Solving Probability Density: Get Free Burger in 10 Mins

johnhuntsman
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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?
 
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johnhuntsman said:
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
 
johnhuntsman said:
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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