Answer:
The probability that a randomly selected call time will be less than 30 seconds is 0.7443.
Step-by-step explanation:
We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.
Let X = caller times at a customer service center
The probability distribution (pdf) of the exponential distribution is given by;
[tex]f(x) = \lambda e^{-\lambda x} ; x > 0[/tex]
Here, [tex]\lambda[/tex] = exponential parameter
Now, the mean of the exponential distribution is given by;
Mean = [tex]\frac{1}{\lambda}[/tex]
So, [tex]22=\frac{1}{\lambda}[/tex] ⇒ [tex]\lambda=\frac{1}{22}[/tex]
SO, X ~ Exp([tex]\lambda=\frac{1}{22}[/tex])
To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;
[tex]P(X\leq x) = 1 - e^{-\lambda x}[/tex] ; x > 0
Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)
P(X < 30) = [tex]1 - e^{-\frac{1}{22} \times 30}[/tex]
= 1 - 0.2557
= 0.7443