Mathematics 9709 · Probability & Statistics 2

Normal ≈ Poisson

Pile up enough rare events and the lumpy Poisson smooths into a bell. Once $\lambda>15$ the bars of $\text{Po}(\lambda)$ are so many and so nearly symmetric that a normal curve traces them: $N(\lambda,\lambda)$ , same mean, same variance. One catch — bars have width but a curve does not, so nudge each boundary by $\tfrac12$ (the continuity correction).

Po(λ) ≈ N(λ, λ) for λ > 15

Po(20): with a big mean the bars hug a N(20, 20) bell — switch to the normal, remembering the ±0.5 continuity correction.

A discrete $k$ spreads over $(k-0.5,\ k+0.5)$ . Push the boundary out for $\le,\ \ge$ (they include $k$ ), in for $<,\ >$ . Then standardise with $\sigma=\sqrt\lambda$ .

Worked example

Start simple: one corrected tail

$X\sim\text{Po}(30)\approx N(30,30)$ . $P(X>40)$ excludes 40, so use $P(Y>40.5)$ : $z=\tfrac{40.5-30}{\sqrt{30}}=1.917$ , $1-\Phi(1.917)$ .

Answer

\approx0.0276

Worked example

Turtle eggs: a single value, then two tails

Nests hold on average 60 eggs, $X\sim\text{Po}(60)\approx N(60,60)$ . Find (a) $P(X=50)$ , (b) $P(X>74)$ , (c) $P(X\le40)$ .

(a) A single value $50$ becomes the band $49.5<Y<50.5$ : $\Phi(-1.226)-\Phi(-1.356)=0.0226$ .
(b) $>74$ excludes 74, so $Y>74.5$ : $1-\Phi(1.872)=0.0307$ .
(c) $\le40$ includes 40, so $Y<40.5$ : $\Phi(-2.517)=0.0059$ .

Answer

0.0226,\quad 0.0307,\quad 0.0059

A single value is NOT zero here — it is a width-1 band, the only time a continuous model gives a point a probability.

Worked example

Which way does the ±0.5 go?

Posts arrive at $\text{Po}(24)\approx N(24,24)$ . Contrast an inclusive and an exclusive range.

$P(24\le X\le26)$ : both ends included, widen to $23.5<Y<26.5$ : $\Phi(0.510)-\Phi(-0.102)=0.236$ .
$P(19<X\le21)$ : 19 excluded, 21 included, so $19.5<Y<21.5$ : $\Phi(-0.510)-\Phi(-0.919)=0.126$ .

Answer

0.236,\quad 0.126

λ = 4·mean = variance = λ·still skewed → stay Poisson

Worked example

Inverse: the least safe count

With $X\sim\text{Po}(42)\approx N(42,42)$ , find the least integer $x$ for which $P(X>x)\le0.1$ .

Upper tail $0.1\Rightarrow z=1.282$ ; with continuity correction $x+0.5\ge42+1.282\sqrt{42}=50.3$ .
So $x\ge49.8$ ; the least integer is $x=50$ (check: $P(Y>50.5)=0.095<0.1$ ).

Answer

x=50

λ > 15 → N(λ, λ) with a ±0.5 continuity correction; standardise with √λ, never λ.

Examiner note

State

\text{Po}(\lambda)\approx N(\lambda,\lambda)

(variance

\lambda

too), show the 0.5 correction, and standardise with

\sqrt\lambda

. (9709/62/M/J/24 Q1) (9709/62/O/N/23 Q1)

Common mistake

Forgetting the continuity correction or pushing it the wrong way; dividing by

\lambda

instead of

\sqrt\lambda

; and using the normal route when

\lambda\le15

(stay Poisson).

PreviousPoisson ≈ binomial NextHypothesis testing with the Poisson

Normal ≈ Poisson

Po(λ) ≈ N(λ, λ) for λ > 15

Po(20): with a big mean the bars hug a N(20, 20) bell — switch to the normal, remembering the ±0.5 continuity correction.

A discrete $k$ spreads over $(k-0.5,\ k+0.5)$ . Push the boundary out for $\le,\ \ge$ (they include $k$ ), in for $<,\ >$ . Then standardise with $\sigma=\sqrt\lambda$ .

Worked example

Start simple: one corrected tail

$X\sim\text{Po}(30)\approx N(30,30)$ . $P(X>40)$ excludes 40, so use $P(Y>40.5)$ : $z=\tfrac{40.5-30}{\sqrt{30}}=1.917$ , $1-\Phi(1.917)$ .

Answer

\approx0.0276

Worked example

Turtle eggs: a single value, then two tails

Nests hold on average 60 eggs, $X\sim\text{Po}(60)\approx N(60,60)$ . Find (a) $P(X=50)$ , (b) $P(X>74)$ , (c) $P(X\le40)$ .

(a) A single value $50$ becomes the band $49.5<Y<50.5$ : $\Phi(-1.226)-\Phi(-1.356)=0.0226$ .
(b) $>74$ excludes 74, so $Y>74.5$ : $1-\Phi(1.872)=0.0307$ .
(c) $\le40$ includes 40, so $Y<40.5$ : $\Phi(-2.517)=0.0059$ .

Answer

0.0226,\quad 0.0307,\quad 0.0059

A single value is NOT zero here — it is a width-1 band, the only time a continuous model gives a point a probability.

Worked example

Which way does the ±0.5 go?

Posts arrive at $\text{Po}(24)\approx N(24,24)$ . Contrast an inclusive and an exclusive range.

$P(24\le X\le26)$ : both ends included, widen to $23.5<Y<26.5$ : $\Phi(0.510)-\Phi(-0.102)=0.236$ .
$P(19<X\le21)$ : 19 excluded, 21 included, so $19.5<Y<21.5$ : $\Phi(-0.510)-\Phi(-0.919)=0.126$ .

Answer

0.236,\quad 0.126

λ = 4·mean = variance = λ·still skewed → stay Poisson

Worked example

Inverse: the least safe count

With $X\sim\text{Po}(42)\approx N(42,42)$ , find the least integer $x$ for which $P(X>x)\le0.1$ .

Upper tail $0.1\Rightarrow z=1.282$ ; with continuity correction $x+0.5\ge42+1.282\sqrt{42}=50.3$ .
So $x\ge49.8$ ; the least integer is $x=50$ (check: $P(Y>50.5)=0.095<0.1$ ).

Answer

x=50

λ > 15 → N(λ, λ) with a ±0.5 continuity correction; standardise with √λ, never λ.

Examiner note

State

\text{Po}(\lambda)\approx N(\lambda,\lambda)

(variance

\lambda

too), show the 0.5 correction, and standardise with

\sqrt\lambda

. (9709/62/M/J/24 Q1) (9709/62/O/N/23 Q1)

Common mistake

Forgetting the continuity correction or pushing it the wrong way; dividing by

\lambda

instead of

\sqrt\lambda

; and using the normal route when

\lambda\le15

(stay Poisson).