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katrin [286]
3 years ago
15

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 80 months, with a s

tandard deviation of 10 months. If he is correct, what is the probability that the mean of a sample of 73 computers would be less than 83.28 months? Round your answer to four decimal places.
Business
1 answer:
dangina [55]3 years ago
3 0

Answer:

0.4998

Explanation:

Given that:

Mean life of computer μ = 80

standard deviation σ  = 8

Mean of sample n = 73

x = 83.28

We have to find P(Mean of sample is less than 83.38 months) =

P(X<x=83.38), to do this we have to first determine the z score.

Since we are dealing with multiple samples, the z score will :

z = (x – μ) / (σ / √n)

z = (83.28 - 80)/ (8/√73)

z = 3.28/0.936

z = 3.5

P(X<x=83.38) = P ( Z < z = 3.5)

Use the standard normal table to find P ( Z < 3.5 )

We will have P (Z < 3.5 ) = 0.4998

The probability that the mean of a sample of 73 computers would be less than 83.28 months is 0.4998

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the supply of a good will be more elastic, the a. more the good is considered a luxury. b. broader is the definition of the mark
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The correct answer is Option D.

The longer the time period under consideration, the more elastic the supply of a good will be.

<h3>What is supply of goods ?</h3>
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To learn more about supply of goods refer to :

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Which of the following statements is INCORRECT. All else equal, 1. If a bond's yield-to-maturity (YTM, i.e., market interest rat
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1

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When the yield to maturity is less than the coupon rate, the bond is selling at a premium.

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How to pass a successful business message​
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Here are a few steps you could do!

Explanation:

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At a time when demand for ready-to-eat cereal was stagnant, a spokesperson for the cereal maker Kellogg’s was quoted as saying,
larisa86 [58]

Answer:

for interest rates equal to or lower than 200%, the firms will use trigger strategies to support the collusive level of advertising

Explanation:

Using the below expression to determine the range of interest rates could these firms use trigger strategies to support the collusive level of advertising; we have:

\frac{current \ period's \ profit \ of \ the \ cheating \ firm \ - \  firm's \  profit \  in \ each \ period \ under \ collision  }{ firm's \ profit \ in \  each \ period \ under \ collision \ - \ profit \ in  \ each \ subsequent \ period \ of \ cheating \ firm  } \leq \frac{1}{i}

where;

the \ current \ period's \ profit \ of \ the \ cheating \ firm \ = \ 49

firm's \  profit \  in \ each \ period \ under \ collision  = \  9

\ profit \ in  \ each \ subsequent \ period \ of \ cheating \ firm  } = \ 1

Then :

= \frac{49-9}{9-1} \leq \frac{1}{i}

= \frac{40}{8}  \leq \frac{1}{i}

i \leq \frac{8}{40}

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Thus; for interest rates equal to or lower than 200%, the firms will use trigger strategies to support the collusive level of advertising

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