Question

a company that receives the majority of its orders by telephone conducted a study to determing how long customers were willing to wait on hold before ordering a product. The length of waiting time was found to be a variable best approximated by an exponential distribution with a mean length of waiting time equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order 0.51342 0.48658

Answer 1

Answer:

0.2231 = 22.31% of customers having to hold more than 4.5 minutes will hang up before placing an order

Step-by-step explanation:

Exponential distribution:

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

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

In which $\mu = \frac{1}{m}$ 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 length of waiting time was found to be a variable best approximated by an exponential distribution with a mean length of waiting time equal to 3 minutes.

This means that $m = 3, \mu = \frac{1}{3}$

What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order?

This is:

$P(X > 4.5) = e^{-\frac{1}{3}*4.5} = e^{-\frac{4.5}{3}} = e^{-1.5} = 0.2231$

0.2231 = 22.31% of customers having to hold more than 4.5 minutes will hang up before placing an order