Every summary statistic you built for a discrete variable now returns, with the sum replaced by an integral. The median and quartiles come out of the cumulative function by solving an equation; the mean and variance come out of the density by integrating. This completes the continuous toolkit.
Commit to a prediction about three different middles.
A distribution has a long tail stretching to the right. Without looking ahead predict the order of its mean, median and mode from smallest to largest, and say which of the three a news report would quote if it wanted the figure to sound large.
Find the median and quartiles by solving F(x) = p.
Quantiles · one equation, three answers
We just saw that the CDF gives the accumulated probability below any value. That raises a question: quantiles ask the reverse — given an amount of accumulated probability, which value sits there? This card answers it → set $F$ equal to the required proportion and solve for $x$.
The median is the value with half the probability below it. The quartiles cut off a quarter and three quarters. Each is a value of $x$, read off by inverting the cumulative function.
The general form is $F(x_p) = p$, so any percentile works the same way. Because $F$ is non-decreasing and continuous, such a value always exists.
Worked through. Take $f(x) = \dfrac{x}{8}$ on $[0, 4]$, whose cumulative function is $F(x) = \dfrac{x^2}{16}$.
Only the positive root is taken in each case, because the domain is $[0, 4]$. Discarding the negative root is a step the marker expects to see, not an automatic one.
Quantiles come from the CDF: solve $F(x) = p$ for $x$. Median $p = 0.5$, $Q_1$ at $0.25$, $Q_3$ at $0.75$.; Reject any root outside the domain, and say that you have.
Pause, copy the quantile rule $F(x_p) = p$ with the three standard values of $p$, and the note about rejecting roots outside the domain, into your book.
Fill the blanks: $X$ is uniformly distributed on $[0, 10]$, so $F(x) = \dfrac{x}{10}$. Give each answer as a decimal.
median $=$
$Q_1 =$
$Q_3 =$
Replace the discrete sum with an integral.
We just saw quantiles obtained from the cumulative function. That raises a question: the mean was a probability-weighted sum in the discrete case, so what weights the values when there is no list to sum over? This card answers it → the density does, and the sum becomes an integral.
Set the two side by side, because the correspondence is the whole idea:
| discrete | continuous |
|---|---|
| $E(X) = \sum x\,p(x)$ | $E(X) = \displaystyle\int_{a}^{b} x\,f(x)\,dx$ |
| $\sum p(x) = 1$ | $\displaystyle\int_{a}^{b} f(x)\,dx = 1$ |
The value $x$ is weighted by the density at $x$, and the integral accumulates those weighted values across the whole domain. Everything you know about the expected value carries over: it is the balance point of the density curve, it is a long-run average, and it need not be a value the variable is especially likely to take.
Worked through. For $f(x) = \dfrac{x}{8}$ on $[0, 4]$:
$E(X) = \displaystyle\int_{0}^{4} x \cdot \frac{x}{8}\,dx = \int_{0}^{4} \frac{x^2}{8}\,dx = \frac{1}{8}\left[\frac{x^3}{3}\right]_{0}^{4} = \frac{1}{8} \cdot \frac{64}{3} = \frac{8}{3} \approx 2.667$
Compare with the median $2\sqrt{2} \approx 2.828$ found above. The mean sits below the median here, because the density is heaviest at the right but drops to zero abruptly at $x = 4$ while trailing to zero gradually at the left.
Quick check: Which integral gives the expected value of a continuous random variable with density $f$ on $[a, b]$?
Compute the variance and standard deviation by integration.
We just saw the mean as $\int x f(x)\,dx$. That raises a question: the variance was the expected squared deviation, so does the same substitution work for it? This card answers it → it does, and the computational shortcut $E(X^2) - \mu^2$ survives unchanged.
The right-hand form is the one to use. It needs $E(X^2) = \displaystyle\int_a^b x^2 f(x)\,dx$, which is no harder than the integral for the mean, and it avoids expanding a squared bracket inside the integral.
As always, $\operatorname{SD}(X) = \sigma = \sqrt{\operatorname{Var}(X)}$.
Worked through, continuing with $f(x) = \dfrac{x}{8}$ on $[0, 4]$, for which $\mu = \dfrac{8}{3}$:
$E(X^2) = \displaystyle\int_{0}^{4} x^2 \cdot \frac{x}{8}\,dx = \frac{1}{8}\left[\frac{x^4}{4}\right]_{0}^{4} = \frac{1}{8} \cdot 64 = 8$
$\operatorname{Var}(X) = 8 - \left(\dfrac{8}{3}\right)^2 = 8 - \dfrac{64}{9} = \dfrac{72 - 64}{9} = \dfrac{8}{9} \approx 0.889$
$\sigma = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3} \approx 0.943$
$E(X) = \int_a^b x f(x)\,dx$ and $\operatorname{Var}(X) = \int_a^b x^2 f(x)\,dx - \mu^2$.; Continuous uniform on $[a,b]$: $E(X) = (a+b)/2$ and $\operatorname{Var}(X) = (b-a)^2/12$.
