Question

asked Jun 1, 2021 180k views

1 Answer

Wrosecrans · Answer 1 · 2021-06-06T03:54:33+0000

Answer:

a) 0.426 = 42.6% probability that a customer waits less than a second for credit card approval.

b) 0.189 = 18.9% probability that a customer waits more than 3 seconds.

c) The minimum approval time for the slowest 5% of transactions is 5.39 seconds.

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:

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

An online retailer has determined that the average time for credit card transactions to be electronically approved is 1.8 seconds.

This means that
$m = 1.8, \mu = (1)/(1.8) = 0.5555$

(a) Use an exponential density function to find the probability that a customer waits less than a second for credit card approval.

This is
$P(X \leq 1)$ .

$P(X \leq 1) = 1 - e^(-0.5555*1) = 0.426$

0.426 = 42.6% probability that a customer waits less than a second for credit card approval.

(b) Find the probability that a customer waits more than 3 seconds.

This is P(X > 3).

$P(X > x) = 1 - P(X \leq 3) = 1 - (1 - e^(-0.5555*3)) = 0.189$

0.189 = 18.9% probability that a customer waits more than 3 seconds.

(c) What is the minimum approval time for the slowest 5% of transactions

This is x for which
$P(X \leq x) = 0.95$

So

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

0.95 = 1 - e^(-0.5555x)

e^(-0.5555x) = 0.05

$\ln{e^(-0.5555x)} = ln(0.05)$

-0.5555x = ln(0.05)

x = -(ln(0.05))/(0.5555)

x = 5.39

The minimum approval time for the slowest 5% of transactions is 5.39 seconds.

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