Mathematics 9709 · Probability & Statistics 2

PDFs & probability as area

A continuous variable has no probability table — it has a curve $f(x)$ , and probability = area under that curve. There is no $P(X=4.76)$ to read off (a single point has no width, so that area is 0); every question is “shade the region, find its area” — by geometry when $f$ is straight lines, otherwise by integrating.

Probability is the area under f(x)

A valid PDF has $f(x)\ge0$ and total area $\int f(x)\,dx=1$ ; then $P(a<X<b)=\int_a^b f(x)\,dx$ (integrate only the slice asked for). Since $P(X=a)=0$ , $<$ and $\le$ are interchangeable.

Probability = area. Total area = 1; a probability is the area of one slice; a single point has probability 0, so < and ≤ are the same.

A piecewise pdf made of straight lines. P(4 < X < 6) is just the shaded area: split at the breakpoint x = 5 and add the two pieces.

Worked example

Easiest first: probability as a shape (geometry)

When every piece of $f$ is a straight line you can skip calculus and read the area off as triangles, trapezia and rectangles. For the taxi pdf above, $P(4<X<6)$ is a trapezium $(4\to5)$ plus a rectangle $(5\to6)$ :

$P(4<X<6)=\tfrac12\cdot1\cdot(0.16+0.20)+0.20\cdot1=0.38$

Answer

0.38

Same answer as integrating each piece — geometry is the faster route whenever $f$ is linear.

Straight-line f? Use geometry (½·base·height, trapezium, rectangle). Curved f? Integrate. Either way you are just finding an area.

Worked example

Find a constant k, then a tail (calculus)

Water erupts to height $X$ m with $f(x)=x^2/k$ for $1\le x\le10$ . Set the total area to 1, then read off the tail.

Total area: $\int_1^{10}\tfrac{x^2}{k}dx=\big[\tfrac{x^3}{3k}\big]_1^{10}=\tfrac{999}{3k}=1\Rightarrow k=333$ .
Tail: $P(X>6)=\int_6^{10}\tfrac{x^2}{333}dx=\big[\tfrac{x^3}{999}\big]_6^{10}=\tfrac{1000-216}{999}=\tfrac{784}{999}$ .

Answer

k=333,\quad P(X>6)\approx0.785

“Show that” / find a constant: set the TOTAL area = 1 and solve exactly (fractions, not decimals).

Worked example

An infinite upper limit

A battery lasts $X$ years with $f(x)=2/x^2$ for $x\ge2$ . An $\infty$ limit is no harder: substitute it and the term goes to 0.

It is a PDF: $\int_2^{\infty}\tfrac{2}{x^2}dx=\big[-\tfrac{2}{x}\big]_2^{\infty}=0-(-1)=1$ .
Tail: $P(X>3)=\big[-\tfrac{2}{x}\big]_3^{\infty}=0-(-\tfrac23)=\tfrac23$ .

Answer

P(X>3)=\tfrac23\approx0.667

f(x) = 2/x² for x ≥ 2 — a long thin right tail. As x → ∞ the area −2/x → 0, so the improper integral is finite.

An ∞ limit just means “substitute infinity”: 1/x, 1/x² and e^(−λx) all vanish, leaving only the finite (lower) term.

Worked example

A two-piece pdf: split, then add to 1

$f(x)=k(x-3)$ on $[3,7)$ and $k(11-x)$ on $[7,11]$ . Integrate each piece, add, set to 1.

$\int_3^7 k(x-3)dx+\int_7^{11}k(11-x)dx=8k+8k=16k=1\Rightarrow k=\tfrac1{16}$

Answer

k=\tfrac1{16}

A constant $k$ / “show that”: set the total area $\int f=1$ .
A probability $P(a<X<b)$ : integrate (or take the area of) only that slice.
Piecewise? split at each breakpoint and add — integrate only the parts inside the interval.

