1 Answer

Jatin Gadhiya · Answer 1 · 2022-01-25T16:00:56+0000

Answer:

The advertisement should use 16 minutes.

Explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^(-\mu x)$

In which
$\mu = (1)/(m)$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^(-\mu x)$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^(-\mu x)) = e^(-\mu x)$

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.

This means that
$m = 3.5, \mu = (1)/(3.5) = 0.2857$

What number of minutes should the advertisement use?

The values of x for which:

P(X > x) = 0.01

So

$e^(-\mu x) = 0.01$

e^(-0.2857x) = 0.01

$\ln{e^(-0.2857x)} = ln(0.01)$

-0.2857x = ln(0.01)

x = -(ln(0.01))/(0.2857)

x = 16.12

Rounding to the nearest number, the advertisement should use 16 minutes.

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