SR22, SR23 - Averages and spread from a pdf | OCR A-Level Further Mathematics B (MEI)

SR22 uses a density to calculate the mean, variance and standard deviation, including the continuous uniform model. SR23 finds the median by splitting probability in half and the mode by locating the density's maximum. The three measures of centre answer different questions.

Weight each value by its density

For the density $f(x)=2x$ on $[0,1]$ , is the average value the point that divides the probability into two equal halves? We will find both. A steadily rising density puts more weight towards $1$ , but it does not make all sensible measures of centre coincide.

The discrete probability-weighted sum becomes an integral: a thin strip near $x$ has probability approximately $f(x)\,dx$ , so its contribution to the mean is approximately $x f(x)\,dx$ .

You already know

A pdf is non-negative with total area one. A mean is a probability-weighted value, and variance measures squared deviations from it. Write the relevant definite integral before calculator evaluation; split it if the density changes rule.

Use the support limits, not the endpoints of a probability interval from an earlier question. Expectation uses the whole distribution.

Continuous mean and variance

For a pdf $f$ on a finite support $[a,b]$ ,

\mu=E(X)=\int_a^b x f(x)\,dx

\operatorname{Var}(X)=\int_a^b(x-\mu)^2f(x)\,dx =\int_a^b x^2f(x)\,dx-\mu^2

The integral of $x^2f(x)$ is $E(X^2)$ , not $E(X)^2$ .

In the formula booklet.

Standard deviation is the non-negative square root of variance. It has the variable's original units; variance has squared units. This square-root relation is in the formula booklet.

Worked example 1: a rising linear density

Question. A continuous random variable has pdf $f(x)=2x$ for $0\le x\le1$ , and zero otherwise. Find its mean, variance and standard deviation.

Plan. Calculate the first two moments, then subtract the square of the mean from the second moment.

\begin{aligned} E(X)&=\int_0^1x(2x)\,dx=\left[\frac{2x^3}{3}\right]_0^1=\frac23&&\text{first moment}\\ E(X^2)&=\int_0^1x^2(2x)\,dx=\left[\frac{x^4}{2}\right]_0^1=\frac12&&\text{second moment}\\ \operatorname{Var}(X)&=\frac12-\left(\frac23\right)^2=\frac1{18}&&\text{subtract mean squared}\\ \operatorname{sd}(X)&=\sqrt{\frac1{18}}=\frac{\sqrt2}{6}&&\text{non-negative square root} \end{aligned}

Why this step? The density already weights the outcomes. Integrating $x$ alone would give the wrong mean.

Answer. Mean $2/3$ , variance $1/18$ , standard deviation $\sqrt2/6$ .

Check. The mean is inside $[0,1]$ and above its midpoint, as the density increases towards $1$ . The variance is positive.

Change only the power in the density, from a linear rise to a quadratic rise; the coefficient changes to keep total area one.

Worked example 2: more weight near the upper end

Question. Find the same measures when $f(x)=3x^2$ on $[0,1]$ , and zero otherwise.

Plan. Integrate $xf(x)$ and $x^2f(x)$ over the unchanged support.

\begin{aligned} E(X)&=\int_0^1 3x^3\,dx=\frac34&&\text{first moment}\\ E(X^2)&=\int_0^1 3x^4\,dx=\frac35&&\text{second moment}\\ \operatorname{Var}(X)&=\frac35-\left(\frac34\right)^2=\frac3{80}&&\text{variance identity}\\ \operatorname{sd}(X)&=\sqrt{\frac3{80}}=\frac{\sqrt{15}}{20}&&\text{simplify the root} \end{aligned}

Why this step? Moving more probability close to $1$ raises the mean but can reduce the spread: the two measures need not rise together.

Answer. Mean $3/4$ , variance $3/80$ , standard deviation $\sqrt{15}/20$ .

Check. $3/4>2/3$ , while $3/80<1/18$ . The steeper density concentrates values near the upper end.

A negative variance signals an error. Check whether you squared the mean, integrated the whole support and retained enough precision before subtracting.

Watch out

Wrong: $\operatorname{Var}(X)=E(X^2)-E(X)$ . Right: subtract $[E(X)]^2$ . Also $E(X^2)$ is usually different from $[E(X)]^2$ .

Increase the power once more and keep the calculations exact.

A uniform interval has a midpoint and a width

For a continuous uniform distribution on $[a,b]$ , every equal-length subinterval has the same probability. The constant density must therefore be $1/(b-a)$ on the interval and zero elsewhere.

Its mean is the midpoint. Its variance depends on the interval's width, so shifting the whole interval changes the mean without changing the spread.

Continuous uniform moments

If $X$ is uniformly distributed over $[a,b]$ , where $a<b$ ,

E(X)=\frac{a+b}{2},\qquad \operatorname{Var}(X)=\frac{(b-a)^2}{12}

In the formula booklet.

The results may still need deriving if the question asks. Starting from the constant density,

\begin{aligned} E(X)&=\frac1{b-a}\int_a^b x\,dx=\frac{a+b}{2}\\ E(X^2)&=\frac1{b-a}\int_a^b x^2\,dx=\frac{a^2+ab+b^2}{3}\\ \operatorname{Var}(X)&=\frac{a^2+ab+b^2}{3}-\frac{(a+b)^2}{4} =\frac{(b-a)^2}{12} \end{aligned}

The density $1/(b-a)$ and this derivation are not printed separately: obtain them from total area one and the supplied moment integrals.

Worked example 3: waiting within a uniform interval

Question. A waiting time $X$ , in minutes, is modelled by a continuous uniform distribution on $[2,8]$ . Find its mean, variance and standard deviation.

Plan. Use the interval midpoint and squared width.

\begin{aligned} E(X)&=\frac{2+8}{2}=5\,\mathrm{min}&&\text{midpoint}\\ \operatorname{Var}(X)&=\frac{(8-2)^2}{12}=3\,\mathrm{min^2}&&\text{squared width}\\ \operatorname{sd}(X)&=\sqrt3\,\mathrm{min}&&\text{original units} \end{aligned}

Why this step? The spread depends on $8-2$ , rather than on the upper endpoint alone.

Answer. Mean $5\,\mathrm{min}$ , variance $3\,\mathrm{min^2}$ , standard deviation $\sqrt3\,\mathrm{min}$ .

Check. Directly, $E(X^2)=\frac16\int_2^8x^2\,dx=28$ , so the variance is $28-5^2=3$ .

Change only the upper endpoint from $8$ to $10$ . The midpoint moves and the interval widens.

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SR22, SR23 - Averages and spread from a pdf

Weight each value by its density

A uniform interval has a midpoint and a width

Continue this lesson.

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