Answer :
Step-by-step explanation:
a)P( a customer has to wait more than 4 minutes )=P(X>4)=e-4/2.5 =0.202
b)probability that a customer is served within the first minute =P(X<1)=1-e-1/2.5 =0.3297
c) here 99th percentile corresponding value =-2.5*ln(1-0.99)=11.51 minutes ~ 12minutes
Answer:
a) 0.2018
b)0.329
c)x=5
Step-by-step explanation:
We are given that This process can be described using an exponential probability density function with a mean of 2.5.
So, [tex]\lambda = 2.5[/tex]
[tex]f(x)=\frac{1}{\lambda}e^{-\frac{1}{\lambda} x}[/tex]
Where x>0 , [tex]\lambda[/tex] > 0
a) Find the probability that a customer has to wait more than 4 minutes.
[tex]P(X>4)=\frac{1}{\lambda} \int\limits^{\infty} _{4} {e^{\frac{-1}{\lambda} x} \, dx[/tex]
[tex]P(X>4)=1-(1-e^{\frac{-4}{\lambda}})\\P(X>4)=e^{\frac{-4}{2.5}}\\P(X>4)=0.2018[/tex]
b)Find the probability that a customer is served within the first minute.
[tex]P(0<x<1)=1-e^{\frac{-1}{2.5}}=0.329[/tex]
c)The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she doesn't want to give away free hamburgers to more than 1% of her customers. What number of minutes should the advertisement use?
[tex]P(X>x) \leq 0.01\\e^{\frac{-x}{2.5}} \leq 0.01\\\frac{-x}{2.5} \leq log (0.01)\\x=-2.5 \times log (0.01)\\x=5[/tex]