Question

The rate at which customers are served at an airport check-in counter is a Poisson process with a rate of 8.1 per hour. The probability that more than 50 customers are served at the counter in the next 5 hours is P(Xp>50). If this is solved as a Poisson variable, the calculations will be tedious. So we use the normal approximation. Now, P(Xp > 50)=P(Z > a), where Z is the standard normal variable. What is the value of a? Please report your answer in 3 decimal places.

Answer 1

Answer:

Value of a is 0.667                      

Step-by-step explanation:

We are given the following in the question:

The rate of customer at an airport follows a Poisson distribution with

$\lambda = 8.1$

The normal approximation of Poisson distribution can be done in the following manner

$\mu = \lambda = 8.1\\\sigma^2 = \lambda = 8.1\\\sigma = \sqrt{8.1} = 2.85$

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of a such that

$P(X_p > 50)=P(Z > a)$

More than 50 customers per 5 hours means more than 10 customer per hour.

Thus, we can write

$P( x > 10) = P( z > \displaystyle\frac{10 - 8.1}{2.85}) = P(z > 0.667)$

Therefore, value of a is 0.667