Respuesta :
Answer:
0.52763 is the probability that the time between the next two calls will be 54 seconds or less.
0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.
Step-by-step explanation:
We are given the following information in the question:
The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.
The distribution function can be written as:
[tex]f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}[/tex]
The probability for exponential distribution is given as:
[tex]P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}[/tex]
a) P( time between the next two calls will be 54 seconds or less)
[tex]P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763[/tex]
0.52763 is the probability that the time between the next two calls will be 54 seconds or less.
b) P(time between the next two calls will be greater than 118.5 seconds)
[tex]p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285[/tex]
0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.
