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At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.

a. Find the probability that the time between two successive visits to the urgent care facility is less than 2minutes.

b. Find the probability that the time between two successive visits to the urgent care facility is more than 15minutes.

c. If 10minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes?

d. Find the probability that more than eight patients arrive during a half-hour period.

Short Answer

Expert verified

a. 0.2845

b. 0.1173

c. 0.4895

d.0.0311

Step by step solution

01

introduction

Probability is just the way that probably something is to occur. Whenever we're uncertain about the result of an occasion, we can discuss the probabilities of specific results — how likely they are. The investigation of occasions represented by likelihood is called insights.

02

Explanation (part a)

Let duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, and the decay constant is

The cdf is

the probability that the time between two successive visits to the urgent care facility is less than 2 minutes:

P(T<2)=1−1−e−27≈0.2485

03

Explanation (part b)

the probability that the time between two successive visits to the urgent care facility is more than 15 minutes is

P(T>15)=1−P(T<15)=1−1−e−157≈e−157≈0.1173mins

04

Explanation (part c)

If 10 minutes have passed since the last arrival, the probability that the next person will arrive within the next five minutes is

P(T>15∣T>10)=P(T>5)=1−1−e−57=e−57≈0.4895mins

05

Explanation (part d)

Let X= no. of patients arriving during a half-hour period.

Then Xhas the Poisson distribution with a mean of

307X~Poisson307

P(X>8)=1−P(X≤8)≈0.0311

the probability that more than eight patients arrive during a half-hour period is0.0311

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