Chapter 2 Section 4 Question 3
Question
Let Z ~ $Exponential(4)$. Compute each of the following.
  1. $P(Z \geq 5)$
  2. $P(Z \geq -5)$
  3. $P(Z^2 \geq 9)$
  4. $P(Z^4-17 \geq 9)$
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 21 2020
From example 2.4.5 in the book we know: $$P(X\geq x)=\int_{x}^{\infty} \lambda e^{-\lambda z}dz=e^{-\lambda x}$$ In this case $\lambda =4$
(a) $P(Z \geq 5)$
$$P(Z \geq 5)=e^{-4*5}$$ $$P(Z \geq 5)=e^{-20}$$ $$P(Z \geq 5)\approx 0.00000000206$$
(b) $P(Z \geq -5)$
Since Example 2.4.5 says that f(x)=0 for any $x < 0$. We will instead take our integral for the domain $[0,\infty)$ this will be the same as solving for $P(Z \geq 0)$ $$P(Z \geq -5)=e^{-4*(0)}$$ $$P(Z \geq -5)=e^{0}$$ $$P(Z \geq -5)=1$$
(c) $P(Z^2 \geq 9)$
If $Z^2 \geq 9$ Then $Z \geq 3$ or $Z\leq -3$
We will solve this as two separate probabilities then add them together. $$P(Z \geq 3)=e^{-4*3}$$ $$P(Z \geq 3)=e^{-12}$$ $$P(Z \geq 3)\approx 0.000006144$$ For the other interval we first solve for $Z\geq 3$ then will do $1-P(Z\geq 3)$ to get $P(Z\leq 3)$
Since Example 2.4.5 says that f(x)=0 for any $x < 0$. We will instead take our integral for the domain $[0,\infty)$ this will be the same as solving for $P(Z \geq 0)$ $$P(Z \geq -3)=e^{-4*0}$$ $$P(Z \geq -3)=e^{0}$$ $$P(Z \geq -3)=1$$ we will now do 1- this to flip the inequality $$P(Z \leq -3)=1-P(Z \geq -3)$$ $$P(Z \leq -3)=1-1$$ $$P(Z \leq -3)=0$$ we will now add up the two probabilities: $$P(Z \geq 3)+P(Z \leq -3)$$ $$=e^{-12}+0$$ $$P(Z \geq 3)+P(Z \leq -3)=e^{-12}$$ $$P(Z \geq 3)+P(Z \leq -3)\approx 0.000006144$$
(d) $P(Z^4-17 \geq 9)$
$$Z^4-17 \geq 9$$ $$Z^4 \geq 26$$ $$Z \geq \sqrt[4]{26}$$ $$Z \leq -\sqrt[4]{26}$$ we will solve for both probabilities separately then add them together $$P(Z\geq \sqrt[4]{26})=e^{-4*(\sqrt[4]{26})}$$ $$P(Z\geq \sqrt[4]{26})\approx 0.00011947$$ For the other interval we first solve for $Z\geq -\sqrt[4]{26}$ then will do $1-P(Z\geq -\sqrt[4]{26})$ to get $P(Z\leq -\sqrt[4]{26})$ Since Example 2.4.5 says that f(x)=0 for any $x < 0$. We will instead take our integral for the domain $[0,\infty)$ this will be the same as solving for $P(Z \geq 0)$ $$P(Z\geq -\sqrt[4]{26})=e^{-4*0}$$ $$P(Z\geq -\sqrt[4]{26})=1$$ we will now do 1- this to flip the inequality $$P(Z\leq -\sqrt[4]{26})=1-P(Z\geq -\sqrt[4]{26})$$ $$P(Z\leq -\sqrt[4]{26})=1-1$$ $$P(Z\leq -\sqrt[4]{26})=0$$ we will now add up the two probabilities: $$P(Z\geq \sqrt[4]{26})+P(Z\leq -\sqrt[4]{26})$$ $$=e^{-4*(\sqrt[4]{26})}+0$$ $$=e^{-4*(\sqrt[4]{26})}$$ $$\approx 0.00011947$$