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aev [14]
4 years ago
14

Suppose a bounty hunter captures a certain type of fugitive 62% of the time. On average, the apprehension of this type of crimin

al costs him $9,000 in expenses, but if he captures the criminal he receives a bounty (monetary reward) of $50,000.
Based ONLY on this information, should the bounty hunter take the job?
A) The expected value is $22,000 -- the bounty hunter should take the job.
B) The expected value is $27,580 -- the bounty hunter should take the job.
Eliminate
C) The expected value is −$22,000 -- the bounty hunter should not take the job.
D) The expected value is −$27,580 -- the bounty hunter should not take the job.
Mathematics
2 answers:
kiruha [24]4 years ago
8 0

Answer:

A) The expected value is $22,000 -- the bounty hunter should take the job.

Step-by-step explanation:

The expected value is  probability*reward- cost

Expected value =  .62* 50000 - cost

                             31000- 9000

                              22000

eimsori [14]4 years ago
4 0

Hi There!

Question - Suppose a bounty hunter captures a certain type of fugitive 62% of the time. On average, the apprehension of this type of criminal costs him $9,000 in expenses, but if he captures the criminal he receives a bounty (monetary reward) of $50,000.

Based ONLY on this information, should the bounty hunter take the job?

Solution:

62% = 0.62

50,000 * 0.62 = 31,000

31,000 - 9,000 = 22,00

Answer:

A) The expected value is $22,000 -- the bounty hunter should take the job.

Hope This Helps :)

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kipiarov [429]
  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

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4 years ago
If he is correct, what is the probability that the mean of a sample of 68 computers would differ from the population mean by les
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Complete Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad

Answer:

P(-1.72

Step-by-step explanation:

From the question we are told that:

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Sample Mean \=x =2.08

Standard Deviation \sigma=10

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From Table

P(|\=x-\mu|

T Test

Z=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }

Z=\frac{2.08}{\frac{10}{\sqrt{68} } }

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P(|\=x-\mu|

P(-1.72

Therefore From Table

P(-1.72

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