Mathematics 9709 · Probability & Statistics 2

Mean & variance of a CRV

The mean is the balance point of the curve. You cannot sum $xP(x)$ (every point has probability 0), so integrate $x\,f(x)$ instead. Variance is the same trick with $x^2$ : the S1 formula with $\int$ replacing $\sum$ .

Integrate x·f and x²·f

$E(X)=\int x\,f(x)\,dx,\qquad \operatorname{Var}(X)=\int x^2 f(x)\,dx-\big[E(X)\big]^2$

E(X) = ∫x·f, Var = ∫x²·f − [E(X)]². E(X²) is a NEW integral, not E(X)²; keep E(X) exact before squaring.

Worked example

Easiest first: mean and variance of one curve

$f(x)=x^2/72$ on $[0,6]$ . Multiply by $x$ for the mean, by $x^2$ for $E(X^2)$ .

$E(X)=\int_0^6\tfrac{x^3}{72}dx=\big[\tfrac{x^4}{288}\big]_0^6=\tfrac{1296}{288}=4.5$ .
$E(X^2)=\int_0^6\tfrac{x^4}{72}dx=\big[\tfrac{x^5}{360}\big]_0^6=\tfrac{7776}{360}=21.6$ .
$\operatorname{Var}(X)=21.6-4.5^2=1.35$ .

Answer

E(X)=4.5,\ \operatorname{Var}(X)=1.35

Worked example

Keep E(X) exact before you square

$f(t)=\tfrac{1}{99}(40-t^2)$ on $[1,4]$ . The mean lands on an ugly fraction — do not round it before squaring.

$E(T)=\tfrac{1}{99}\int_1^4(40t-t^3)dt=\tfrac{105}{44}=2.386$ .
$E(T^2)=\tfrac{1}{99}\int_1^4(40t^2-t^4)dt=\tfrac{353}{55}=6.418$ .
$\operatorname{Var}(T)=\tfrac{353}{55}-\big(\tfrac{105}{44}\big)^2=0.723$ .

Answer

E(T)\approx2.39,\ \operatorname{Var}(T)\approx0.723

Worked example

Mean of a piecewise pdf, then judge a claim

Taxi arrival pdf: $x/25$ on $[0,5]$ , $\tfrac15$ on $[5,7]$ , $(8-x)/5$ on $[7,8]$ . The firm claims “a taxi arrives within 5 minutes on average”. True?

Multiply each piece by $x$ , integrate, add: $E(X)=\int_0^5\tfrac{x^2}{25}dx+\int_5^7\tfrac{x}{5}dx+\int_7^8\tfrac{x(8-x)}{5}dx$ .
$E(X)=\tfrac{125}{75}+\tfrac{24}{10}+\tfrac{11}{15}=4.8$ .
The claim is about the mean: $4.8<5$ , so it is true.

Answer

E(X)=4.8<5\Rightarrow\text{claim true.}

Piecewise mean = sum of the per-piece integrals of x·f (then x²·f for the variance) — do each piece on its own limits and add.

Worked example

Turn a probability into an expected count (×N)

A pond-quality index has $f(x)=12(x-2x^2+x^3)$ on $[0,1]$ . Frogspawn thrive when $X>0.8$ . Out of 150 ponds, how many are ideal?

$P(X>0.8)=\int_{0.8}^{1}12(x-2x^2+x^3)dx=\tfrac{17}{625}=0.0272$ .
Expected count $=N\times P=150\times0.0272=4.08$ .
Round to a whole number of ponds.

Answer

\approx4\text{ ponds}

“Expected number out of N” = N × P(event). Find the probability by integrating, then multiply and round.

Mean vs median, and skew (drag the widget)

Mean and median coincide only when $f$ is symmetric. For $x^2/72$ the mean $(4.5)$ sits below the median $(4.76)$ : the long thin left tail drags the mean toward it. The slider below makes the rule physical — pull the outlier out and the mean chases it while the median barely moves.

outlier

mean = 22.0·median = 20·mean > median → positive skew

Examiner note

A “show that”

E(X)

stays exact (fractions); some pdfs need integration by parts, and a

1/x^2

pdf gives

E(X)=\int\tfrac1x\,dx=\ln

(a log, not a power). (9709/62/M/J/24 Q7) (9709/62/O/N/23 Q5)

Common mistake

Squaring

E(X)

and calling it

E(X^2)

(it is a new integral of

x^2f

); integrating

f

instead of

xf

for the mean; and rounding

E(X)

before squaring it — keep it exact, the subtraction loses precision otherwise.

Your turn— tap to reveal the worked answer (expected count)

Components have acceptable length $f(x)=\tfrac{1}{20}(10x-x^3)$ on $[1,3]$ . Find the mean.

