Mathematics 9709 · Probability & Statistics 2

Sample means of discrete distributions

When you average counts (Poisson or binomial), the underlying values are still discrete — so you still nudge by a continuity correction. But you're averaging over $n$ items, so that half-step is shared out: the correction shrinks to $\tfrac{1}{2n}$ , not $\tfrac12$ .

The sample mean of counts: ±1/(2n)

Get the per-item mean and variance (Poisson: both $\lambda$ ; binomial: $Np$ and $Np(1-p)$ ), apply $\bar X\sim N(\text{mean},\ \text{var}/n)$ , then shift the boundary by $\tfrac{1}{2n}$ (out for $\le,\ge$ , in for $<,>$ ) before standardising.

Mean of n counts: X̄ ~ N(mean, var/n) with a ±1/(2n) continuity correction — the half-step shared over n, not ±½.

Worked example

Poisson sample mean (one tail)

Threats at 12.2/day → 85.4/week, $\text{Po}(85.4)$ . Over 100 weeks $\bar X\sim N(85.4,\tfrac{85.4}{100})=N(85.4,0.854)$ . $P(\bar X<86)$ uses $86-\tfrac{1}{200}=85.995$ : $z=\tfrac{85.995-85.4}{\sqrt{0.854}}=0.644$ .

Answer

\Phi(0.644)\approx0.74

Worked example

Binomial sample mean

$X\sim B(60,0.25)$ : mean 15, variance 11.25. For the mean of 50, $\bar X\sim N(15,\tfrac{11.25}{50})=N(15,0.225)$ . $P(\bar X\le16)$ uses $16+\tfrac{1}{100}=16.01$ : $z=\tfrac{16.01-15}{\sqrt{0.225}}=2.129$ .

Answer

\Phi(2.129)\approx0.983

Worked example

A two-sided interval (watch each end)

$X\sim\text{Po}(2)$ ; mean of 50. Find $P(1.5<\bar X\le2.2)$ , where $\bar X\sim N(2,\tfrac{2}{50})=N(2,0.04)$ and the correction is $\tfrac{1}{2\cdot50}=0.01$ .

Lower end $>1.5$ excludes 1.5: move in to $1.51$ ; $z=\tfrac{1.51-2}{0.2}=-2.45$ .
Upper end $\le2.2$ includes 2.2: move out to $2.21$ ; $z=\tfrac{2.21-2}{0.2}=1.05$ .
$\Phi(1.05)-\Phi(-2.45)=0.8531-0.0071$ .

Answer

\approx0.846

Examiner note

The correction for a mean of

n

counts is

\tfrac{1}{2n}

(use

\tfrac12

only for a single count or a total); standardise with

\sqrt{\text{var}/n}

. (9709/62/M/J/22 Q7) (9709/62/O/N/22 Q5)

Why exactly 1/(2n)

A total of $n$ counts gets the ordinary $\pm\tfrac12$ ; but the mean is that total divided by $n$ , so the half-step is divided by $n$ too — $\tfrac12\div n=\tfrac{1}{2n}$ . (In the threats example, $P(\bar X<86)$ with $\pm\tfrac{1}{200}$ equals $P(\text{total}<8600)$ with the usual $\pm\tfrac12$ .)

Common mistake

Using

\pm\tfrac12

instead of

\pm\tfrac{1}{2n}

(the

\tfrac12

is for a single count or a total, not a mean of

n

); correcting the wrong direction; and binomial variance as

Np

instead of

Np(1-p)

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Sample means of discrete distributions

The sample mean of counts: ±1/(2n)

Mean of n counts: X̄ ~ N(mean, var/n) with a ±1/(2n) continuity correction — the half-step shared over n, not ±½.

Worked example

Poisson sample mean (one tail)

Answer

\Phi(0.644)\approx0.74

Worked example

Binomial sample mean

Answer

\Phi(2.129)\approx0.983

Worked example

A two-sided interval (watch each end)

$X\sim\text{Po}(2)$ ; mean of 50. Find $P(1.5<\bar X\le2.2)$ , where $\bar X\sim N(2,\tfrac{2}{50})=N(2,0.04)$ and the correction is $\tfrac{1}{2\cdot50}=0.01$ .

Lower end $>1.5$ excludes 1.5: move in to $1.51$ ; $z=\tfrac{1.51-2}{0.2}=-2.45$ .
Upper end $\le2.2$ includes 2.2: move out to $2.21$ ; $z=\tfrac{2.21-2}{0.2}=1.05$ .
$\Phi(1.05)-\Phi(-2.45)=0.8531-0.0071$ .

Answer

\approx0.846

Examiner note

The correction for a mean of

n

counts is

\tfrac{1}{2n}

(use

\tfrac12

only for a single count or a total); standardise with

\sqrt{\text{var}/n}

. (9709/62/M/J/22 Q7) (9709/62/O/N/22 Q5)

Why exactly 1/(2n)

Common mistake

Using

\pm\tfrac12

instead of

\pm\tfrac{1}{2n}

(the

\tfrac12

is for a single count or a total, not a mean of

n

); correcting the wrong direction; and binomial variance as

Np

instead of

Np(1-p)