math 1710 class 8pi.math.cornell.edu/~web1710/slides/sep12_v2.pdf · 4 for betty. 3 for carla. but...

Math 1710Class 8

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Binomial

Two Ways of“Randomly”Flipping 2Coins

SobrietyCheckpoints

NormalDistribution

Working WithNormalDistributions

NormalApproximation

MakingNormalApproximationAccurate

Math 1710 Class 8

Dr. Allen Back

Sep. 12, 2016

Math 1710Class 8

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Binomial


SobrietyCheckpoints

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NormalApproximation


Three Girls Out of Five Children

Suppose P(Girl)=.6 and gender of births independent.

Math 1710Class 8

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?

Math 1710Class 8

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)=?

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)= P(GGGBB) = .63.42 = .03456.But this undercounts the answer.

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)= P(GGGBB) = .63.42 = .03456.But this undercounts the answer.Other orders also possible; e.g. BBGGG .

Math 1710Class 8

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)= P(GGGBB) = .63.42 = .03456.But this undercounts the answer.Other orders also possible; e.g. BBGGG .How many such orders?

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)= P(GGGBB) = .63.42 = .03456.But this undercounts the answer.10 Possible Orders:

First child a girl: GGGBB GGBGB GGBBG GBGGBGBGBG GBBGG

First child a boy: BGGGB BGGBG BGBGG BBGGG

Math 1710Class 8

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Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)= P(GGGBB) = .63.42 = .03456.But this undercounts the answer.10 Possible Orders:



So answer is 10(.6)3(.4)2 = .3456.

Math 1710Class 8

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Why 10 Possible Orders?

5 birth positions, 3 of which girls

Math 1710Class 8

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5 birth positions, 3 of which girlsSo

C5,3 =

(53

)

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C5,3 =

(53

)=

5 · 4 · 31 · 2 · 3

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5 birth positions, 3 of which girls(53

)=

5 · 4 · 31 · 2 · 3

=5 · 41 · 2

= 10

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)=

5 · 4 · 31 · 2 · 3

=5 · 41 · 2

= 10

Above showed (53

)=

(52

)which works in general as well.

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)=

5 · 4 · 31 · 2 · 3

=5 · 41 · 2

= 10

If one tacks a 2! onto both the numerator and denominator,

5 · 4 · 31 · 2 · 3

=5 · 4 · 3 · 2 · 1

3!2!

showing (53

)=

5!

3!2!

which is also a general formula.

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Why C5,3 = 5·4·31·2·3?


Math 1710Class 8

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Why C5,3 = 5·4·31·2·3?

5 birth positions, 3 of which girlsSuppose the three girls are named Abby, Betty, and Carla.

Math 1710Class 8

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Why C5,3 = 5·4·31·2·3?

5 birth positions, 3 of which girlsSuppose the three girls are named Abby, Betty, and Carla.Then 5 birth positions for Abby.4 for Betty.3 for Carla.

Math 1710Class 8

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Why C5,3 = 5·4·31·2·3?

5 birth positions, 3 of which girlsSuppose the three girls are named Abby, Betty, and Carla.Then 5 birth positions for Abby.4 for Betty.3 for Carla.But each set of 3 birth positions for the girls shows up 3! timesdepending on the order of births among Abby, Betty, and Carla.

Math 1710Class 8

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Why C5,3 = 5·4·31·2·3?

5 birth positions, 3 of which girlsSuppose the three girls are named Abby, Betty, and Carla.Then 5 birth positions for Abby.4 for Betty.3 for Carla.But each set of 3 birth positions for the girls shows up 3! timesdepending on the order of births among Abby, Betty, and Carla.

So

C5,3 =

(53

)=

5 · 4 · 31 · 2 · 3

Math 1710Class 8

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SobrietyCheckpoints

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Pascal’s Triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

(a + b)0 = 1(a + b)1 = 1a + 1b(a + b)2 = 1a2 + 2ab + 1b2.(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3.. . .

Math 1710Class 8

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Pascal’s Triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

10 Possible Orders: (Two points of view.)



Math 1710Class 8

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Pascal’s Triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

10 Possible Orders: (Three points of view.)



(53

)=

(42

)+

(43

).

