Random variables

Unit 9 Random variables

Key unit competence: Explore the parameters of a random variable (discrete or continuous) for a given context.

Key ideas

• A random variable $X$ assigns a number to each outcome. It is discrete if it takes isolated values, continuous if it takes all values in an interval.
• Discrete: probability function $p(x)=P(X=x)$ with $p(x)\ge 0$ and $\sum p(x)=1$. Cumulative distribution function $F(x)=P(X\le x)$.
• Expected value $E(X)=\sum xp(x)$; variance $\mathrm{Var}(X)=E(X^2)-[E(X){]}^2$; standard deviation $\sigma =\sqrt{\mathrm{Var}(X)}$. $E(aX+b)=aE(X)+b$, $\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)$.
• Binomial $X\sim B(n,p)$: $n$ independent trials, success probability $p$: $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$ with $q=1-p$; $E(X)=np$, $\mathrm{Var}(X)=npq$.
• Poisson $X\sim \mathcal{P}(\lambda )$ (number of events in an interval): $P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$; $E(X)=\mathrm{Var}(X)=\lambda$.
• Continuous: probability density function $f(x)\ge 0$, ${\int }_{-\infty }^{+\infty }f(x)dx=1$, $P(a\le X\le b)={\int }_a^{b}f(x)dx$, $P(X=a)=0$. $F(x)={\int }_{-\infty }^{x}f(t)dt$ and $f=F^{\prime }$.
• $E(X)=\int xf(x)dx$, $\mathrm{Var}(X)=\int x^{2}f(x)dx-{\mu }^2$. Uniform on $[a,b]$: $f=\frac{1}{b-a}$, $E=\frac{a+b}{2}$, $\mathrm{Var}=\frac{(b-a{)}^2}{12}$.

Formulas to know

Common misconceptions

• "$\mathrm{Var}(3X)=3\mathrm{Var}(X)$." No: $\mathrm{Var}(aX)=a^2\mathrm{Var}(X)$.
• "For a continuous variable, $P(X=a)=f(a)$." No: $P(X=a)=0$; $f(a)$ is a density, not a probability.
• "$\mathrm{Var}(X)=E(X^2)$." Subtract $[E(X){]}^2$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Two dice are thrown and $X$ is the number of sixes. Give the distribution of $X$ and $E(X)$.

   Show answer

   $P(0)=\frac{25}{36}$, $P(1)=\frac{10}{36}$, $P(2)=\frac{1}{36}$. $E(X)=\frac{10+2}{36}=\frac{1}{3}$.

   Common mistake: Taking $P(1)=\frac{5}{36}$ (forgetting the two orders).

2. A bus comes at a random time in the next 20 minutes (uniform). Find the probability of a delay of more than 15 minutes, and the mean and variance of the delay.

   Show answer

   $P(X>15)=\frac{5}{20}=\frac{1}{4}$. $E(X)=10$ min, $\mathrm{Var}(X)=\frac{400}{12}=\frac{100}{3}$.

   Common mistake: Using $\frac{(b-a{)}^2}{2}$ for the variance.

3. $p(x)=\frac{x}{15}$ for $x=1,2,3,4,5$. Check it is a probability function and find $E(X)$ and $\mathrm{Var}(X)$.

   Show answer

   $\frac{15}{15}=1$. $E(X)=\frac{55}{15}=\frac{11}{3}$. $E(X^2)=\frac{225}{15}=15$. $\mathrm{Var}(X)=15-\frac{121}{9}=\frac{14}{9}$.

   Common mistake: Forgetting to square $E(X)$.

4. $f(x)=2x$ on $[0,1]$. Find the cumulative distribution function $F$ and the median.

   Show answer

   $F(x)=0$ for $x<0$, $F(x)=x^2$ for $0\le x\le 1$, $F(x)=1$ for $x>1$. Median: $x^2=\frac{1}{2}$, $x=\frac{1}{\sqrt{2}}\approx 0.707$.

   Common mistake: Giving only the middle piece of $F$.

5. A die is thrown 6 times. Find the probability of at least one six.

   Show answer

   $1-{\left(\frac{5}{6}\right)}^6\approx 0.665$.

   Common mistake: Answering $6\times \frac{1}{6}=1$.

6. $f(x)=k(4-x^2)$ on $[0,2]$. Find $k$ and $P(X\le 1)$.

   Show answer

   $k\left(8-\frac{8}{3}\right)=1$, $k=\frac{3}{16}$. $P(X\le 1)=\frac{3}{16}\left(4-\frac{1}{3}\right)=\frac{11}{16}$.

   Common mistake: Integrating over $[-2,2]$.

7. A test has 10 questions with 4 options each. A student guesses every answer. Find the expected score and $P(X=5)$.

   Show answer

   $X\sim B(10,0.25)$: $E(X)=2.5$. $P(X=5)=\left(\genfrac{}{}{0ex}{}{10}{5}\right)(0.25{)}^5(0.75{)}^5\approx 0.0584$.

   Common mistake: Leaving out $\left(\genfrac{}{}{0ex}{}{10}{5}\right)$.

8. Calls arrive at an average of 4 per hour (Poisson). Find $P(\text{no call})$ and $P(\text{at most 2 calls})$ in one hour.

   Show answer

   $P(0)=e^{-4}\approx 0.0183$. $P(X\le 2)=e^{-4}\left(1+4+\frac{16}{2}\right)=13e^{-4}\approx 0.238$.

