FS1-2 Poisson and binomial distributions — printable coverage pack

FS1-2.1 · The Poisson distribution. The additive property of Poisson distributions.

A Poisson model counts events occurring independently at a constant mean rate, with $P(X=x)=e^{-\lambda}\dfrac{\lambda^x}{x!}$ for $x=0,1,2,\ldots$ .
Scale $\lambda$ with the length of the interval, and add parameters for independent Poisson counts: $X+Y\sim\operatorname{Po}(\lambda+\mu)$ .
For example, a rate of $1.6$ events per hour gives a mean of $4$ events in $2.5$ hours.
A common error is to add Poisson parameters for overlapping intervals or dependent counts; the additive result requires independence.

1. A sensor records a Poisson-distributed number of amber flashes with mean $2.7$ per minute. Find the probability of no amber flashes in one minute.[2 marks]
Answer
- $P(X=0)=0.0672$ to $4$ significant figures
Method: With $X\sim\operatorname{Po}(2.7)$ , $P(X=0)=e^{-2.7}\dfrac{2.7^0}{0!}=e^{-2.7}=0.0672055\ldots$ .

1. Independent counts $A$ and $B$ have distributions $\operatorname{Po}(3.2)$ and $\operatorname{Po}(1.7)$ . Find $P(A+B\geq7)$ .[3 marks]
Answer
- $P(A+B\geq7)=0.2233$ to $4$ significant figures
Method: Independence gives $A+B\sim\operatorname{Po}(4.9)$ . Hence $P(A+B\geq7)=1-P(A+B\leq6)=0.223345\ldots$ .

If $X\sim\operatorname{Bin}(n,p)$ , then $E(X)=np$ and $\operatorname{Var}(X)=np(1-p)$ .
If $Y\sim\operatorname{Po}(\lambda)$ , then both $E(Y)$ and $\operatorname{Var}(Y)$ equal $\lambda$ .
For a linear transformation, $E(aX+b)=aE(X)+b$ and $\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$ .
A common error is to use $np$ as the binomial variance or to add the constant $b$ when transforming a variance.

1. Given $X\sim\operatorname{Bin}(80,0.15)$ , find $E(X)$ and $\operatorname{Var}(X)$ .[2 marks]
Answer
- $E(X)=12$
- $\operatorname{Var}(X)=10.2$
Method: $E(X)=np=80(0.15)=12$ . Also $\operatorname{Var}(X)=np(1-p)=80(0.15)(0.85)=10.2$ .

1. The random variable $Y$ is Poisson and $\operatorname{Var}(2Y+3)=28$ . Find the parameter of $Y$ and $E(2Y+3)$ .[4 marks]
Answer
- $Y\sim\operatorname{Po}(7)$
- $E(2Y+3)=17$
Method: If $Y\sim\operatorname{Po}(\lambda)$ , then $\operatorname{Var}(2Y+3)=4\lambda$ . Hence $4\lambda=28$ and $\lambda=7$ . Therefore $E(2Y+3)=2E(Y)+3=2(7)+3=17$ .

1. A binomial random variable $X$ has mean $18$ and variance $13.5$ . Determine $n$ and $p$ .[5 marks]
Answer
- $n=72$
- $p=0.25$
Method: Use $np=18$ and $np(1-p)=13.5$ . Dividing the second equation by the first gives $1-p=13.5/18=0.75$ , so $p=0.25$ . Then $n=18/0.25=72$ .

When $n$ is large and $p$ is small, $\operatorname{Bin}(n,p)$ can be approximated by $\operatorname{Po}(np)$ .
Keep the probability event unchanged and use $\lambda=np$ ; a continuity correction is not used for a binomial-to-Poisson approximation.
For example, $\operatorname{Bin}(900,0.004)$ is approximated by $\operatorname{Po}(3.6)$ .
A common error is to quote only $np$ without checking that $n$ is large and $p$ is small.

1. Let $X\sim\operatorname{Bin}(900,0.004)$ . State a suitable Poisson approximation and use it to estimate $P(X=0)$ .[3 marks]
Answer
- $X\mathrel{\dot\sim}\operatorname{Po}(3.6)$
- $P(X=0)\approx0.02732$
Method: Here $n$ is large, $p$ is small and $np=3.6$ , so use $Y\sim\operatorname{Po}(3.6)$ . Then $P(X=0)\approx P(Y=0)=e^{-3.6}=0.0273237\ldots$ .

1. The number of flawed seals in a batch has distribution $X\sim\operatorname{Bin}(500,0.008)$ . Use a Poisson approximation to estimate $P(X\geq6)$ .[4 marks]
Answer
- $P(X\geq6)\approx0.2149$
Method: $np=500(0.008)=4$ , so $X\mathrel{\dot\sim}Y$ where $Y\sim\operatorname{Po}(4)$ . Thus $P(X\geq6)\approx1-P(Y\leq5)=0.2148696\ldots$ .