It is known that the number of people who enter a bank during a time interval of

t minutes is a Poisson random variable with the parameter �t. The bank opens at 8am and you
arrive at the bank at uniformly random time between 8am and 9am. Let X be the number of
people who entered the bank before you. Find the expectation and the variance of X.

To find the expectation and variance of X, we need to understand the concept of a Poisson random variable and apply it to this situation.

A Poisson random variable represents the number of events occurring in a fixed interval of time or space. It is characterized by a single parameter, lambda (λ), which represents the average rate of events occurring in that interval.

In this case, the parameter λ is given as λ = �t, where t is the time interval in minutes.

The first step is to convert the given time interval of one hour (60 minutes) into the corresponding λ value for a one-minute interval. To do this, we divide 60 (minutes) by t.

Let's assume t = 1 minute (as we are given the arrival time is uniformly random between 8am and 9am).

Thus, λ = �t = �(1) = �.

Now, X represents the number of people who entered the bank before you, and X is a Poisson random variable with parameter λ.

The expectation (E[X]) of a Poisson random variable is equal to its parameter λ, which in this case is �.

Therefore, E[X] = �.

The variance (Var[X]) of a Poisson random variable is also equal to its parameter λ.

Therefore, Var[X] = �.

In summary:
- The expectation (average) of X is equal to the parameter λ, so E[X] = �.
- The variance of X is also equal to the parameter λ, so Var[X] = �.

Since we don't know the specific value of � from the given information, we cannot find the exact numerical values of expectation and variance. However, we now know how to calculate them based on the given parameter.