Asked	Do
find k / 'show that'	∫f over the whole domain = 1
P(a<X<b)	∫ from a to b only (or area by geometry)
median / quartile	area from the left = 0.5 / 0.25
E(X)	∫ x·f
Var(X)	∫ x²·f − [E(X)]²

Examiner note

A “show that” constant must be done exactly (fractions, not decimals); keep both limits when substituting

[F(x)]_a^b

. (9709/62/M/J/22 Q5)

Common mistake

Integrating the whole pdf when only a slice is asked; forgetting to split a piecewise

f

at the breakpoint; and dropping the lower-limit term when substituting into

[F(x)]_a^b

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PDFs & probability as area

Probability is the area under f(x)

A valid PDF has $f(x)\ge0$ and total area $\int f(x)\,dx=1$ ; then $P(a<X<b)=\int_a^b f(x)\,dx$ (integrate only the slice asked for). Since $P(X=a)=0$ , $<$ and $\le$ are interchangeable.

Probability = area. Total area = 1; a probability is the area of one slice; a single point has probability 0, so < and ≤ are the same.

A piecewise pdf made of straight lines. P(4 < X < 6) is just the shaded area: split at the breakpoint x = 5 and add the two pieces.

Worked example

Easiest first: probability as a shape (geometry)

$P(4<X<6)=\tfrac12\cdot1\cdot(0.16+0.20)+0.20\cdot1=0.38$

Answer

0.38

Same answer as integrating each piece — geometry is the faster route whenever $f$ is linear.

Straight-line f? Use geometry (½·base·height, trapezium, rectangle). Curved f? Integrate. Either way you are just finding an area.

Worked example

Find a constant k, then a tail (calculus)

Water erupts to height $X$ m with $f(x)=x^2/k$ for $1\le x\le10$ . Set the total area to 1, then read off the tail.

Total area: $\int_1^{10}\tfrac{x^2}{k}dx=\big[\tfrac{x^3}{3k}\big]_1^{10}=\tfrac{999}{3k}=1\Rightarrow k=333$ .
Tail: $P(X>6)=\int_6^{10}\tfrac{x^2}{333}dx=\big[\tfrac{x^3}{999}\big]_6^{10}=\tfrac{1000-216}{999}=\tfrac{784}{999}$ .

Answer

k=333,\quad P(X>6)\approx0.785

“Show that” / find a constant: set the TOTAL area = 1 and solve exactly (fractions, not decimals).

Worked example

An infinite upper limit

A battery lasts $X$ years with $f(x)=2/x^2$ for $x\ge2$ . An $\infty$ limit is no harder: substitute it and the term goes to 0.

It is a PDF: $\int_2^{\infty}\tfrac{2}{x^2}dx=\big[-\tfrac{2}{x}\big]_2^{\infty}=0-(-1)=1$ .
Tail: $P(X>3)=\big[-\tfrac{2}{x}\big]_3^{\infty}=0-(-\tfrac23)=\tfrac23$ .

Answer

P(X>3)=\tfrac23\approx0.667

f(x) = 2/x² for x ≥ 2 — a long thin right tail. As x → ∞ the area −2/x → 0, so the improper integral is finite.

An ∞ limit just means “substitute infinity”: 1/x, 1/x² and e^(−λx) all vanish, leaving only the finite (lower) term.

Worked example

A two-piece pdf: split, then add to 1

$f(x)=k(x-3)$ on $[3,7)$ and $k(11-x)$ on $[7,11]$ . Integrate each piece, add, set to 1.

$\int_3^7 k(x-3)dx+\int_7^{11}k(11-x)dx=8k+8k=16k=1\Rightarrow k=\tfrac1{16}$

Answer

k=\tfrac1{16}

A constant $k$ / “show that”: set the total area $\int f=1$ .
A probability $P(a<X<b)$ : integrate (or take the area of) only that slice.
Piecewise? split at each breakpoint and add — integrate only the parts inside the interval.

Asked	Do
find k / 'show that'	∫f over the whole domain = 1
P(a<X<b)	∫ from a to b only (or area by geometry)
median / quartile	area from the left = 0.5 / 0.25
E(X)	∫ x·f
Var(X)	∫ x²·f − [E(X)]²

Examiner note

A “show that” constant must be done exactly (fractions, not decimals); keep both limits when substituting

[F(x)]_a^b

. (9709/62/M/J/22 Q5)

Common mistake

Integrating the whole pdf when only a slice is asked; forgetting to split a piecewise

f

at the breakpoint; and dropping the lower-limit term when substituting into

[F(x)]_a^b