Q.The number of calls arriving at a call centre in a given minute follows a Poisson distribution with a mean of 4 calls per minute. Find the probability that exactly 3 calls arrive in a particular minute. (Use e^{-4} ≈ 0.0183.)

Telephone calls, misprints and bacteria-per-volume are standard Poisson counts (rare, independent events at a constant rate); customer arrivals at a petrol station vary with peak/lean demand, so it is the odd one out.

The Poisson distribution models the number of times a rare event occurs in a fixed interval of time or space, under the assumptions that events are independent and the average rate $\lambda$ is constant.

• (a) bacteria in a cubic foot of soil — a count in a fixed volume of space: a standard Poisson example.
• (b) telephone calls in a ten-minute interval — a count in a fixed time interval at a steady rate: a classic Poisson example.
• (c) misprints per page — a count in a fixed unit (a page): the textbook Poisson example.
• (d) customers arriving at a petrol station — arrivals swing sharply between rush hours and quiet periods, so the average rate is not constant over time, breaking the key Poisson assumption.

Hence (d) is the case that does not generate a Poisson distribution as cleanly as the others.

Poisson, $\lambda=np=1$; $P(X=3)=\dfrac{e^{-1}1^3}{3!}=0.0613$.

With $n=100$ toys and defect rate $p=0.01$, the expected number of defectives is $\lambda=np=100\times 0.01=1$. For large $n$ and small $p$ the Binomial is approximated by the Poisson distribution:

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

For $x=3$:

$P(X=3)=\frac{e^{-1}(1)^3}{3!}=\frac{0.3679}{6}=0.0613.$

The Poisson distribution is the one whose mean equals its variance, both being $\lambda$.

For a Poisson distribution with parameter $\lambda$:

$\text{Mean}=\lambda,\qquad \text{Variance}=\lambda.$

Hence mean $=$ variance. (For a binomial, variance $=npq<np=$ mean; for a normal the two parameters $\mu$ and $\sigma^2$ are independent.) So the described parametric distribution is Poisson.

$P(X=1)=P(X=2)$ gives $m = \dfrac{m^2}{2}$, so $m = 2$, not $1$ — the statement is False.

For $X\sim P(m)$, the Poisson probability is

$P(X=x) = \dfrac{e^{-m} m^{x}}{x!}$.

Apply the given condition $P(X=1) = P(X=2)$:

$e^{-m} m = \dfrac{e^{-m} m^{2}}{2}$.

Divide both sides by $e^{-m} m$ (with $m\neq 0$):

$1 = \dfrac{m}{2} \;\Rightarrow\; m = 2$.

Since the condition forces $m = 2$, the claim that $m = 1$ is incorrect.

$P(\text{at most 2 defective}) = 2.5\,e^{-1}$.

Defectives are rare ($p = 0.02$) over many switches ($n = 50$), so use the Poisson approximation with mean

$\lambda = np = 50 \times 0.02 = 1.$

The Poisson probability is $P(X = r) = \dfrac{e^{-\lambda}\lambda^{r}}{r!}$. "At most two" means $r = 0, 1, 2$:

$P(X \le 2) = e^{-1}\left(\frac{1^0}{0!} + \frac{1^1}{1!} + \frac{1^2}{2!}\right) = e^{-1}\left(1 + 1 + \frac{1}{2}\right).$

$= e^{-1}\times 2.5 = 2.5\,e^{-1}.$

Small $p$ and large $n$ give a Poisson approximation with $\lambda = np = 1$. Then $P(X \le 2) = P(0) + P(1) + P(2) = e^{-1}(1 + 1 + 0.5) = 2.5e^{-1}$.

Step 1 — Parameters. $n = 50$, $p = 0.02$, so $\lambda = np = 50 \times 0.02 = 1$. Since $p$ is small and $n$ large, use the Poisson distribution $P(X = r) = \dfrac{e^{-\lambda}\lambda^{r}}{r!}$.

Step 2 — At most two defective.

$P(X \le 2) = e^{-1}\left(\frac{1^0}{0!} + \frac{1^1}{1!} + \frac{1^2}{2!}\right) = e^{-1}\left(1 + 1 + \frac{1}{2}\right).$

Step 3 — Simplify.

$P(X \le 2) = e^{-1}(2.5) = 2.5e^{-1}.$

Equating $P(X=2)$ and $P(X=3)$ for a Poisson distribution gives $\lambda=3$.

1. The Poisson probability mass function is $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$.
2. So $P(X=2)=\dfrac{e^{-\lambda}\lambda^2}{2!}=\dfrac{e^{-\lambda}\lambda^2}{2}$ and $P(X=3)=\dfrac{e^{-\lambda}\lambda^3}{3!}=\dfrac{e^{-\lambda}\lambda^3}{6}$.
3. Setting $P(X=2)=P(X=3)$: $\dfrac{e^{-\lambda}\lambda^2}{2}=\dfrac{e^{-\lambda}\lambda^3}{6}$.
4. Cancel $e^{-\lambda}$ (never zero) and multiply both sides by $6$: $3\lambda^2=\lambda^3$.
5. Rearrange: $\lambda^3-3\lambda^2=0 \Rightarrow \lambda^2(\lambda-3)=0$.
6. So $\lambda=0$ or $\lambda=3$. Since a Poisson parameter must be positive ($\lambda>0$), reject $\lambda=0$.
7. Hence $\lambda=3$.

Since $P(X=0)=e^{-\lambda}=k$ gives $\lambda=\log(1/k)$, and the Poisson variance equals $\lambda$, the variance is $\log(1/k)$.

1. The Poisson pmf is $P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$.
2. At $x=0$: $P(X=0)=e^{-\lambda}$.
3. Given $P(X=0)=k$, so $e^{-\lambda}=k$.
4. Taking natural log of both sides: $-\lambda = \log k \Rightarrow \lambda = -\log k = \log\dfrac{1}{k}$.
5. For a Poisson distribution, mean $=$ variance $=\lambda$.
6. Therefore variance $=\log\dfrac{1}{k}$.