Respuesta :
Answer:
B) 0.60653
Step-by-step explanation:
The exponential distribution has one parameter which is lambda denoted as λ.
λ = 1/β; where β represents the population mean.
The cumulative distribution function is:
F(X;λ) = [tex]1 - e^{-lambdax}[/tex]
[tex]=1 - e^{-\frac{1}{\beta}x}[/tex]
The average waiting time is β = 3 min.
The proportion of customers willing to wait on hold for more than 1.5 minutes is:
P(X>1.5) = 1-P(X[tex]\leq[/tex]1.5)
[tex]= 1-(1-e^{-1.5*(1/3)} )\\= 1-1+e^{-1.5/3} \\=0.60653[/tex]
The proportion of customers having to hold more than 1.5 minutes will hang up before placing an order is option b.
Given information:
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 (i.e. the mean number of calls answered in a minute is ).
Calculation of the proportion:
[tex]P(X>1.5) = 1 - P(X \leq 1.5)[/tex]
[tex]= 1 - (1 - e^{1.5\times (1\div 3)}\\\\= 1 -1 + e^{-1.5\div 3}[/tex]
= 0.60653
learn more about the time here: https://brainly.com/question/24335111
