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Hunter-Best [27]
3 years ago
15

It has been observed that the average number of traffic accidents requiring medical assistance on the Hollywood Freeway between

7 and 8 a.m. on Wednesdays is 1. An ambulance can only service one accident requiring medical assistance at a time. The hospital dispatcher needs to know the chance there will be a need for exactly 2 ambulances on the Hollywood Freeway between 7 and 8 a.m. on Wednesdays. What will you, the statistics consultant hired by the hospital, tell the dispatcher
Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

0.184

Step-by-step explanation:

As the statistician consultant, I would have to calculate the chance of having 2 ambulances on the freeway at this hours and relay the message to the dispatcher.

Probability of having the need for 2 ambulances.

We will have a poisson distribution:

Lambda = 1

P(x=2) = e^-2*1/2!

= 2.71828^-1/2

= 0.368/2

= 0.184

I would tell the dispatch rider that the possibility that 2 ambulances would be required is 0.184.

Thank you

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Part A

<h3>Answer: T(n) = 58n+1800</h3>

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Work Shown:

1800 = initial cost or starting cost

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<h3>Answer: D(T) = 0.85T</h3>

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D = discounted cost

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Part C

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Work Shown:

We combine parts A and B.

T(n) = 58n+1800 from part A

D(T) = 0.85*T from part B

D(T(n)) = 0.85*( T(n) )

D(T(n)) = 0.85*( 58n+1800 ) ... plug in T(n) = 58n+1800

D(T(n)) = 0.85*( 58n+1800 )

D(T(n)) = 0.85*(58n)+0.85*(1800) ... distribute

D(T(n)) = 49.3n + 1530

If we plugged in n = 0, then,

D(T(n)) = 49.3n + 1530

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Or we could plug n = 0 into T(n) to get T(0) = 1800

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Either way you'll get the same answer.

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======================================

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