Question 2.4.1
Let \(W \sim \text{Uniform}[1,4]\). Compute each of the following.
\(\mathbb{P}(W \geq 5)\)
\(\mathbb{P}(W \geq 2)\)
\(\mathbb{P}(W^2 \leq 9)\) (Hint: If \(W^2 \leq 9\), what must \(W\) be?)
\(\mathbb{P}(W^2 \leq 2)\)
\[
\mathbb{P}(W \geq 5) = 0
\]
\[
\mathbb{P}(W \geq 2) = 2/3
\]
\[
\mathbb{P}(W^2 \leq 9) = \mathbb{P}(W \leq 3) = 2/3
\]
\[
\mathbb{P}(W^2 \leq 2) = \mathbb{P}(W \leq \sqrt{2}) = (\sqrt{2} - 1)/3
\]
Question 2.4.2
Establish for which constants \(c\) the following functions are densities.
\(f(x) = cx\) on \((0,1)\) and \(0\) otherwise.
\(f(x) = cx^{1/2}\) on \((0,2)\) and \(0\) otherwise.
\[
1 = \int_0^1 cx \, dx = c/2
\] so \(c = 2\).
\[
1 = \int_0^2 cx^{1/2} \, dx = c(2/3)x^{3/2}\Big|_0^2 = c(2/3)2^{3/2}
\] so \(c = (3/2)2^{-3/2} = 3/(4\sqrt{2})\).
Question 2.4.3
Let \(X \sim \text{Exponential}(3)\). Compute each of the following.
\(\mathbb{P}(0 < X < 1)\)
\(\mathbb{P}(0 < X < 3)\)
\(\mathbb{P}(2 < X < 5)\)
\(\mathbb{P}(X > 2)\)
\[
\mathbb{P}(0 < X < 1) = \int_0^1 3e^{-3t} \, dt= 1 - e^{-3} = 0.95021
\]
\[
\mathbb{P}(0 < X < 3) = \int_0^3 3e^{-3t} \, dt = 1 - e^{-9} = 0.99988
\]
\[
\begin{aligned}
\mathbb{P}(2 < X < 5) &= \int_2^5 3e^{-3t} \, dt \\
&= - e^{-15} - (- e^{-6}) \\
&\approx 0.002
\end{aligned}
\]
\[
\begin{aligned}
\mathbb{P}(X > 2)
&= \int_2^{\infty} 3e^{-3t} \, dt \\
&= e^{-6} \\
&\approx 0.002
\end{aligned}
\]
Could also solve for \(1 - \mathbb{P}(0 < X \leq 2)\).
Question: 2.4.4
A customer service call center receives on average, 4 calls per hour.
What is the probability that the time until the next call is less than 15 minutes?
What is the probability that the time until the next call is more than 30 minutes (i.e., 0.5 hours)?
Let \(T\) be the time (in hours) until the next call. Then \(T \sim \text{Exponential}(\lambda = 4)\). 15 minutes = 0.25 hours.
We want \(\mathbb{P}(T < 0.25)\): \[
\begin{aligned}
\mathbb{P}(T < 0.25) &= \int_0^{0.25} 4e^{-4t} dt \\
&= 1 - e^{-1} \\
&\approx 0.6321
\end{aligned}
\] Therefore, the probability that the next call arrives within 15 minutes is \(\boldsymbol{\approx 0.6321}\).
We want \(\mathbb{P}(T > 0.5)\): \[
\begin{aligned}
\mathbb{P}(T > 0.5) &= 1 - \mathbb{P}(T \leq 0.5) \\
&= 1 - \int_0^{0.5} 4e^{-4t} dt \\
&= 1 - ( 1 - e^{-2}) \\
&\approx 0.1353
\end{aligned}
\] Therefore, the probability that no call arrives in the next 30 minutes is \(\boldsymbol{\approx 0.1353}\).
Question 2.4.5
Suppose that the time (in hours) it takes to complete a customer’s order at a tailor follows a Gamma distribution parameters \(\alpha = 3\) and rate parameter \(\lambda = 2\).
What is the probability that the tailor’s next order will take less than 2 hours to complete?
Let \(T\) be the time (in hours) to complete an order. Then \(T \sim \text{Gamma}(\alpha = 3, \lambda = 2)\).
Since \(\alpha = 3\) is a positive integer, the cumulative distribution function has the closed form \[
\mathbb{P}(T \leq t) = 1 - \sum_{n=0}^{\alpha - 1} \frac{(\lambda t)^n e^{-\lambda t}}{n!}
\]
We want \(\mathbb{P}(T < 2)\), so with \(\lambda t = (2)(2) = 4\): \[
\begin{aligned}
\mathbb{P}(T < 2) &= 1 - \sum_{n=0}^{2} \frac{(4)^n e^{-4}}{n!} \\
&= 1 - e^{-4}\left(1 + 4 + \frac{4^2}{2}\right) \\
&= 1 - e^{-4}(1 + 4 + 8) \\
&= 1 - 13e^{-4} \\
&= 1 - 13(0.018316) \\
&\approx 1 - 0.2381 \\
&\approx 0.7619
\end{aligned}
\] Therefore, the probability that the tailor’s next order takes less than 2 hours to complete is \(\boldsymbol{\approx 0.7619}\).