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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 4.8 / Problem 117E

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

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The time between calls to a plumbing supply business is

Chapter 4, Problem 117E

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QUESTION:

Problem 117E

The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes.

(a) What is the probability that there are no calls within a 30-minute interval? (b) What is the probability that at least one call arrives within a 10-minute interval?

(c) What is the probability that the first call arrives within 5 and 10 minutes after opening?

(d) Determine the length of an interval of time such that the probability of at least one call in the interval is 0.90.

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QUESTION:

Problem 117E

The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes.

(a) What is the probability that there are no calls within a 30-minute interval? (b) What is the probability that at least one call arrives within a 10-minute interval?

(c) What is the probability that the first call arrives within 5 and 10 minutes after opening?

(d) Determine the length of an interval of time such that the probability of at least one call in the interval is 0.90.

ANSWER:

Solution :

Step 1 of 4:

We know that the mean time between calls of 15 minutes.

Let X denotes the time until first call.

Then the exponential with mean $\lambda{}$ is

$\lambda{}=$ $\frac{1}{E(X)}$

$\lambda{}=$ $\frac{1}{15}$ call per min.

Our goal is:

a). We need to find the probability of no calls within a 30-minute interval.

b). We need to find the probability of at least one call arrives within a 10-minute interval.

c). We need to find the probability of the first call arrives within 5 and 10 minutes.

d). We need to determine the length of an interval of time .

a). The probability of no calls within a 30-minute interval is

P(X>30) = $\int\limits_{30}^{\infty{}}F(x)\ dx$

P(X>30) = $\int\limits_{30}^{\infty{}}\lambda{}{e}^{-\lambda{}x}\ dx$

P(X>30) = $\lambda{}\int\limits_{30}^{\infty{}}\ {e}^{-\lambda{}x}\ dx$

P(X>30) = $\frac{1}{15}\int\limits_{30}^{\infty{}}{e}^{-\frac{x}{15}}\ dx$

P(X>30) = $\frac{1}{15}\int\limits_{30}^{\infty{}}-15\ {e}^{-\frac{x}{15}}\ dx$

P(X>30) = ${\left[{-{e}^{-\frac{x}{15}}}\right]}_{30}^{\infty{}}$

P(X>30) = $\left[{{e}^{-\frac{30}{15}}}\right]$

P(X>30) = ${e}^{-2}$

P(X>30) = 0.1353

Therefore, the probability of no calls within a 30-minute interval is 0.1353.

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