Question

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? Step 1 We need to find a so that P(X ≥ a) =

Answer 1

Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

Step 1

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

$f(t)=\left \{ {{0 ,\-t$

Consider the second function:

$f(t)=\frac{e^{-t/\mu}}{\mu}$

Where Average waiting time = μ = 2.5

The function f(t) becomes

$f(t)=0.4e^{-0.4t}$

Step 2

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

The probability that a costumer has to wait for more than x minutes is:

$\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt$

which is equal to 0.01

Step 3

Solve the equation for x

$\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01$

Take natural log on both sides

$ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53$

Results

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger