Question

At a checkout counter customers arrive at an average of 1.5 per minute. Find the
probabilities that (a) at most 4 will arrive at any given time (b) at least 3 will arrive during
an interval of 2 min time (c) at most 15 will arrive during an interval of 6 min.

Answer 1

Assume that customers arrival has Poisson distribution with $\lambda = 1.5$

Let $P_k=$ the number of arrivals during the given interval.

a)

Let $\lambda_t$ be arrival rate during $t$ minutes, $\lambda_t=1.5\cdot t$;

$P(k\le4)=$

$P_0+P_1+P_2+P_3+P_4=$

$\sum_0^4{\frac{(1.5t)^k}{k!}e^{-1.5t}}$

b)

$\lambda=1.5\cdot 2=3$

$P(k\ge3)=1-P_0-P_1-P_2=$

$1-\frac{3^0}{0!}\cdot e^{-3}-\frac{3^1}{1!}\cdot e^{-3}-\frac{3^2}{2!}\cdot e^{-3}=$

$1-e^{-3}(1+3+9/2)=1-\frac{17}{2}e^{-3}\approx 0.577.$

c)

$\lambda=1.5\cdot6=9$

Use Normal distribution $N(\lambda,\lambda)$ to approximate Poisson distribution:

$F_{poisson}(x,\lambda)\approx F_{normal}(x+1/2,\lambda,\lambda)$

$F_{poisson}(x,9)\approx F_{normal}(x+1/2,9,9)$

$F_{normal}(x+1/2,9,9)=Pr(X_{normal}\le15.5)=$

$Pr(\frac{X_{normal}-9}{3}\le\frac{15.5-9}{3})=Pr(Z\le2.1667)\approx0.985$

Another option is to use cumulative Poisson distribution table, this method gives 0.978.

Answer: a) $\sum_0^4{\frac{(1.5t)^k}{k!}e^{-1.5t}}$, b) 0.423, c) 0.978.