:• ± 1σ = 0.683 • ± 2σ = 0.955

For larger σ, bell shaped curve becomes wider and lower (since area =1 for any σ)



Normal Distribution

22)(21 2

xexf

122)(

21 2

dxe

dxxf

x

Mathematically• Where

– σ2 = Variance– µ = Mean

The Area Under the Curve



68-95-99.7 Rule for Normal Dist

68% of the data

95% of the data

99.7% of the data

σσ

2σ2σ3σ 3σ



68-95-99.7 Rule in Math terms… Using Definite-Integral Calculus

99.7% 997.2

1

95% 95.2

1

68% 68.2

1

3

3

)(21

2

2

)(21

)(21

2

2

2

dxe

dxe

dxe

x

x

x



How Good is the Rule for Real? Check some example data: The mean, µ, of the weight of a large

group of women Cross Country Runners = 127.8 lbs

The standard deviation (σ) for this Group = 15.5 lbs



80 90 100 110 120 130 140 150 160 0

5

10

15

20

25

P e r c e n t

POUNDS

127.8 143.3112.3

68% of 120 = .68x120 = ~ 82 runners

In fact, 79 runners fall within 1σ (15.5 lbs) of the mean



80 90 100 110 120 130 140 150 160 0

5

10

15

20

25

P e r c e n t

POUNDS

127.896.8

95% of 120 = .95 x 120 = ~ 114 runners

In fact, 115 runners fall within 2σ of the mean

158.8



80 90 100 110 120 130 140 150 160 0

5

10

15

20

25

P e r c e n t

POUNDS

127.881.3

99.7% of 120 = .997 x 120 = 119.6 runners

In fact, all 120 runners fall within 3σ of the mean

174.3



Estimating µ & σ (1) The Location &

Width Parameters, µ & σ, are Calculated from the ENTIRE POPULATION• Mean, µ

NxN

kk

1

NxN

kk

1

22

• Variance, σ2

• Standard Deviation, σ2

For LARGE Populations it is usually impractical to measure all the xk

In this case we take a Finite SAMPLE to ESTIMATE µ & σ



Estimating µ & σ (2) Say we want to

characterize Miles/Yr driven by Every Licensed Driver in the USA

We assume that this is Normally Distributed, so we take a Sample of N = 1013 Drivers

We Take the Mean of the SAMPLE

NxxµN

kn

1

Use the SAMPLE-Mean to Estimate the POPULATION-Mean

NxxN

kn

1



Estimating µ & σ (3) Now Calc the

SAMPLE Variance & StdDev

S Estimate

1

1

2

2

N

xxS

N

kk

• Number decreased from N to (N – 1) To Account for case where N = 1– In this case x-bar = x1,

and the S2 result is meaningless

• standard deviation: positive square root of the variance– small std dev:

observations are clustered tightly around a central value

– large std dev: observations are scattered widely about the mean



Sample Mean and StdDev

For a series of N observations, the most probable estimate of the mean µ is the average

x of the observations. We refer to this as the sample mean

x to distinguish it from the population mean µ.

x1N

xi Sample Mean

Calculate the Population Variance, σ2, from:

2 s2 1

N 1x i x

2

Sample Variance

But we cannot know the true population mean µ so the practical estimate for the sample variance and standard deviation would be:

2222

22 121

iii

i xNNN

xNx

xN



Error Function (erf) & Probability Guass’s Defining

Eqn dyezerf

z y 0

22

This looks a lot Like the normal dist

dxeI xG

22)(21 2

Now Let

Consider the Gaussian integral

22)(21 2

xexf

OrdxeI

x

G

2

221

dydx

dxdy

xy

2

Or2

12



Error Function (erf) & Probability Subbing for x &

dx

dyezerfz y

0

22

As

dxeIx

G

2

2

21

dyeI

dyeI

yG

yG

2

2

1

221

ReArranging

erf

dye

dyeI

y

yG

21

221

1

2

2



Error Function (erf) & Probability Now the Limits Plotting

This Fcn is Symmetrical about y = 0

Recall

-3 -2 -1 0 1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y

f(y) =

exp

(-y2 )

