Question

The rate at which customers are served at an airport check-in counter is a Poisson process with a rate of 10.4 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

SOLUTION:

From given data,

Poisson process with a rate of 10.4 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.

Let  Xp be a Poisson random variable which denotes the number of customers that are served at the counter in the next 5 hours.

Then : k = t

= 10.4 * 5

= 52

Now let Y be a normal random variable which we will use to approximate  Xp .

The continuity correction that we must introduce to facilitate this approximation is:

P( Xp > k) = P(Y > k - 0.5)

P( Xp > 50) =  P( Xp > 49 ) = P( Y >48.5 )

P( Y >48.5 ) = P[( Y - 52 ) / > (48.5 - 52 ) / ]

= P(Z > -3.5 / )

   = P(Z > -0.48536 )

= > a = -0.485