Solving Probability Density: Get Free Burger in 10 Mins

[johnhuntsman]

The manager of a fast food restaurant determines that the average time that her customers wait for their food is 2.5 minutes. The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. She doesn't want to give away free hamburgers to more than 2% of her customers. What should the advertisement say?

I'm solving for t:

\int_ 0.4e^{-t/2.5}~dt=0.02

-e^{-t/2.5}=0.02

0=0.02+e^{-t/2.5}

Take the natural logs and add -t/2.5 to get it back on the other side.

t/2.5=-3.91

t=-1.56

The answer is ten minutes. What did I do wrong?

 

[Ray Vickson]

The manager of a fast food restaurant determines that the average time that her customers wait for their food is 2.5 minutes. The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. She doesn't want to give away free hamburgers to more than 2% of her customers. What should the advertisement say?

I'm solving for t:

\int_ 0.4e^{-t/2.5}~dt=0.02

-e^{-t/2.5}=0.02

0=0.02+e^{-t/2.5}

Take the natural logs and add -t/2.5 to get it back on the other side.

t/2.5=-3.91

t=-1.56

The answer is ten minutes. What did I do wrong?

You cannot have exp(-t/2.5) = -0.02, since the exponential function is always > 0. You did the integration incorrectly.

RGV

 

[ehild]

The manager of a fast food restaurant determines that the average time that her customers wait for their food is 2.5 minutes. The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. She doesn't want to give away free hamburgers to more than 2% of her customers. What should the advertisement say?

I'm solving for t:

\int_ 0.4e^{-t/2.5}~dt=0.02

-e^{-t/2.5}=0.02

You need to show the limits of the integral and calculate with them. The probability that somebody is not served for x minutes is

\int _x^\infty{0.4 e^{-0.4 t}dt}=0.02

ehild