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Natali5045456 [20]
3 years ago
5

2-55 Nite Time Inn has a toll-free telephone number so that customers can call at any time to make a reservation. A typical call

takes about 4 minutes to complete, and the time required follows an exponential distribution. Find the probability that a call takes
Mathematics
1 answer:
Korvikt [17]3 years ago
7 0

The question is incomplete. Here is the complete question.

Nite Time Inn has a toll-free telephone number so that customers can call at any time to make a reservation. A typical call takes about 4 minutes to complete, and the time required follows an exponential distribution. find the probability that a call takes

a) 3 minutes or less

b) 4 minutes of less

c) 5 minutes of less

d) Longer than 5 minutes

e) Longer than 7 minutes

Answer: a) P(X<3) = 0.882

              b) P(X<4) = 0.908

              c) P(X<5) = 0.928

              d) P(X>5) = 0.286

              e) P(X>7) = 0.174

Step-by-step explanation: <u>Exponential</u> <u>distribution</u> is related with teh amount of time until some specific event happens.

If X is a continuous random variable, probability is calculated as:

P(X

in which:

m is decay parameter, given by: m=\frac{1}{mean}

For the Nite Time Inn calls:

m=\frac{1}{4}

m = 0.25

(a) P(X<3)

P(X

P(X

P(X

P(X < 3) = 0.882

<u>The probability the call takes less than 3 minutes is 0.882.</u>

(b) P(X<4)

P(X

P(X

P(X < 4) = 0.908

<u>The probability the call takes less tahn 4 minutes is 0.908.</u>

(c) P(X<5)

P(X

P(X

P(X < 5) = 0.928

<u>The probability of calls taking less than 5 minutes is 0.928.</u>

(d) P(X>5)

Knowing that the sum of probabilities of less than and more than has to equal 1:

P(X<x) + P(X>x) = 1

P(X>x) = 1 - P(PX<x)

P(X>x) = 1-(1-me^{-m*x})

P(X>x)=me^{-mx}

For P(X>5):

P(X>5) = 0.25e^{-1.25}

P(X > 5) = 0.286

<u>The probability of calls taking more than 5 minutes is 0.286.</u>

(e) P(X>7)

P(X>7)=0.25e^{-0.25.7}

P(X>7)=0.25e^{-1.75}

P(X > 7) = 0.174

<u>The probability of calls taking more than 7 minutes is 0.174.</u>

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Answer:

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