Respuesta :
Answer:
The probability that the time between the next two calls is between 3 minutes and 7 minutes is 0.2442.
Step-by-step explanation:
Let X = time between calls made to Amazon's customer service.
The average time between calls is, β = 10 minutes.
The random variable X follows an Exponential distribution with parameter [tex]\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10[/tex]
The probability distribution function of X is:
[tex]f_{X}(x)=\left \{ {{\lambda e^{-\lambda x};\ x>0}} \atop {0;\ otherwise}} \right.[/tex]
Compute the probability that the time between the next two calls is between 3 minutes and 7 minutes as follows:
[tex]P(3<X<7)=\int\limits^{7}_{3} {\lambda e^{-\lambda x}} \, dx \\=\lambda \int\limits^{7}_{3} {e^{-\lambda x}} \, dx\\=\lambda \times |\frac{e^{-\lambda x}}{-\lambda}|^{7}_{3}\\=|-e^{-\lambda x}|^{7}_{3}\\=-e^{-0.10\times 7}+e^{-0.10\times 3}\\=-0.49659+0.74081\\=0.24422\\\approx0.2442[/tex]
Thus, the probability that the time between the next two calls is between 3 minutes and 7 minutes is 0.2442.
Answer:
Probability that the time between the next two calls is between 3 minutes and 7 minutes is 0.2442.
Step-by-step explanation:
We are given that the time between calls made to Amazon's customer service is exponentially distributed with a mean of 10 minutes.
Let X = time between calls made to Amazon's customer service
The probability distribution function of exponential distribution is given by;
[tex]f(x) = \lambda e^{-\lambda x} , x >0[/tex] where, [tex]\lambda[/tex] = parameter of distribution.
Now, the mean of exponential distribution is = [tex]\frac{1}{\lambda}[/tex] which is given to us as 10 minutes that means [tex]\lambda = \frac{1}{10}[/tex] .
So, X ~ Exp( [tex]\lambda = \frac{1}{10}[/tex] )
Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;
[tex]F(x) = P(X \leq x) = 1 - e^{-\lambda x}[/tex] , x > 0
Now, Probability that the time between the next two calls is between 3 minutes and 7 minutes is given by = P(3 min < X < 7 min)
P(3 min < X < 7 min) = P(X < 7 min) - P(X [tex]\leq[/tex] 3 min)
P(X < 7) = [tex]1 - e^{-\frac{1}{10} \times 7}[/tex] = 1 - 0.4966 = 0.5034 {Using CDF}
P(X [tex]\leq[/tex] 3) = [tex]1 - e^{-\frac{1}{10} \times 3}[/tex] = 1 - 0.7408 = 0.2592
Therefore, P(3 min < X < 7 min) = 0.5034 - 0.2592 = 0.2442.
Hence, probability that the time between the next two calls is between 3 minutes and 7 minutes is 0.2442.
