Respuesta :
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:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
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 [tex]m = 3, \mu = \frac{1}{3}[/tex]
What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order?
This is:
[tex]P(X > 4.5) = e^{-\frac{1}{3}*4.5} = e^{-\frac{4.5}{3}} = e^{-1.5} = 0.2231[/tex]
0.2231 = 22.31% of customers having to hold more than 4.5 minutes will hang up before placing an order