Pause, copy the two integral formulas for $E(X)$ and $\operatorname{Var}(X)$, together with the uniform results $(a+b)/2$ and $(b-a)^2/12$, into your book.
True or false: For a continuous random variable, the median is always equal to the mean.
Work three examples, then read the traps.
Worked examples · 3 in a row, reveal as you go
A density is $f(x) = \dfrac{3x^2}{8}$ on $0 \leq x \leq 2$, with cumulative function $F(x) = \dfrac{x^3}{8}$. Find the mode, the median and the mean, and comment on their order.
For the same density $f(x) = \dfrac{3x^2}{8}$ on $[0, 2]$ with $\mu = \dfrac{3}{2}$, find $\operatorname{Var}(X)$ and $\sigma$.
Show by integration that a variable uniform on $[a, b]$ has $E(X) = \dfrac{a+b}{2}$ and $\operatorname{Var}(X) = \dfrac{(b-a)^2}{12}$.
Know what to do when the integral is beyond the course.
We just saw $E(X)$ and $\operatorname{Var}(X)$ evaluated for polynomial densities. That raises a question: plenty of real densities produce integrals no Mathematics Advanced technique can do, so what then? This card answers it → the syllabus expects you to set the integral up correctly and evaluate it with a computational application.
The formulas are unchanged. What changes is who does the antidifferentiation. The syllabus divides the work in two:
The normal distribution, which the next lessons cover, is the standard example. Its density is $f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$, and even the plain area $\int f$ has no antiderivative expressible in elementary functions. That is precisely why the normal distribution is handled with tables and the empirical rule rather than by integration.
Odd one out: Three of these correctly describe a summary statistic of a continuous random variable. Which one is wrong?
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · integrate, solve, compare
$X$ is uniform on $[4, 10]$. Write down $E(X)$ and $\operatorname{Var}(X)$ using the standard results.
$f(x) = \dfrac{1}{2}x$ on $[0, 2]$. Find $E(X)$ by integration.
$F(x) = \dfrac{x^2}{36}$ on $[0, 6]$. Find the median and the lower quartile exactly.
A student computes $E(X) = 3$ and $E(X^2) = 8$ for a continuous variable. Explain why one of these numbers must be wrong.
Explain in one sentence why the mean and median of a symmetric density must coincide.
Match each equation or integral to the statistic it produces:
Earlier you were asked to order the mean, median and mode for a density with a long tail to the right, and which one a report would quote to sound large. The order is mode < median < mean. The mode sits at the peak, near the bulk of the data; the mean is dragged furthest toward the tail because the integral $\int x f(x)\,dx$ weights extreme values by how far out they are, while the median only counts how many lie beyond it. A report wanting a large figure quotes the mean — which is exactly why average income is usually reported as a mean and typical income as a median.
The continuous toolkit is now complete: density, cumulative function, interval probability, quantiles, mean and variance. The remaining lessons apply all of it to one specific density, the normal, which models an enormous range of natural measurements. Its integral is beyond this course, which is why it is approached through the empirical rule and standardised scores rather than by integration.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 3 for the checkpoint question bank.
Q1. A continuous random variable has density $f(x) = \dfrac{x}{2}$ for $0 \leq x \leq 2$, and zero elsewhere.
(a) Find the cumulative distribution function $F(x)$ on $[0, 2]$. (b) Find the median exactly. (c) Find $E(X)$ and $\operatorname{Var}(X)$. (3 marks)
Q2. The lifetime $L$ hours of a component is modelled by the density $f(l) = \dfrac{3(4 - l)^2}{64}$ for $0 \leq l \leq 4$, and zero elsewhere.
(a) Verify that $f$ is a valid density. (b) Show that $F(l) = 1 - \dfrac{(4-l)^3}{64}$ and hence find the median lifetime, to two decimal places. (c) Find the mode, and explain in context what the gap between the mode and the median tells you about component lifetimes. (3 marks)
Q3. Two councils report on household water use. Council A reports a mean of 640 litres per day; Council B reports a median of 480 litres per day. Both are drawing on the same continuous distribution, which is strongly skewed to the right.
(a) Explain, referring to the integral $\int x f(x)\,dx$, why the mean exceeds the median for a right-skewed distribution. (b) Each council claims its figure is "the typical household". Evaluate both claims. (c) A third analyst proposes reporting the interquartile range alongside the median. Explain what that adds, and describe how each quartile would be obtained from the distribution. (3 marks)
Drill 1: $E(X) = \tfrac{4 + 10}{2} = 7$ and $\operatorname{Var}(X) = \tfrac{(10-4)^2}{12} = \tfrac{36}{12} = 3$.
