The arrival of customers over fixed time intervals in a bank follow a Poisson distribution with an average of 30 customers/hour. The probability that the time between successive customer arrival is between 1 and 3 minutes is _______ (correct to two decimal places).

The correct answer is 0.38.

Step-by-step Explanation:

We are given that the arrival of customers in a bank follows a Poisson distribution with an average arrival rate of 30 customers per hour.

Let $\lambda$ be the average arrival rate. Since we need to find the probability of the time between successive arrivals in minutes, we first convert the hourly rate into a per-minute rate:
$\lambda =\frac{30\text{ customers}}{60\text{ minutes}}=0.5\text{ customers/minute}$

For a Poisson process with arrival rate $\lambda$, the time interval $T$ between successive arrivals follows an exponential distribution. The cumulative distribution function (CDF) of this exponential distribution is given by:
$P(T\le t)=1-e^{-\lambda t}$

We want to find the probability that the time between successive customer arrivals is between 1 and 3 minutes, which is represented by $P(1<T<3)$.

Using the CDF, this probability can be calculated as:
$P(1<T<3)=P(T<3)-P(T<1)$

Now, we compute each term individually using $\lambda =0.5$:
For $t=1$:
$P(T<1)=1-e^{-0.5\times 1}=1-e^{-0.5}\approx 1-0.6065=0.3935$

For $t=3$:
$P(T<3)=1-e^{-0.5\times 3}=1-e^{-1.5}\approx 1-0.2231=0.7769$

Subtracting the two values gives the required probability:
$P(1<T<3)=0.7769-0.3935=0.3834$

Rounding to two decimal places, we get:
$P(1<T<3)\approx 0.38$