![Page 1: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/1.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe
that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 2: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/2.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure.
Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 3: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/3.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1
box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 4: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/4.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81
we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 5: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/5.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 6: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/6.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 7: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/7.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90)
=
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 8: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/8.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 9: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/9.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation
page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 10: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/10.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 11: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/11.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)
=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 12: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/12.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(
90 + 12− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 13: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/13.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 +
12− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 14: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/14.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2−
0.45 · 200√200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 15: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/15.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200
√200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 16: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/16.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 17: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/17.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
=
Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 18: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/18.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07)
= 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 19: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/19.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1
= 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 20: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/20.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 21: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/21.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 22: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/22.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure)
= P(∪200
i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 23: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/23.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)
=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 24: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/24.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(
200 + 12− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 25: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/25.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 +
12− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 26: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/26.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2−
0.45 · 200√200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 27: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/27.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200
√200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 28: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/28.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681
![Page 29: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/29.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)=
1−Φ(1.49) = 0.0681
![Page 30: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/30.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49)
= 0.0681
![Page 31: BFN/bfnIMM - DTU 02405 Probability 2004-5-16 BFN/bfn We introduce the events O i to describe that i voters in the survey oppose the measure. From section 2.1 box at bottom of page](https://reader036.vdocuments.site/reader036/viewer/2022070214/61171bd2d0136f67d7159c43/html5/thumbnails/31.jpg)
IMM - DTU 02405 Probability
2004-5-16BFN/bfn
We introduce the events Oi to describe that i voters in the survey oppose the measure. Fromsection 2.1 box at bottom of page 81 we deduce that X is binomially distributed.
Question a) The probability in question is (page 81)
P (O90) =
(200
90
)0.45900.55110
We evaluate this probability by approximation page 99.
P (O90)=̃Φ
(90 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)− Φ
(90− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)
= Φ(0.07)− Φ(−0.07) = 2 · Φ(0.07)− 1 = 0.056
Question b) The probability in question is
P (more than 100 voters oppose the measure) = P(∪200i=101Oi
)=̃
Φ
(200 + 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)−Φ
(101− 1
2− 0.45 · 200√
200 · 0.45 · (1− 0.45)
)= 1−Φ(1.49) = 0.0681