Continuous Random Variables and Probability Density Functions

The queuing time $T$ (in minutes) at a customer service desk has probability density function $f(t) = k\,t(16 - t^2)$ for $0 \le t \le 4$, and $f(t) = 0$ elsewhere, for some $k \in R$. Show that $k = \dfrac{1}{64}$.

1. 1

   The total probability is $1$. Set the integral of the density over $[0, 4]$ equal to $1$.

   $$\int_0^4 k t(16 - t^2)\,dt = 1$$
2. 2

   Expand and antidifferentiate, then evaluate.

   $$k\left[8t^2 - \frac{t^4}{4}\right]_0^4 = k\left(128 - 64\right) = 64k$$
3. 3

   Set the result equal to $1$ and solve for the constant. The report notes this show that required explicit working, with the omitted $dt$ a common slip.

   $$64k = 1 \implies k = \frac{1}{64}$$

VCAA 2023 Mathematical Methods Exam 1, Q8a

The mass of each piece of luggage (kg) is a continuous random variable $X$ with density $f(x) = \dfrac{1}{67500}x^2(30 - x)$ for $0 \le x \le 30$, and $f(x) = 0$ elsewhere. A piece is labelled heavy if its mass exceeds $23$ kg. Find the mean of $X$, the standard deviation of $X$, and, given that a piece is heavier than the mean, the probability it is labelled as heavy (to three decimal places).

1. 1

   The mean is $\mu = \int_0^{30} x f(x)\,dx$, and the standard deviation is $\sigma = \sqrt{\int_0^{30} x^2 f(x)\,dx - \mu^2}$.

   $$\mu = 18, \quad \sigma = \sqrt{36} = 6$$
2. 2

   The condition heavier than the mean means $X > 18$. Set up the conditional probability as a ratio of two areas.

   $$\Pr(X > 23 \mid X > 18) = \frac{\Pr(X > 23)}{\Pr(X > 18)}$$
3. 3

   Evaluate each integral and divide. The report warns against finding the variance instead of the standard deviation, and against using the wrong denominator here.

   $$\frac{\int_{23}^{30} f(x)\,dx}{\int_{18}^{30} f(x)\,dx} \approx 0.446$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q4b

A continuous random variable $X$ has density $f(x) = kx$ for $0 \le x \le 4$, and $f(x) = 0$ elsewhere. Find $k$, then find $\Pr(X > 2)$.

1. 1

   The total area under any density must be $1$. Integrate over the interval where $f$ is non zero and set it equal to $1$.

   $$\int_0^4 kx \,dx = k\left[\frac{x^2}{2}\right]_0^4 = k\cdot 8 = 1$$
2. 2

   Solve for the constant.

   $$k = \frac{1}{8}$$
3. 3

   For $\Pr(X > 2)$, find the area to the right of $2$, which means integrate from $2$ to the top of the support, $4$.

   $$\Pr(X > 2) = \int_2^4 \frac{x}{8}\,dx = \frac{1}{8}\left[\frac{x^2}{2}\right]_2^4$$
4. 4

   Evaluate.

   $$\Pr(X > 2) = \frac{1}{8}\left(8 - 2\right) = \frac{6}{8} = \frac{3}{4}$$

A continuous random variable $X$ has density $f(x) = \dfrac{3}{8}x^2$ for $0 \le x \le 2$, and $f(x) = 0$ elsewhere. Find the mean $E(X)$ and the median $m$.

1. 1

   The mean is $E(X) = \int x\,f(x)\,dx$ over the support. Multiply the density by $x$, then integrate.

   $$E(X) = \int_0^2 x\cdot\frac{3}{8}x^2 \,dx = \frac{3}{8}\int_0^2 x^3 \,dx = \frac{3}{8}\left[\frac{x^4}{4}\right]_0^2$$
2. 2

   Evaluate the mean.

   $$E(X) = \frac{3}{8}\cdot\frac{16}{4} = \frac{3}{2}$$
3. 3

   The median $m$ splits the area in half. Set the area from the start of the support up to $m$ equal to $\tfrac{1}{2}$.

   $$\int_0^m \frac{3}{8}x^2 \,dx = \frac{1}{8}\big[x^3\big]_0^m = \frac{m^3}{8} = \frac{1}{2}$$
4. 4

   Solve for $m$, keeping the value exact.

   $$m^3 = 4 \quad\Rightarrow\quad m = \sqrt[3]{4} \approx 1.587$$

A continuous random variable $X$ is uniform on $[0, 2]$, so $f(x) = \dfrac{1}{2}$ for $0 \le x \le 2$ and $f(x) = 0$ elsewhere. Find $E(X)$, then $\operatorname{Var}(X)$ and the standard deviation.

1. 1

   First the mean, $E(X) = \int x\,f(x)\,dx$.

   $$E(X) = \int_0^2 x\cdot\frac{1}{2}\,dx = \frac{1}{2}\left[\frac{x^2}{2}\right]_0^2 = \frac{1}{2}\cdot 2 = 1$$
2. 2

   Now $E(X^2) = \int x^2 f(x)\,dx$, where you square $x$ but keep the same density.

   $$E(X^2) = \int_0^2 x^2\cdot\frac{1}{2}\,dx = \frac{1}{2}\left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2}\cdot\frac{8}{3} = \frac{4}{3}$$
3. 3

   Variance is the average of the squares minus the square of the average.

   $$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{4}{3} - 1^2 = \frac{1}{3}$$
4. 4

   The standard deviation is the square root of the variance. Do not stop at the variance.

   $$sd(X) = \sqrt{\operatorname{Var}(X)} = \frac{1}{\sqrt{3}} \approx 0.577$$