The Poisson formula and scaling λ | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Recognise a count model with one parameter

Definition

2 marks

What is meant by a Poisson distribution?

Model answer: A probability model for the number of events in a fixed interval when events occur singly, independently and at random, at a constant mean rate.

P(X=x)=e^{-\lambda}\frac{\lambda^x}{x!}

Symbols

$\lambda$ = mean number of events in the chosen interval (events)
$x$ = observed count: 0, 1, 2, ... (events)

E(X)=\lambda,\qquad \operatorname{Var}(X)=\lambda

$\lambda$ is an average count, so it need not be an integer.
The random variable is discrete: $x=0,1,2,\ldots$ .
For real data, a sample mean close to its sample variance supports a Poisson model, but context and independence must still make sense.

Scale lambda before touching the calculator

Longer interval → larger λ

5 min

λ = 0.9

10 min

λ = 1.8

15 min

λ = 2.7

time × k ⇒ λ × k

The Poisson probability rule stays the same.

Scale the mean count in direct proportion to the interval length.

Worked example

Current-paper rate and strict inequalities

Used computers are donated at a mean rate of 2.4 per week. Find the probability that more than 6 and fewer than 9 are donated in 4 weeks.

\lambda=4(2.4)=9.6

P(6<X<9)

=P(X=7)+P(X=8)

Answer

P(6<X<9)=0.222

(9709/62/O/N/25 Q1(b))

Examiner note

Calculator route: use Poisson PDF for one exact count and Poisson CDF for a lower tail. Translate the words into included integers before entering a bound.

Add independent Poisson counts

Key idea

X\sim\operatorname{Po}(\lambda_1)

and

Y\sim\operatorname{Po}(\lambda_2)

are independent, then

X+Y\sim\operatorname{Po}(\lambda_1+\lambda_2)

. A difference is not Poisson.

Worked example

Current-paper combined count

Let $X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ be independent. Find $P(X+Y>2)$ .

X+Y\sim\operatorname{Po}(5)

P(X+Y>2)

=1-P(X+Y\le2)

Answer

P(X+Y>2)=0.875

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

Neighbouring probabilities can be compared without calculating both:

\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}

Therefore $P(X=n)=P(X=n+1)$ when $n+1=\lambda$ . For $X\sim\operatorname{Po}(15)$ , this gives $n=14$ . (9709/62/M/J/25 Q3(b))

Your turnIndependent Poisson-count check [3 marks]

$X\sim\operatorname{Po}(1.5)$ . Find the probability that the sum of three independent values of X is between 3 and 5 inclusive.

Show worked answer

T\sim\operatorname{Po}(3\times1.5)

P(3\le T\le5)

=P(T\le5)-P(T\le2)

Answer

P(3\le T\le5)=0.529

Common mistake

Do not add rates unless the counts are independent and refer to the same overall interval. Scale time first, then combine.

Recognise a count model with one parameter

Definition

2 marks

What is meant by a Poisson distribution?

Model answer: A probability model for the number of events in a fixed interval when events occur singly, independently and at random, at a constant mean rate.

P(X=x)=e^{-\lambda}\frac{\lambda^x}{x!}

Symbols

$\lambda$ = mean number of events in the chosen interval (events)
$x$ = observed count: 0, 1, 2, ... (events)

E(X)=\lambda,\qquad \operatorname{Var}(X)=\lambda

$\lambda$ is an average count, so it need not be an integer.
The random variable is discrete: $x=0,1,2,\ldots$ .
For real data, a sample mean close to its sample variance supports a Poisson model, but context and independence must still make sense.

Scale lambda before touching the calculator

Longer interval → larger λ

5 min

λ = 0.9

10 min

λ = 1.8

15 min

λ = 2.7

time × k ⇒ λ × k

The Poisson probability rule stays the same.

Scale the mean count in direct proportion to the interval length.

Worked example

Current-paper rate and strict inequalities

Used computers are donated at a mean rate of 2.4 per week. Find the probability that more than 6 and fewer than 9 are donated in 4 weeks.

\lambda=4(2.4)=9.6

P(6<X<9)

=P(X=7)+P(X=8)

Answer

P(6<X<9)=0.222

(9709/62/O/N/25 Q1(b))

Examiner note

Calculator route: use Poisson PDF for one exact count and Poisson CDF for a lower tail. Translate the words into included integers before entering a bound.

Add independent Poisson counts

Key idea

X\sim\operatorname{Po}(\lambda_1)

and

Y\sim\operatorname{Po}(\lambda_2)

are independent, then

X+Y\sim\operatorname{Po}(\lambda_1+\lambda_2)

. A difference is not Poisson.

Worked example

Current-paper combined count

Let $X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ be independent. Find $P(X+Y>2)$ .

X+Y\sim\operatorname{Po}(5)

P(X+Y>2)

=1-P(X+Y\le2)

Answer

P(X+Y>2)=0.875

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

Neighbouring probabilities can be compared without calculating both:

\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}

Therefore $P(X=n)=P(X=n+1)$ when $n+1=\lambda$ . For $X\sim\operatorname{Po}(15)$ , this gives $n=14$ . (9709/62/M/J/25 Q3(b))

Your turnIndependent Poisson-count check [3 marks]

$X\sim\operatorname{Po}(1.5)$ . Find the probability that the sum of three independent values of X is between 3 and 5 inclusive.

Show worked answer

T\sim\operatorname{Po}(3\times1.5)

P(3\le T\le5)

=P(T\le5)-P(T\le2)

Answer

P(3\le T\le5)=0.529

Common mistake

Do not add rates unless the counts are independent and refer to the same overall interval. Scale time first, then combine.