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zlopas [31]
3 years ago
10

Jamal has a utility function u=w12, where w is his wealth in millions of dollars and u is the utility he obtains from that wealt

h. in the final stage of a game show, the host offers jamal a choice between (a) $4 million for sure, or (b) a gamble that pays $1 million with probability 0.6 and $9 million with probability 0.4.
Business
2 answers:
Zielflug [23.3K]3 years ago
8 0

Answer:

The expected value of both offers are:

Offer A's expected price = $4 million

Offer B's expected price = ($1 million x 0.6) + ($9 million x 0.4) = $4.2 million

Jamal's utility function U = W¹/²  or  U = √W

Offer A's expected utility = √$4,000,000 = 2,000 utils

Offer B's expected utility = √$4,200,000 = 2,049 utils

Both the difference in expected value and utility is not that large, but the difference in risk is great, so if I was Jamal I would choose option A.

GREYUIT [131]3 years ago
6 0

Answer: Jamal Should choose option B

Explanation:

The question is unclear with regards to the requirements, we will assume the question wants us to find an option that will maximize jamal's wealth.

Utility Function reflects Jamal's satisfaction that he derives from his wealth.

U = w12

a. Wealth (w) = $4million

U = 12w = (4000 000) x 12

U = 48000 000

b. Wealth =  $1 million with probability 0.6 and $9 million with probability 0.4.

We first need to calculate expected wealth before we can calculate how much utility will Jamal derive from this option

Expected Utility = 1 000 000 x 0.4 + 9000 000 x 0.4 = $4200 000

Utility = w12 = (4200 000) x 12 = 50 400 000

Jamal will have a utility of 48000 000 if he chooses option A and Option B provides Jamal with a Utility of 50 400 000. Option B provides a higher utility than option A, therefore the option that will maximize Jamal's Utility is Option B.

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The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almana
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Answer:

a) P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

b) P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

c) P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the scores for critical reading of a population, and for this case we know the distribution for X is given by:

X \sim N(502,100)  

Where \mu=502 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(502-10 < \bar X < 502+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

Part b

Let X the random variable that represent the scores for Math of a population, and for this case we know the distribution for X is given by:

X \sim N(515,100)  

Where \mu=515 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(515-10 < \bar X < 515+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

Part c

Let X the random variable that represent the scores for Writing of a population, and for this case we know the distribution for X is given by:

X \sim N(494,100)  

Where \mu=494 and \sigma=100

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)

And for this case we want this probability:

P(494-10 < \bar X < 494+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

3 0
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