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General Terminology

In general we speak about a sequence of Bernoulli Trials:

2 outcomes, conventionally called success and failure.

constant probablility p of success.

the successive trials are independent.

So, for each trial, the number of successes (0 or 1) is aBernoulli(p) RV.

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General Terminology

Binomial(n,p) RV Y describes the number of successes in nBernoulli trials.For Y we know µ = np, σ =

√npq, and

P(Y = k) =

(nk

)pkqn−k .

(Here q = 1− p.)

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General Terminology

Binomial(n,p) RV Y describes the number of successes in nBernoulli trials.For Y we know µ = np, σ =

√npq, and

P(Y = k) =

(nk

)pkqn−k .

(Here q = 1− p.)A substitute for a big table giving the prob. dist. of Y .

Math 1710Class 8

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General Terminology

Why µ = np, σ =√

npq for a Binomial(n,p) RV Y ?

Math 1710Class 8

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General Terminology


npq for a Binomial(n,p) RV Y ?First confirm for a Bernoulli(p) RV, µ = p and the the varianceis pq.

Math 1710Class 8

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General Terminology


npq for a Binomial(n,p) RV Y ?Now let Xi be a Bernoulli(p) RV counting the number ofsuccesses (1 or 0) on the i’th trial.

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General Terminology



Y = X1 + X2 + . . .+ Xn.

(Remember Y is the total number of successes.)

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General Terminology



Y = X1 + X2 + . . .+ Xn.

(Remember Y is the total number of successes.)Since both means and variances add for the sum of independentRV’s, we obtain the formulas for the binomial case.

Math 1710Class 8

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Suppose 70% approve the President . . .

You poll 100 people.What is the probability that exactly 65 report approval?

Math 1710Class 8

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You poll 100 people.What is the probability that exactly 65 report approval?Solution: Y = Binomial(100, .7)

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P(Y = 65) =?

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P(Y = 65) =

(10065

).765.335.

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P(Y = 65) =

(10065

).765.335.

A calculator could help with

(10065

).

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SobrietyCheckpoints

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Calculating Combinations


(10065

).

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(10065

).

TI-83,84 100 Math→ Prb → nCr 65(Math is at the left of row 3.)

TI-89 Math→ Probability → nCr(100, 65)(Math is above the 5.)

TI-30 100 nCr 65(nCr is above the 8 on my TI-30.)

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(10065

).

TI-83,84 100 Math→ Prb → nCr 65(Math is at the left of row 3.)

TI-89 Math→ Probability → nCr(100, 65)(Math is above the 5.)

TI-30 100 nCr 65(nCr is above the 8 on my TI-30.)

An answer like 1.095067153E27 means 1.095× 1027 and so

P(Y = 65) =

(10065

).765.335 = .04678.

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Tedious

Notice that calculating P(60 ≤ Y ≤ 65) by the above methodwould not be pleasant.

We’ll see that an important technique called normalapproximation will get us quickly to that kind of answer.

TI-8x calculators have a binomialcdf function which can do this.Please don’t use that function to supply any homework orexam answers in this course.

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X1 + X2 vs. 2X

Tossing one fair coin is described by Bernoulli(.5):

X probability0 .51 .5

Math 1710Class 8

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Binomial


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X1 + X2 vs. 2X



The RV 2X ?

Math 1710Class 8

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X1 + X2 vs. 2X



The RV 2X :

2X probability0 .52 .5

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X1 + X2 vs. 2X



The RV 2X :


If X1 and X2 are independent copies of X , then X1 + X2 cancome out to 0,1, or 2.

Math 1710Class 8

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X1 + X2 vs. 2X



The RV 2X :


The RV X1 + X2 ?

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X1 + X2 vs. 2X



The RV 2X :


The RV X1 + X2:

X1 + X2 probability0 .251 .52 .25

Math 1710Class 8

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How many heads in total?

How can these two possibilities come up in tossing 2 coins?

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How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.

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This is X1 + X2.

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This is X1 + X2.

Method 2: Toss one coin.Then turn the other coin over to the same result.

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This is X1 + X2.


Note the second coin is still random.

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This is X1 + X2.


This is 2X .