$E(X)=\tfrac{1}{20}\int_1^3(10x^2-x^4)dx$ — multiply by $x$ , integrate, evaluate.

Answer

E(X)=\tfrac{287}{150}\approx1.91

PreviousMedian, quartiles & percentiles NextUniform & exponential models

Mean & variance of a CRV

Integrate x·f and x²·f

$E(X)=\int x\,f(x)\,dx,\qquad \operatorname{Var}(X)=\int x^2 f(x)\,dx-\big[E(X)\big]^2$

E(X) = ∫x·f, Var = ∫x²·f − [E(X)]². E(X²) is a NEW integral, not E(X)²; keep E(X) exact before squaring.

Worked example

Easiest first: mean and variance of one curve

$f(x)=x^2/72$ on $[0,6]$ . Multiply by $x$ for the mean, by $x^2$ for $E(X^2)$ .

$E(X)=\int_0^6\tfrac{x^3}{72}dx=\big[\tfrac{x^4}{288}\big]_0^6=\tfrac{1296}{288}=4.5$ .
$E(X^2)=\int_0^6\tfrac{x^4}{72}dx=\big[\tfrac{x^5}{360}\big]_0^6=\tfrac{7776}{360}=21.6$ .
$\operatorname{Var}(X)=21.6-4.5^2=1.35$ .

Answer

E(X)=4.5,\ \operatorname{Var}(X)=1.35

Worked example

Keep E(X) exact before you square

$f(t)=\tfrac{1}{99}(40-t^2)$ on $[1,4]$ . The mean lands on an ugly fraction — do not round it before squaring.

$E(T)=\tfrac{1}{99}\int_1^4(40t-t^3)dt=\tfrac{105}{44}=2.386$ .
$E(T^2)=\tfrac{1}{99}\int_1^4(40t^2-t^4)dt=\tfrac{353}{55}=6.418$ .
$\operatorname{Var}(T)=\tfrac{353}{55}-\big(\tfrac{105}{44}\big)^2=0.723$ .

Answer

E(T)\approx2.39,\ \operatorname{Var}(T)\approx0.723

Worked example

Mean of a piecewise pdf, then judge a claim

Taxi arrival pdf: $x/25$ on $[0,5]$ , $\tfrac15$ on $[5,7]$ , $(8-x)/5$ on $[7,8]$ . The firm claims “a taxi arrives within 5 minutes on average”. True?

Multiply each piece by $x$ , integrate, add: $E(X)=\int_0^5\tfrac{x^2}{25}dx+\int_5^7\tfrac{x}{5}dx+\int_7^8\tfrac{x(8-x)}{5}dx$ .
$E(X)=\tfrac{125}{75}+\tfrac{24}{10}+\tfrac{11}{15}=4.8$ .
The claim is about the mean: $4.8<5$ , so it is true.

Answer

E(X)=4.8<5\Rightarrow\text{claim true.}

Piecewise mean = sum of the per-piece integrals of x·f (then x²·f for the variance) — do each piece on its own limits and add.

Worked example

Turn a probability into an expected count (×N)

A pond-quality index has $f(x)=12(x-2x^2+x^3)$ on $[0,1]$ . Frogspawn thrive when $X>0.8$ . Out of 150 ponds, how many are ideal?

$P(X>0.8)=\int_{0.8}^{1}12(x-2x^2+x^3)dx=\tfrac{17}{625}=0.0272$ .
Expected count $=N\times P=150\times0.0272=4.08$ .
Round to a whole number of ponds.

Answer

\approx4\text{ ponds}

“Expected number out of N” = N × P(event). Find the probability by integrating, then multiply and round.

Mean vs median, and skew (drag the widget)

outlier

mean = 22.0·median = 20·mean > median → positive skew

Examiner note

A “show that”

E(X)

stays exact (fractions); some pdfs need integration by parts, and a

1/x^2

pdf gives

E(X)=\int\tfrac1x\,dx=\ln

(a log, not a power). (9709/62/M/J/24 Q7) (9709/62/O/N/23 Q5)

Common mistake

Squaring

E(X)

and calling it

E(X^2)

(it is a new integral of

x^2f

); integrating

f

instead of

xf

for the mean; and rounding

E(X)

before squaring it — keep it exact, the subtraction loses precision otherwise.

Your turn— tap to reveal the worked answer (expected count)

Components have acceptable length $f(x)=\tfrac{1}{20}(10x-x^3)$ on $[1,3]$ . Find the mean.

$E(X)=\tfrac{1}{20}\int_1^3(10x^2-x^4)dx$ — multiply by $x$ , integrate, evaluate.

Answer

E(X)=\tfrac{287}{150}\approx1.91