Question

Customers arrive at a shop according to the Poisson distribution with a
mean of 10 customers/hour. The manager notes that no customer arrives
for the first 3 minutes after the shop opens. The probability that a
customer arrives within the next 3 minutes is

Answer 1

Answer:

For the Poisson process of rate$\lambda , and for any t > 0$, the Probability mass function for$\mathrm{N}(\mathrm{t})$ (i.e., the number of arrivals in $(0, \mathrm{t}])$is given by the Poisson PMF

$P_{N(t)}(n)=\frac{(\lambda t)^n \exp (-\lambda t)}{n !}$

Step-by-step explanation:

Step :1 If the arrivals of a Poisson process are split into two new arrival processes, each new process is independent.

Calculation:

Given:

Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour.

$\Rightarrow \lambda=10 customers / 60 \mathrm{~min}=1 customer / 6 \mathrm{~min}=1 / 6 customers per minute;$

The manager notes that no customer arrives for the first 3 minutes after the shop opens,

$\Rightarrow \mathrm{t}=3 \mathrm{~min} and \mathrm{n}=0 From the Poisson's PMF,The probability of zero customers in 3 minutes will be$

$P_{N(3)}(0)=\frac{\left(\frac{1}{6} \cdot 3\right)^0 \exp \left(-\frac{1}{6} \cdot 3\right)}{0 !}=e^{-0.5}=0.606$

$P N(3)(0)=fracleft(frac16cdot3right)0 exp left(-frac16cdot3right)0!=e-0.5=0.606$

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