Consider the following exponential probability density function. \[ f(x)=\frac{1}{4} e^{-x / 4} \quad \text { for } x \geq 0 \] (a) Write the formula for $P\left(x \leq x_{0}\right)$. (b) Find $P(x \leq 3)$. (Round your answer to four decimal places.) (c) Find $P(x \geq 4)$. (Round your answer to four decimal places.) (d) Find $P(x \leq 6)$. (Round your answer to four decimal places.) (e) Find $P(3 \leq x \leq 6)$. (Round your answer to four decimal places.)

Step 1 ：The formula for \(P\left(x \leq x_{0}\right)\) is given by the integral of the probability density function from 0 to \(x_0\). Therefore, we have: \[P\left(x \leq x_{0}\right) = \int_{0}^{x_0} f(x) dx = \int_{0}^{x_0} \frac{1}{4} e^{-x / 4} dx\]

Step 2 ：To find \(P(x \leq 3)\), we substitute \(x_0 = 3\) into the formula from the previous step: \[P(x \leq 3) = \int_{0}^{3} \frac{1}{4} e^{-x / 4} dx = 1 - e^{-3 / 4}\] The result is approximately \(\boxed{0.5276}\)

Step 3 ：To find \(P(x \geq 4)\), we note that this is equivalent to \(1 - P(x \leq 4)\). Therefore, we substitute \(x_0 = 4\) into the formula from the first step and subtract the result from 1: \[P(x \geq 4) = 1 - \int_{0}^{4} \frac{1}{4} e^{-x / 4} dx = 1 - (1 - e^{-4 / 4}) = e^{-1}\] The result is approximately \(\boxed{0.3679}\)

Step 4 ：To find \(P(x \leq 6)\), we substitute \(x_0 = 6\) into the formula from the first step: \[P(x \leq 6) = \int_{0}^{6} \frac{1}{4} e^{-x / 4} dx = 1 - e^{-6 / 4}\] The result is approximately \(\boxed{0.7135}\)

Step 5 ：To find \(P(3 \leq x \leq 6)\), we subtract \(P(x \leq 3)\) from \(P(x \leq 6)\): \[P(3 \leq x \leq 6) = P(x \leq 6) - P(x \leq 3) = (1 - e^{-6 / 4}) - (1 - e^{-3 / 4}) = e^{-3 / 4} - e^{-6 / 4}\] The result is approximately \(\boxed{0.1859}\)