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Experimentally X1 + X2 and 2X

X1 + X2 frequency0 ?1 ?2 ?

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x̄ ,s?

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These x̄ ,s should be close to µ = 1, σ =√

2(.5)(.5) = .707resp.

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2X frequency0 ?2 ?

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2X frequency0 ?2 ?

x̄ ,s?

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2X frequency0 ?2 ?

These x̄ ,s should be close to µ = 1, σ = 2(.5) = 1 resp.

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Sobriety Checkpoints

Officers ask questions, then maybe detain for a breathylyzertest.Suppose 12% of drivers nationally drink. (Inappropriately interms of driving.)Officers have right idea about drinking or not drinking about80% of the time.

1) P(someone not drinking is detained for test)?

2) P(being detained)?

3) P(someone detained has been drinking)?

4) P(someone released has not been drinking)?

Math 1710Class 8

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SobrietyCheckpoints

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A Continuous Distribution

There’s a normal distribution with any mean µ or σ > 0.

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NormalApproximation



There’s a normal distribution with any mean µ or σ > 0.

N(µ, σ)

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NormalApproximation



N(µ, σ)

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N(µ, σ)

Area corresponds to probability.

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NormalApproximation



N(µ, σ)

The entire area under the curve is 1.

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N(µ, σ)

The area between a and b is the probability of a value x fallingwithin that range.

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The 68− 95− 99.7 Rule

68% within 1 standard deviation of the mean.

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The 68− 95− 99.7 Rule

68% within 1 standard deviation of the mean.

N(µ, σ)

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The 68− 95− 99.7 Rule

95% within 2 standard deviations of the mean.

N(µ, σ)

Math 1710Class 8

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The 68− 95− 99.7 Rule

99.7% within 3 standard deviations of the mean.

N(µ, σ)

Math 1710Class 8

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The Standard Normal Distribution

Standard normal is the case µ = 0 and σ = 1.

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Standard normal is the case µ = 0 and σ = 1.

N(0,1)

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A general normal distribution:

N(µ, σ)

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N(µ, σ)

Can convert from a general N(µ, σ) to N(0, 1) via the Z-score.

z =x − µσ

Math 1710Class 8

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N(µ, σ)

z =x − µσ

The Z score is just the offset from the mean in standarddeviation units.

Math 1710Class 8

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z = x−µσ

This transformation preserves area and probability.

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UsingTable Z

N(0,1)

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UsingTable Z

N(0,1)

Table Z gives us the area to the left on the standard normal.

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UsingTable Z

N(0,1)

Table Z gives us the area to the left on the standard normal.i.e. P(Z < z)

Math 1710Class 8

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UsingTable Z

Table Z gives us the area to the left on the standard normal.For example P(Z < 1) = .8413 since

Math 1710Class 8

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SobrietyCheckpoints

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UsingTable Z

Table Z gives us the area to the left on the standard normal.For example P(Z < 1) = .8413 since

And P(Z < 1.16) = .8770.

Math 1710Class 8

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UsingTable Z

Table Z gives us the area to the left on the standard normal.Note you use row 1.1 and column .06 for this!

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UsingTable Z

We can also get the area under the curve between two values.

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UsingTable Z

For example P(−.67 < Z < 1) =?

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UsingTable Z

For example P(−.67 < Z < 1) =?

N(0,1)

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UsingTable Z

Table Z tells us P(−.67 < Z < 1) = .8413− .2514 = .5899.

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UsingTable Z

The reason for the subtractionP(−.67 < Z < 1) = P(Z < 1)− P(Z < −.67) is:

-

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UsingTable Z


=

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Mileage Example

Suppose a normal model N(24 mpg, 6 mpg) describes fuelefficiency of cars in a region:

Math 1710Class 8

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Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Percent of cars with mileage below 15?

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SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

% between 20 and 30?

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SobrietyCheckpoints

NormalDistribution


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Mileage Example

% of cars above 40?

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SobrietyCheckpoints

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Mileage Example

Worst 20% of cars?

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Mileage Example

Third Quartile?

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Mileage Example

Gas mileage of the 5% most efficient?

