Respuesta :
Answer: 0.3679
Step-by-step explanation:
The cumulative distribution function for exponential distribution is :-
[tex]P(x)=1-e^{-\lambda x}[/tex], where [tex]\lambda[/tex] is the mean of the distribution.
Given : A mean length of waiting time equal [tex] 2.8[/tex] minutes
i.e. Mean number of calls answered in a minute [tex]\lambda=\dfrac{1}{2.8}[/tex]
Now, the proportion of callers is put on hold longer than 2.8 minutes is given by :_
[tex]P(x>2.8)=1-P(x<2.8)\\\\=1-(1-e^{-\frac{1}{2.8}\times2.8})\\\\=1-1+e^{-1}\\\\=e^{-1}\\\\=0.367879441171\approx0.3679[/tex]
Hence, the proportion of callers is put on hold longer than 2.8 minutes = 0.3679
Using the exponential distribution, it is found that 0.3679 = 36.79% of callers is put on hold longer than 2.8 minutes.
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 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]
In this problem, the mean is of 2.8 minutes, hence:
[tex]m = 2.8, \mu = \frac{1}{2.8}[/tex]
The proportion that is put on hold longer than 2.8 minutes is given by:
[tex]P(X > 2.8) = e^{-\frac{1}{2.8} \times 2.8} = e^{-1} = 0.3679[/tex]
0.3679 = 36.79% of callers is put on hold longer than 2.8 minutes.
You can learn more about the exponential distribution at https://brainly.com/question/18596455
