Use the following information to answer the next six exercises: On average, a clothing store gets$120$ customers per day.

What is the probability of getting $35$customers in the first four hours? Assume the store is open $12$hours each day.

Poisson distribution is a discrete likelihood conveyance that communicates the likelihood of a given number of occasions happening in a decent timespan or space assuming these occasions happen with a known steady mean rate and freely of the time since the last occasion.

Numberof shoppersper hour $=\frac{120}{12}=10$
Number of customers within very first four hours $=10\times 4=40$
Therefore, during this case average number of consumers every four hours localid="1649711283194" $=40$
This implies,localid="1649711288878" $Y~P\left(40\right)$.
We want to search out the probability of getting localid="1649711303347" $35$customers within the first four hours i.e.,

localid="1649711309395" $P$localid="1649711296747" $(=35)$.

We know,

$P(Y=y)=\frac{e^{-\lambda }{\lambda }^y}{y!};y=0,1,2,\dots ..\mathrm{\&}\lambda >0$

On putting the values, the specified probability is obtained as:
localid="1649711320840" $P(Y=35)=0.0485$