Chapter 2 Section 4 Question 2
Question
U ~ $Uniform[1, 4]$. Compute each of the following.
1. $P(W \geq 5)$
2. $P(W \geq 2)$
3. $P(W^2 \geq 9)$ (Hint: If $W^2 \leq 9$, what must W be?)
4. $P(W^2 \leq 2)$
From example 2.4.3 in the book we know: $$\textit{f}(x)=\begin{cases}\frac{1}{R-L}&L\leq x\leq R \\ 0& otherwise\end{cases}$$ In this case $R=4,L=1$
(a) $P(W \geq 5)$
since our probability will be equal to the area of a rectangle (base*height) where $W \geq 5$. In this case our base will be the intersection of $[5,\infty)\cap [1,4]=\emptyset$ since none of our functions domain intersects with the uniform distribution domain. We can conclude $P(W \geq 5)=0$
(b) $P(W \geq 2)$
since our probability will be equal to the area of a rectangle (base*height) where $W \geq 5$. In this case our base will be the intersection of $[2,\infty)\cap [1,4]=[2,4]$ The length of our base is therefore $4-2=2$. Our height is given by $\frac{1}{4-1}=\frac{1}{3}$ $$area=base \times height$$ $$P(W \geq 2)=2\times\frac{1}{3}$$ $$P(W \geq 2)=\frac{2}{3}$$
(c) $P(W^2 \geq 9)$ (Hint: If $W^2 \leq 9$, what must W be?)
If $W^2 \geq 9$ Then $W \geq 3$ or $W\leq -3$
Since our probability will be equal to the area of a rectangle (base*height) where $W^2 \geq 9$. In this case our base will be the intersection of $((-\infty,-3]\cup [3,\infty))\cap[1,4]=[3,4]$ The length of our base is therefore $4-3=1$. Our height is given by $\frac{1}{4-1}=\frac{1}{3}$ $$area=base \times height$$ $$P(W^2 \geq 9)=1\times\frac{1}{3}$$ $$P(W^2 \geq 9)=\frac{1}{3}$$
(d) $P(W^2 \leq 2)$
If $W^2 \leq 2$ Then $W \leq \sqrt{2}$ and $W\geq -\sqrt{2}$ $$W\in (-\infty,\sqrt{2}]\cap[-\sqrt{2},\infty)$$ $$W\in [-\sqrt{2},\sqrt{2}]$$ Since our probability will be equal to the area of a rectangle (base*height) where $W^2 \leq 2$. In this case our base will be the intersection of $[-\sqrt{2},\sqrt{2}]\cap[1,4]=[1,\sqrt{2}]$ The length of our base is therefore $\sqrt{2}-1\approx 0.4142$. Our height is given by $\frac{1}{4-1}=\frac{1}{3}$ $$area=base \times height$$ $$P(W^2 \leq 2)=0.4142 \times\frac{1}{3}$$ $$P(W^2 \leq 2)\approx 0.138$$