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Mileage Example

Suppose a normal model N(24 mpg, 6 mpg) describes fuelefficiency of cars in a region:

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SobrietyCheckpoints

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NormalApproximation


Mileage Example


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Mileage Example


N(24,6)

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Mileage Example


N(24,6)

We understand a general N(µ, σ) by reducing to N(0, 1).

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Mileage Example


N(24,6)

We understand a general N(µ, σ) by reducing to N(0, 1).The Z score of 15 is ?

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Mileage Example

Percent of cars with mileage below 15?The Z score of 15 is ?

z =15− 24

6= −1.5

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Mileage Example


z =15− 24

6= −1.5

N(0,1)

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Mileage Example


N(0,1)

Table Z tells us P(Z < −1.5) = .0668.6.7% of cars will have mileage below 15 mpg.

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SobrietyCheckpoints

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Mileage Example


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Mileage Example


N(24,6)

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Mileage Example


N(24,6)

The Z score of 20 is ?

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Mileage Example


N(24,6)

z1 =20− 24

6= −.67

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Mileage Example


N(24,6)

The Z score of 20 is z1 = −.67?

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Mileage Example


N(24,6)

The Z score of 20 is z1 = −.67?The Z score of 30 is ?

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SobrietyCheckpoints

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Mileage Example


N(24,6)

The Z score of 20 is z1 = −.67?The Z score of 30 is ?

z2 =30− 24

6= 1

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Mileage Example


N(24,6)

The Z score of 20 is z1 = −.67?The Z score of 30 is z2 = 1.

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SobrietyCheckpoints

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Mileage Example

% between 20 and 30?The Z score of 20 is z1 = −.67?The Z score of 30 is z2 = 1.

N(0,1)

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Mileage Example

% between 20 and 30?The Z score of 20 is z1 = −.67?The Z score of 30 is z2 = 1.

N(0,1)

Table Z tells us P(−.67 < Z < 1) = .8413− .2514 = .5899.59.0% of cars will have mileage between 20 and 30 mpg.

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Mileage Example


-

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Mileage Example


=

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Mileage Example

% of cars above 40?

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Mileage Example

% of cars above 40?

N(24,6)


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SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

% of cars above 40?

N(24,6)


z =40− 24

6= 2.67

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

% of cars above 40?

N(24,6)

z =40− 24

6= 2.67

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

% of cars above 40?

N(0,1)

Table Z tells us P(Z < 2.67) = .9962.Therefore P(Z > −2.67) = 1− .9962 = .0038.0.38% of cars will have mileage above 40 mpg.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?Now we need to start with the N(0, 1) picture.

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?

N(0,1)

P(Z <?) = .20Table Z suggests −.84. to 2 decimal places.P(Z < −.84) = .2005.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?Translating to N(24, 6) :

Having a Z score of −.84 meansbeing .84 std. dev. to the left of 24.So the value is 24 + (−.84)(6) = 24− 5.04 = 18.96.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?N(24, 6) :

The worst 20% cars are those with mileage below 18.96 mpg.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Worst 20% of cars?Instead of viewing

x = 24 + (−.84)(6)

as being clear from the picture, some people prefer to solve

x − 24

6= −.84.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?Again start with the N(0, 1) picture.

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?

N(0,1)

P(Z <?) = .75Table Z suggests .67. to 2 decimal places.P(Z < .67) = .7486.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?Translating to N(24, 6) :

Having a Z score of .67 meansbeing .67 std. dev. to the right of 24.So the value is 24 + (.67)(6) = 28.02.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Third Quartile?N(24, 6) :

The 3rd quartile of cars are those with a mileage of 28.02 mpg.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example


Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Gas mileage of the 5% most efficient?Again start with the N(0, 1) picture.

N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example


N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example


N(0,1)

P(Z >?) = .05. That means P(Z <?) = .95.Table Z suggests 1.65. to 2 decimal places.P(Z < 1.65) = .9505. (1.64 is an equally good choice.)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example


N(0,1)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Gas mileage of the 5% most efficient?Translating to N(24, 6) :

Having a Z score of 1.65 meansbeing 1.65 std. dev. to the right of 24.So the value is 24 + (1.65)(6) = 33.90.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


Mileage Example

Gas mileage of the 5% most efficient?N(24, 6) :

The top 5% of cars are those with a mileage of at least 22.90mpg.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation



You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation



You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)Solution: Let X = Binomial(100, .7)

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation




X has meanµ = (100)(.7) = 70

andσ =

√(100)(.7)(.3) =

√21 = 4.58.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation




X has meanµ = (100)(.7) = 70

andσ =

√(100)(.7)(.3) =

√21 = 4.58.

