Normal approximation to Poisson | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Match both the mean and the variance

X\sim\operatorname{Po}(\lambda)\approx Y\sim N(\lambda,\lambda)

Symbols

$\lambda$ = Poisson mean; also the normal mean and variance (count)
$\sqrt\lambda$ = normal standard deviation (events)

Cambridge's guide is $\lambda>15$ .
Write the normal distribution before standardising.
Apply a continuity correction because Poisson is discrete.

Find the last out and first in

last excluded

50.5

first included

X > 50 ⇒ 51, 52, … ⇒ Y > 50.5

Use the half-unit correction only for the normal approximation.

Put the normal boundary halfway between the last excluded and first included integer.

Build the total, then correct its boundary

Worked example

Current-paper total of independent counts

$X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ are independent. T is the total of 100 values of X and 150 values of Y. Approximate $P(T<560)$ .

T\sim\operatorname{Po}(100(3)+150(2))

T\approx N(600,600)

P(T<560)\approx P(Y<559.5)

z=\frac{559.5-600}{\sqrt{600}}=-1.653

Answer

P(T<560)=0.0492

(9709/61/O/N/25 Q1(c))

Examiner note

A sum of independent Poisson counts is still exactly Poisson. The approximation happens only when that large-λ Poisson total is changed to a normal distribution.

Translate every endpoint before using z

$X>k$ becomes $Y>k+0.5$ .
$X\le k$ becomes $Y<k+0.5$ .
$a\le X\le b$ becomes $a-0.5<Y<b+0.5$ .

Your turnIndependent normal-approximation check [4 marks]

$X\sim\operatorname{Po}(42)$ . Use a normal approximation to find $P(35\le X\le50)$ .

Show worked answer

Y\sim N(42,42)

After the continuity correction,

35\le X\le50

34.5<Y<50.5

z_1=-1.157,\qquad z_2=1.312

Answer

P(35\le X\le50)=0.782

Common mistake

The second parameter in

N(\lambda,\lambda)

is the variance. Divide by

\sqrt\lambda

when standardising.

Match both the mean and the variance

X\sim\operatorname{Po}(\lambda)\approx Y\sim N(\lambda,\lambda)

Symbols

$\lambda$ = Poisson mean; also the normal mean and variance (count)
$\sqrt\lambda$ = normal standard deviation (events)

Cambridge's guide is $\lambda>15$ .
Write the normal distribution before standardising.
Apply a continuity correction because Poisson is discrete.

Find the last out and first in

last excluded

50.5

first included

X > 50 ⇒ 51, 52, … ⇒ Y > 50.5

Use the half-unit correction only for the normal approximation.

Put the normal boundary halfway between the last excluded and first included integer.

Build the total, then correct its boundary

Worked example

Current-paper total of independent counts

$X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ are independent. T is the total of 100 values of X and 150 values of Y. Approximate $P(T<560)$ .

T\sim\operatorname{Po}(100(3)+150(2))

T\approx N(600,600)

P(T<560)\approx P(Y<559.5)

z=\frac{559.5-600}{\sqrt{600}}=-1.653

Answer

P(T<560)=0.0492

(9709/61/O/N/25 Q1(c))

Examiner note

A sum of independent Poisson counts is still exactly Poisson. The approximation happens only when that large-λ Poisson total is changed to a normal distribution.

Translate every endpoint before using z

$X>k$ becomes $Y>k+0.5$ .
$X\le k$ becomes $Y<k+0.5$ .
$a\le X\le b$ becomes $a-0.5<Y<b+0.5$ .

Your turnIndependent normal-approximation check [4 marks]

$X\sim\operatorname{Po}(42)$ . Use a normal approximation to find $P(35\le X\le50)$ .

Show worked answer

Y\sim N(42,42)

After the continuity correction,

35\le X\le50

34.5<Y<50.5

z_1=-1.157,\qquad z_2=1.312

Answer

P(35\le X\le50)=0.782

Common mistake

The second parameter in

N(\lambda,\lambda)

is the variance. Divide by

\sqrt\lambda

when standardising.