The average time to get an order at a restaurant is 15 minutes. We need to find the probability that you will receive your order in the first 10 minutes. The probability is given by the formula $P(a \le X \le b) = \int_a^b \frac{1}{\mu}e^{-\frac{x}{\mu}} dx$, where $\mu$ is the average value.

1. Problem Description

The average time to get an order at a restaurant is 15 minutes. We need to find the probability that you will receive your order in the first 10 minutes. The probability is given by the formula $P(a \le X \le b) = \int_a^b \frac{1}{\mu}e^{-\frac{x}{\mu}} dx$, where $\mu$ is the average value.

2. Solution Steps

The average time $\mu$ is given as 15 minutes. We want to find the probability that the order is received within the first 10 minutes. This means we want to find $P(0 \le X \le 10)$. So $a = 0$ and $b = 10$.

Using the given formula, we have:

$P(0 \le X \le 10) = \int_0^{10} \frac{1}{15}e^{-\frac{x}{15}} dx$

To solve the integral, let $u = -\frac{x}{15}$, so $du = -\frac{1}{15} dx$, which means $dx = -15 du$.

When $x = 0$, $u = -\frac{0}{15} = 0$.

When $x = 10$, $u = -\frac{10}{15} = -\frac{2}{3}$.

So the integral becomes:

$P(0 \le X \le 10) = \int_0^{-\frac{2}{3}} \frac{1}{15}e^{u} (-15) du$

$P(0 \le X \le 10) = -\int_0^{-\frac{2}{3}} e^{u} du$

$P(0 \le X \le 10) = -[e^{u}]_0^{-\frac{2}{3}}$

$P(0 \le X \le 10) = -(e^{-\frac{2}{3}} - e^0)$

$P(0 \le X \le 10) = -(e^{-\frac{2}{3}} - 1)$

$P(0 \le X \le 10) = 1 - e^{-\frac{2}{3}}$

$P(0 \le X \le 10) = 1 - e^{-0.6667}$

$P(0 \le X \le 10) \approx 1 - 0.5134$

$P(0 \le X \le 10) \approx 0.4866$

The closest answer is 0.

4

8

7.

3. Final Answer

C. 0.487