X will then be approximated by a normal distribution Y withthe same mean and standard deviation.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation




X has meanµ = (100)(.7) = 70

andσ =

√(100)(.7)(.3) =

√21 = 4.58.


Y = N(70, 4.58).

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation




X has meanµ = (100)(.7) = 70

andσ =

√(100)(.7)(.3) =

√21 = 4.58.


Y = N(70, 4.58).We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation



You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)


Y = N(70, 4.58).We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).The Z score of 60 is −10

4.58 = −2.18.The Z score of 65 is −5

4.58 = −1.09.So P(60 ≤ Y ≤ 65) = P(Z < −1.09)− P(Z < −2.18) =.1379− .0146 = .1233.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation



You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).The Z score of 60 is −10

4.58 = −2.18.The Z score of 65 is −5

4.58 = −1.09.So P(60 ≤ Y ≤ 65) = P(Z < −1.09)− P(Z < −2.18) =.1379− .0146 = .1233.Actually P(60 ≤ X ≤ 65) = .15036.So .1233 is not that good an approximation.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation



You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).The Z score of 60 is −10

4.58 = −2.18.The Z score of 65 is −5

4.58 = −1.09.So P(60 ≤ Y ≤ 65) = P(Z < −1.09)− P(Z < −2.18) =.1379− .0146 = .1233.Actually P(60 ≤ X ≤ 65) = .15036.So .1233 is not that good an approximation.P(60 < X < 65) = .09509. would also be approximated by.1233!

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509

P(60 ≤ Y ≤ 65) = .1233is not that good an approximation to either.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509

P(60 ≤ Y ≤ 65) = .1233is not that good an approximation to either.

The problem is that for a continuous model like Y ,P(Y = 65) = 0.But P(X = 65) = .04678.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509

P(60 ≤ Y ≤ 65) = .1233So Y doesn’t care about < vs. ≤ . But X does!

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233Looking closely at the picture provides the key:

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233

It is really the area under the normal curve between 64.5 and65.5 which approximates P(X=65).

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233So P(60 ≤ X ≤ 65) = .15036 should be approximated byP(59.5 ≤ Y ≤ 65.5).

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233So P(60 ≤ X ≤ 65) = .15036 should be approximated byP(59.5 ≤ Y ≤ 65.5).The z-scores of 59.5 and 65.5 are−10.54.58 = −2.29 and −4.5

4.58 = −.98 resp.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233So P(60 ≤ X ≤ 65) = .15036 should be approximated byP(59.5 ≤ Y ≤ 65.5).The z-scores of 59.5 and 65.5 are−10.54.58 = −2.29 and −4.5

4.58 = −.98 resp.So P(59.5 ≤ Y ≤ 65.5) = P(−2.29 ≤ Z ≤ −.98) =.1635− .0110 = .1525.Much closer to P(60 ≤ X ≤ 65) = .15036!

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233So P(60 ≤ X ≤ 65) = .15036 should be approximated byP(59.5 ≤ Y ≤ 65.5).Similarly P(60.5 ≤ Y ≤ 64.5)= P(Z < −1.20)− P(Z < −2.07) = .1151− .0192 = .0959 isa great approximation to P(60 < X < 65) = .09509.

Math 1710Class 8

V2

Binomial


SobrietyCheckpoints

NormalDistribution


NormalApproximation


X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509P(60 ≤ Y ≤ 65) = .1233So P(60 ≤ X ≤ 65) = .15036 should be approximated byP(59.5 ≤ Y ≤ 65.5).This is called the continuity approximation.You needn’t use it on exams or homework!(We suggest you not.)

math 1710 class 8pi.math.cornell.edu/~web1710/slides/sep12_v2.pdf · 4 for betty. 3 for carla. but...

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