Drill 2: $E(X) = \int_0^2 x \cdot \tfrac{x}{2}\,dx = \tfrac{1}{2}\left[\tfrac{x^3}{3}\right]_0^2 = \tfrac{1}{2} \cdot \tfrac{8}{3} = \tfrac{4}{3} \approx 1.333$.
Drill 3: $F(m) = \tfrac{m^2}{36} = 0.5$ gives $m^2 = 18$, so $m = 3\sqrt{2} \approx 4.243$ (rejecting the negative root, which is outside $[0, 6]$). $F(Q_1) = 0.25$ gives $Q_1^2 = 9$, so $Q_1 = 3$.
Drill 4: $\operatorname{Var}(X) = E(X^2) - \mu^2 = 8 - 3^2 = -1$, and a variance can never be negative because it is an average of squared quantities. So at least one of the two values is wrong; in fact $E(X^2) \geq [E(X)]^2$ always holds, with equality only for a constant.
Drill 5: If the density is symmetric about $x = c$, then exactly half the area lies on each side of $c$, so $F(c) = 0.5$ and $c$ is the median; and the symmetric pairing of values equidistant from $c$ makes their contributions to $\int x f(x)\,dx$ average to $c$, so $c$ is also the mean.
Q1 (3 marks): (a) $F(x) = \int_0^x \tfrac{t}{2}\,dt = \tfrac{1}{2}\left[\tfrac{t^2}{2}\right]_0^x = \tfrac{x^2}{4}$ for $0 \leq x \leq 2$. Check $F(0) = 0$ ✓ and $F(2) = \tfrac{4}{4} = 1$ ✓ [1]. (b) $\tfrac{m^2}{4} = 0.5$ gives $m^2 = 2$, so $m = \sqrt{2} \approx 1.414$, rejecting $m = -\sqrt{2}$ which lies outside the domain [1]. (c) $E(X) = \int_0^2 x \cdot \tfrac{x}{2}\,dx = \tfrac{1}{2}\left[\tfrac{x^3}{3}\right]_0^2 = \tfrac{4}{3}$. $E(X^2) = \int_0^2 x^2 \cdot \tfrac{x}{2}\,dx = \tfrac{1}{2}\left[\tfrac{x^4}{4}\right]_0^2 = 2$. So $\operatorname{Var}(X) = 2 - \left(\tfrac{4}{3}\right)^2 = 2 - \tfrac{16}{9} = \tfrac{2}{9} \approx 0.222$ [1].
Q2 (3 marks): (a) $(4 - l)^2 \geq 0$ for all $l$, so $f \geq 0$ on $[0,4]$ ✓. And $\int_0^4 \tfrac{3(4-l)^2}{64}\,dl = \tfrac{3}{64}\left[-\tfrac{(4-l)^3}{3}\right]_0^4 = \tfrac{1}{64}\left[-(4-l)^3\right]_0^4 = \tfrac{1}{64}\left(0 + 64\right) = 1$ ✓ [1]. (b) $F(l) = \int_0^l \tfrac{3(4-t)^2}{64}\,dt = \tfrac{1}{64}\left[-(4-t)^3\right]_0^l = \tfrac{64 - (4-l)^3}{64} = 1 - \tfrac{(4-l)^3}{64}$ ✓. Setting $F(m) = 0.5$: $\tfrac{(4-m)^3}{64} = 0.5$, so $(4-m)^3 = 32$, giving $4 - m = \sqrt[3]{32} \approx 3.1748$ and $m \approx 0.83$ hours [1]. (c) $f'(l) = -\tfrac{6(4-l)}{64} < 0$ for $l < 4$, so $f$ is strictly decreasing and the mode is the left endpoint, $l = 0$. The mode at 0 with a median of only $0.83$ says most components fail very early, while a thin tail of survivors stretches out to 4 hours. Half of all components fail within the first 50 minutes, so a "typical" lifetime quoted as the mean would badly overstate what most users experience [1].
Q3 (3 marks): (a) $E(X) = \int x f(x)\,dx$ weights every value by how large it is as well as by its density. A household using 3000 litres contributes six times as much to the integral as one using 500, however few such households there are. The median counts only how many observations lie either side and is indifferent to how extreme they are. So a long right tail pulls the mean up while leaving the median almost unmoved [1]. (b) Council B's figure is the better description of a typical household: with strong right skew the median is where the middle household actually sits. Council A's mean of 640 is not wrong, but it is not typical — it is inflated by a minority of very heavy users, and more than half of all households use less than it. Council A's figure is the right one for a different question, namely total demand, since mean $\times$ number of households gives the total the network must supply [1]. (c) The IQR reports the spread of the middle half, so it says how variable typical use is, which a median alone cannot. It is also resistant to the extreme tail that distorts the standard deviation. $Q_1$ is obtained by solving $F(x) = 0.25$ and $Q_3$ by solving $F(x) = 0.75$, then $\text{IQR} = Q_3 - Q_1$ [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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