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Karo-lina-s [1.5K]
3 years ago
9

Consider a simple example of moral hazard. Suppose that Woodrow goes into a casino to make one bet a day. The casino is very bas

ic; it has two bets: a safe bet and a risky bet. In the safe bet, a nickel is flipped. If the nickel lands on heads, Woodrow wins $ 100 . If it lands on tails, Woodrow loses $ 100 . The risky bet is similar: a silver dollar is flipped. If the silver dollar lands on heads, Woodrow wins $ 5,000 . If it lands on tails, Woodrow loses $ 10,000 . Each coin has a 50 % chance of landing on each side. What is the expected value of the safe bet?
Mathematics
1 answer:
Neko [114]3 years ago
6 0

Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

If  

S=\left\{s_1,s_2,...,s_n\right\}

is a finite numeric sample space and

P(X=s_k)=p_k for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X)

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,  

P(X=100)=0.5

P(X=-100)=0.5

and the expected value is

E(X) = 100*0.5 - 100*0.5 = 0

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Perpendicular lines have negative reciprocal slopes. All that means is " flip " the slope and change the sign.
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so the perpendicular slope would be 1/2.
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3 years ago
In studies for a​ medication, 3 percent of patients gained weight as a side effect. Suppose 643 patients are randomly selected.
timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

p = 0.03

q = 1 - 0.03 = 0.97

The mean is

\mu  = np

\mu = 643 \times 0.03 = 19.29

The standard deviation is

\sigma =  \sqrt{npq}

\sigma =  \sqrt{643 \times 0.03 \times 0.97}

\sigma =4.33

We want to find the probability that exactly 24 patients will gain weight as side effect.

P(X=24)

We apply the Continuity Correction Factor(CCF)

P(24-0.5<X<24+0.5)=P(23.5<X<24.5)

We convert to z-scores.

P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \:  \frac{24.5 - 19.29}{4.33} ) \\  = P( 0.97\: < \: z \: < \:  1.20) \\  = 0.051

Part b) We want to find the probability that 24 or fewer patients will gain weight as a side effect.

P(X≤24)

We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

P(X \: < \: 24.5) = P(z \: < \:  \frac{24.5 - 19.29}{4.33} )  \\ =   P(z \: < \: 1.20)  \\  = 0.8849

Part c)

We want to find the probability that

11 or more patients will gain weight as a side effect.

P(X≥11)

Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

P(X \: > \: 10.5) = P(z \: > \:  \frac{10.5 - 19.29}{4.33} )  \\ = P(z \: > \:  - 2.03)

= 0.9788

Part d)

We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

P(23.5≤X≤28.5)

Convert to z-scores:

P(23.5  \:  <  \: X \:  <  \: 28.5) = P( \frac{23.5 - 19.29}{4.33}   \:  <  \: z \:  <  \:  \frac{28.5 - 19.29}{4.33} ) \\  = P( 0.97\:  <  \: z \:  <  \: 2.13) \\  = 0.1494

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3 years ago
What is 4 times more than 4
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3 years ago
A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that n = 15, p = \frac{1}{5} = 0.2

What is the probability that he answers at least 10 questions correctly?

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

0.0111% probability that he answers at least 10 questions correctly

3 0
3 years ago
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