math 1710 class 8pi.math.cornell.edu/~web1710/slides/sep12_v2.pdf · 4 for betty. 3 for carla. but...
Post on 25-Aug-2020
0 Views
Preview:
TRANSCRIPT
Math 1710Class 8
V2
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
Suppose P(Girl)=.6 and gender of births independent.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
Suppose P(Girl)=.6 and gender of births independent.P(3 Girls)?P(first 3 G, last 2 B)=?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
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?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Three Girls Out of Five Children
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
So answer is 10(.6)3(.4)2 = .3456.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girls
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girlsSo
C5,3 =
(53
)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girls
C5,3 =
(53
)=
5 · 4 · 31 · 2 · 3
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girls(53
)=
5 · 4 · 31 · 2 · 3
=5 · 41 · 2
= 10
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girls(53
)=
5 · 4 · 31 · 2 · 3
=5 · 41 · 2
= 10
Above showed (53
)=
(52
)which works in general as well.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why 10 Possible Orders?
5 birth positions, 3 of which girls(53
)=
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Why C5,3 = 5·4·31·2·3?
5 birth positions, 3 of which girls
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.)
First child a girl: GGGBB GGBGB GGBBG GBGGBGBGBG GBBGG
First child a boy: BGGGB BGGBG BGBGG BBGGG
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.)
First child a girl: GGGBB GGBGB GGBBG GBGGBGBGBG GBBGG
First child a boy: BGGGB BGGBG BGBGG BBGGG
(53
)=
(42
)+
(43
).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
General Terminology
Why µ = np, σ =√
npq for a Binomial(n,p) RV Y ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
General Terminology
Why µ = np, σ =√
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
General Terminology
Why µ = np, σ =√
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
General Terminology
Why µ = np, σ =√
npq for a Binomial(n,p) RV Y ?
Y = X1 + X2 + . . .+ Xn.
(Remember Y is the total number of successes.)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
General Terminology
Why µ = np, σ =√
npq for a Binomial(n,p) RV Y ?
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that exactly 65 report approval?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that exactly 65 report approval?Solution: Y = Binomial(100, .7)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that exactly 65 report approval?Solution: Y = Binomial(100, .7)
P(Y = 65) =?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that exactly 65 report approval?Solution: Y = Binomial(100, .7)
P(Y = 65) =
(10065
).765.335.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that exactly 65 report approval?Solution: Y = Binomial(100, .7)
P(Y = 65) =
(10065
).765.335.
A calculator could help with
(10065
).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Calculating Combinations
A calculator could help with
(10065
).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Calculating Combinations
A calculator could help with
(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.)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Calculating Combinations
A calculator could help with
(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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
The RV 2X ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
The RV 2X :
2X probability0 .52 .5
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
The RV 2X :
2X probability0 .52 .5
If X1 and X2 are independent copies of X , then X1 + X2 cancome out to 0,1, or 2.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
The RV 2X :
2X probability0 .52 .5
The RV X1 + X2 ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X1 + X2 vs. 2X
Tossing one fair coin is described by Bernoulli(.5):
X probability0 .51 .5
The RV 2X :
2X probability0 .52 .5
The RV X1 + X2:
X1 + X2 probability0 .251 .52 .25
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.
This is X1 + X2.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.
This is X1 + X2.
Method 2: Toss one coin.Then turn the other coin over to the same result.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.
This is X1 + X2.
Method 2: Toss one coin.Then turn the other coin over to the same result.
Note the second coin is still random.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
How many heads in total?
How can these two possibilities come up in tossing 2 coins?Method 1: Just toss them.
This is X1 + X2.
Method 2: Toss one coin.Then turn the other coin over to the same result.
This is 2X .
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
X1 + X2 frequency0 ?1 ?2 ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
X1 + X2 frequency0 ?1 ?2 ?
x̄ ,s?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
X1 + X2 frequency0 ?1 ?2 ?
These x̄ ,s should be close to µ = 1, σ =√
2(.5)(.5) = .707resp.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
2X frequency0 ?2 ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
2X frequency0 ?2 ?
x̄ ,s?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Experimentally X1 + X2 and 2X
2X frequency0 ?2 ?
These x̄ ,s should be close to µ = 1, σ = 2(.5) = 1 resp.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
There’s a normal distribution with any mean µ or σ > 0.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
There’s a normal distribution with any mean µ or σ > 0.
