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kari74 [83]
3 years ago
10

Guests at a Beaumont hotel stay for an average of 9 days with a standard deviation of 2.4 days. Assume that the length of stay i

s normally distributed. Assume that we select randomly 1000 guests. How many of those guests can be expected to stay between 7 and 14 days
Mathematics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

778

Step-by-step explanation:

We have the length of stay as X

X~N (9,(2.4))

The sample is given as n = 1000

P(7<x<14) is what we are trying to get

P(7-9/2.4 < x-9/2.4 < 14-9/2.4)

= P(-0.83<z<2.08)

P(z<2.08)-p(z<-0.83)

When we go to the z table,

0.981-0.0203

= 0.778

0.778x1000

= 778

So in conclusion, 778 of these guests can be expected to stay between 7 and 14 days.

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Your 71,1482 doesn't make sense. But regardless,

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You know the total spent and the cost of one car. So logically, if you subtract the two you'd get the price of the second car.

You can do $38,295 + c = $71,1482.

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The picture of the question in the attached figure

Let

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