The average lifetime of a light bulb is 6000 hours. We want to find the probability that the bulb will last for more than 3000 hours. The probability density function is given by $P(a \le X \le b) = \int_a^b \frac{1}{\mu} e^{-\frac{x}{\mu}} dx$, where $\mu$ is the average value.

1. Problem Description

The average lifetime of a light bulb is 6000 hours. We want to find the probability that the bulb will last for more than 3000 hours. The probability density function is given by $P(a \le X \le b) = \int_a^b \frac{1}{\mu} e^{-\frac{x}{\mu}} dx$, where $\mu$ is the average value.

2. Solution Steps

We are given that the average lifetime of a light bulb is $\mu = 6000$ hours. We want to find the probability that the bulb lasts more than 3000 hours. That is, we want to find $P(X > 3000)$. We can write this as $P(3000 \le X < \infty)$. Therefore, we have $a = 3000$ and $b = \infty$.

Using the given formula, we have:

$P(X > 3000) = \int_{3000}^{\infty} \frac{1}{6000} e^{-\frac{x}{6000}} dx$

To evaluate this integral, we find the antiderivative of $\frac{1}{6000} e^{-\frac{x}{6000}}$:

$\int \frac{1}{6000} e^{-\frac{x}{6000}} dx = -e^{-\frac{x}{6000}} + C$

Now we evaluate the definite integral:

$P(X > 3000) = \lim_{b \to \infty} \int_{3000}^{b} \frac{1}{6000} e^{-\frac{x}{6000}} dx = \lim_{b \to \infty} \left[ -e^{-\frac{x}{6000}} \right]_{3000}^{b} = \lim_{b \to \infty} \left( -e^{-\frac{b}{6000}} - (-e^{-\frac{3000}{6000}}) \right)$

$P(X > 3000) = \lim_{b \to \infty} \left( -e^{-\frac{b}{6000}} + e^{-\frac{1}{2}} \right)$

Since $\lim_{b \to \infty} e^{-\frac{b}{6000}} = 0$, we have:

$P(X > 3000) = 0 + e^{-\frac{1}{2}} = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \approx 0.6065$

3. Final Answer

The probability that the bulb will last for more than 3000 hours is $e^{-1/2} \approx 0.6065$.