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
N(µ, σ)
Area corresponds to probability.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
N(µ, σ)
The entire area under the curve is 1.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
A Continuous Distribution
N(µ, σ)
The area between a and b is the probability of a value x fallingwithin that range.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The 68− 95− 99.7 Rule
68% within 1 standard deviation of the mean.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The 68− 95− 99.7 Rule
68% within 1 standard deviation of the mean.
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The 68− 95− 99.7 Rule
95% within 2 standard deviations of the mean.
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The 68− 95− 99.7 Rule
99.7% within 3 standard deviations of the mean.
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
Standard normal is the case µ = 0 and σ = 1.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
Standard normal is the case µ = 0 and σ = 1.
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
A general normal distribution:
N(µ, σ)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
N(µ, σ)
z =x − µσ
The Z score is just the offset from the mean in standarddeviation units.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
The Standard Normal Distribution
z = x−µσ
This transformation preserves area and probability.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
N(0,1)
Table Z gives us the area to the left on the standard normal.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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!
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
We can also get the area under the curve between two values.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
For example P(−.67 < Z < 1) =?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
For example P(−.67 < Z < 1) =?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
Table Z tells us P(−.67 < Z < 1) = .8413− .2514 = .5899.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
The reason for the subtractionP(−.67 < Z < 1) = P(Z < 1)− P(Z < −.67) is:
-
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
UsingTable Z
The reason for the subtractionP(−.67 < Z < 1) = P(Z < 1)− P(Z < −.67) is:
=
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Suppose a normal model N(24 mpg, 6 mpg) describes fuelefficiency of cars in a region:
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% of cars above 40?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Worst 20% of cars?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Gas mileage of the 5% most efficient?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Suppose a normal model N(24 mpg, 6 mpg) describes fuelefficiency of cars in a region:
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
N(24,6)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
N(24,6)
We understand a general N(µ, σ) by reducing to N(0, 1).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
N(24,6)
We understand a general N(µ, σ) by reducing to N(0, 1).The Z score of 15 is ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?The Z score of 15 is ?
z =15− 24
6= −1.5
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
z =15− 24
6= −1.5
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Percent of cars with mileage below 15?
N(0,1)
Table Z tells us P(Z < −1.5) = .0668.6.7% of cars will have mileage below 15 mpg.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
The Z score of 20 is ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
z1 =20− 24
6= −.67
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
The Z score of 20 is z1 = −.67?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
The Z score of 20 is z1 = −.67?The Z score of 30 is ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
The Z score of 20 is z1 = −.67?The Z score of 30 is ?
z2 =30− 24
6= 1
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% between 20 and 30?
N(24,6)
The Z score of 20 is z1 = −.67?The Z score of 30 is z2 = 1.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
The reason for the subtractionP(−.67 < Z < 1) = P(Z < 1)− P(Z < −.67) is:
-
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
The reason for the subtractionP(−.67 < Z < 1) = P(Z < 1)− P(Z < −.67) is:
=
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% of cars above 40?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% of cars above 40?
N(24,6)
The Z score of 40 is ?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% of cars above 40?
N(24,6)
The Z score of 40 is ?
z =40− 24
6= 2.67
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
% of cars above 40?
N(24,6)
z =40− 24
6= 2.67
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Worst 20% of cars?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Worst 20% of cars?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Worst 20% of cars?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?Again start with the N(0, 1) picture.
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?Again start with the N(0, 1) picture.
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Third Quartile?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Gas mileage of the 5% most efficient?
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Gas mileage of the 5% most efficient?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Gas mileage of the 5% most efficient?
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Mileage Example
Gas mileage of the 5% most efficient?
N(0,1)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)Solution: Let X = Binomial(100, .7)
X has meanµ = (100)(.7) = 70
andσ =
√(100)(.7)(.3) =
√21 = 4.58.
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)Solution: Let X = Binomial(100, .7)
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)Solution: Let X = Binomial(100, .7)
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.
Y = N(70, 4.58).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)Solution: Let X = Binomial(100, .7)
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.
Y = N(70, 4.58).We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
You poll 100 people.What is the probability that between 60 and 65 reportapproval?(including 60 and 65.)
X will then be approximated by a normal distribution Y withthe same mean and standard deviation.
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
Suppose 70% approve the President . . .
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
X = Binomial(100, .7) and Y = N(70, 4.58).
P(60 ≤ X ≤ 65) = .15036P(60 < X < 65) = .09509
Math 1710Class 8
V2
Binomial
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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
Two Ways of“Randomly”Flipping 2Coins
SobrietyCheckpoints
NormalDistribution
Working WithNormalDistributions
NormalApproximation
MakingNormalApproximationAccurate
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.)
top related