   Common mistake: Forgetting the term $k=0$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $f(x)=\frac{x}{2}$ on $[0,2]$. Find $P(X\le 1)$.	a. $1.3$ and $4.41$
   2. For the same $X$ (values 0, 1, 2; probabilities 0.2, 0.5, 0.3), find $\mathrm{Var}(X)$.	b. $0.49$
   3. $E(X)=1.1$ and $\mathrm{Var}(X)=0.49$. Find $E(3X-2)$ and $\mathrm{Var}(3X-2)$.	c. $0.135$
   4. $X\sim \mathcal{P}(2)$. Find $P(X=0)$ (3 d.p.).	d. $\frac{1}{4}$

   Show answer

   1 → d ($\frac{1}{4}$) · 2 → b ($0.49$) · 3 → a ($1.3$ and $4.41$) · 4 → c ($0.135$)

   • Step 1. $P(X\le 1)={\int }_0^1\frac{x}{2}dx$.
     Step 2. ${\left[\frac{x^2}{4}\right]}_0^1=\frac{1}{4}$.
   • Step 1. $E(X^2)=0+0.5+4(0.3)=1.7$.
     Step 2. $\mathrm{Var}(X)=1.7-1.1^2=1.7-1.21=0.49$.
   • Step 1. $E(aX+b)=aE(X)+b=3.3-2=1.3$.
     Step 2. $\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)=9\times 0.49=4.41$.
   • Step 1. $P(X=0)=\frac{e^{-2}{2}^0}{0!}=e^{-2}$.
     Step 2. $e^{-2}\approx 0.135$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{3}{5}$  ·  $\frac{4}{3}$  ·  $\frac{3}{80}$  ·  $n=16$, $p=\frac{1}{4}$
   1. A binomial variable has mean 4 and variance 3. Then .
   2. $f(x)=\frac{x}{2}$ on $[0,2]$. Find $E(X)$. →
   3. $f(x)=3x^2$ on $[0,1]$. $\mathrm{Var}(X)=$ →
   Show answer

   1: $n=16$, $p=\frac{1}{4}$ · 2: $\frac{4}{3}$ · 3: $\frac{3}{80}$

   • Step 1. $npq=3$ and $np=4$, so $q=\frac{3}{4}$ and $p=\frac{1}{4}$.
     Step 2. $n=\frac{4}{p}=16$.
   • Step 1. $E(X)={\int }_0^{2}x\cdot \frac{x}{2}dx={\int }_0^2\frac{x^2}{2}dx$.
     Step 2. ${\left[\frac{x^3}{6}\right]}_0^2=\frac{8}{6}=\frac{4}{3}$.
   • Step 1. $E(X)={\int }_0^{1}3x^{3}dx=\frac{3}{4}$; $E(X^2)={\int }_0^{1}3x^{4}dx=\frac{3}{5}$.
     Step 2. $\mathrm{Var}(X)=\frac{3}{5}-\frac{9}{16}=\frac{48-45}{80}=\frac{3}{80}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $X\sim B(5,0.2)$. Find $P(X\le 1)$ (4 d.p.).	a. $\frac{1}{2}$
   2. $X\sim B(4,\frac{1}{2})$. Find $P(X=2)$.	b. $\frac{3}{8}$
   3. $f(x)=kx$ on $[0,2]$ (0 elsewhere) is a density. Find $k$.	c. $\frac{64}{729}$
   4. $X\sim B(6,\frac{1}{3})$. Find $P(X=0)$.	d. $0.7373$

   Show answer

   1 → d ($0.7373$) · 2 → b ($\frac{3}{8}$) · 3 → a ($\frac{1}{2}$) · 4 → c ($\frac{64}{729}$)

   • Step 1. $P(X=0)=0.8^5=0.32768$; $P(X=1)=5(0.2)(0.8^4)=0.4096$.
     Step 2. Sum: $0.73728\approx 0.7373$.
   • Step 1. $P(X=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right){\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2$.
     Step 2. $6\times \frac{1}{16}=\frac{3}{8}$.
   • Step 1. ${\int }_0^{2}kxdx=k{\left[\frac{x^2}{2}\right]}_0^2=2k$.
     Step 2. $2k=1\Rightarrow k=\frac{1}{2}$.
   • Step 1. $P(X=0)={\left(\frac{2}{3}\right)}^6$.
     Step 2. $\frac{2^6}{3^6}=\frac{64}{729}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $k=0.3$, $E(X)=1.1$  ·  3  ·  2.5  ·  $0.594$
   1. $X$ takes the values 0, 1, 2 with probabilities 0.2, 0.5 and $k$. Find $k$ and $E(X)$. →
   2. $X\sim \mathcal{P}(2)$. Find $P(X\ge 2)$ (3 d.p.). →
   3. $X$ takes the values 1, 2, 3, 4 with $P(X=x)=kx$. $E(X)=$ →
   Show answer

   1: $k=0.3$, $E(X)=1.1$ · 2: $0.594$ · 3: 3

   • Step 1. $0.2+0.5+k=1\Rightarrow k=0.3$.
     Step 2. $E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$.
   • Step 1. $P(X\le 1)=e^{-2}(1+2)=3e^{-2}\approx 0.406$.
     Step 2. $P(X\ge 2)=1-0.406=0.594$.
   • Step 1. $k(1+2+3+4)=1\Rightarrow k=0.1$.
     Step 2. $E(X)=0.1(1+4+9+16)=3$.

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