2yeyf

dyezerfz y

0

22

And the erf properties• erf(0) = 0• erf(h) = 1



Error Function (erf) & Probability By Symmetry about y = 0 for

122 0

0

22

dyedye yy

Thus

dyedyedyeB yyB y

0

0 222 222

So Finally integrating −h to B

)(12 2

BerfdyeB y

2ye



Error Function (erf) & Probability Note That for a

Continuous PDF• Probability that x

is Less or Equal to b

• Probability that x is between a & b

The probability for the Normal Dist

b

dxxfbxP

b

a

dxxfbxaP

But

dxebxaP

dxebxP

b

a

x

bx

22)(21

22)(21

2

2

2

2

21

22)(21

2

2

xerf

dxeI xG



Error Function (erf) & Probability If We Scale this

Properly we can Cast these Eqns into the ½erf Form

MATLAB has the erf built-in, so if we have the sample Mean & StdDev We can Calc Probabilities for Normally Distributed Quantities

2

121

µberfbxP

222

1

µaerfµberfbxaP



All Done for Today

Gaussian?Or

Normal?

Normal distribution was introduced by French mathematician A. De Moivre in 1733.• Used to approximate

probabilities of coin tossing• Called it the exponential

bell-shaped curve 1809, K.F. Gauss, a German

mathematician, applied it to predict astronomical entities… it became known as the Gaussian distribution.

Late 1800s, most believe majority of physical data would follow the distribution called normal distribution

Recall De Moivre’s Theorem

kjkRz

jRRzkk sincos

sincos



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engr/Math/Physics 25

Appendix 6972 23 xxxxf



Basic Fitting Demo File% Bruce Mayer, PE

% ENGR25 * 11Apr10

% file = Demo_Basic_Fitting_Stockton_Temps_1004.m

%

TmaxSTK = [94, 98, 93, 94, 91, 96, 93, 87, 89, 94, 100, 99, 103, 103, 103, 97, 91, 83, 84, 90, 89, 95, 94, 99, 97, 94, 102, 103, 107, 98, 86, 89, 95, 91, 84, 93, 98, 104, 105, 107, 103, 91, 90, 96, 93, 86, 92, 93, 95, 95, 86, 81, 93, 97, 96, 97, 101, 92, 89, 92, 93, 94]

Ntot = length(TmaxSTK)

nday = [1:Ntot];

plot(nday, TmaxSTK, '-dk'), xlabel('No. Days after 31Jun08'), ylabel('Max. Temp (°F)'), title('Stockton, CA - Jul-Aug08')



Normal or Gaussian? Normal distribution was introduced by French

mathematician A. De Moivre in 1733.• Used to approximate probabilities of coin tossing• Called it exponential bell-shaped curve

1809, K.F. Gauss, a German mathematician, applied it to predict astronomical entities… it became known as Gaussian distribution.

Late 1800s, most believe majority data would follow the distribution called normal distribution



Carl Friedrich G

auss



NormalDist Data

Ht (in) No. Area (BW*No.) No./TotArea BW*(No./TotArea)

64 1 0.5 0.0200 1.00%

64.5 0 0 0.0000 0.00%

65 0 0 0.0000 0.00%

65.5 0 0 0.0000 0.00%

66 2 1 0.0400 2.00%

66.5 4 2 0.0800 4.00%

67 5 2.5 0.1000 5.00%

67.5 4 2 0.0800 4.00%

68 8 4 0.1600 8.00%

68.5 11 5.5 0.2200 11.00%

69 12 6 0.2400 12.00%

69.5 10 5 0.2000 10.00%

70 9 4.5 0.1800 9.00%

70.5 8 4 0.1600 8.00%

71 7 3.5 0.1400 7.00%

71.5 5 2.5 0.1000 5.00%

72 4 2 0.0800 4.00%

72.5 4 2 0.0800 4.00%

73 3 1.5 0.0600 3.00%

73.5 1 0.5 0.0200 1.00%

74 1 0.5 0.0200 1.00%

74.5 0 0 0.0000 0.00%

75 1 0.5 0.0200 1.00%

50.0 100.00%



SPICE Circuit

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

Documents

pe engineeringmathphysics

temp f

max temp

maxtemp data

xwhere n

n s j o h n s o n c

titleoakland airport

titlestockton airport