MHB Probability : minutes that customers get served

[mathmari]

Hey! :giggle:

The average time of customer service at the cash registers of a department store is $30$ people per hour. If a new customer arrives at the checkout, then calculate the probabilities:

a) to be served in less than 15 minutes

b) to need to be served from 6 to 8 minutes
a) Do we have Poisson disrtibution with $\lambda=30$ ?
But then we consider "minutes" instead of "number of customers".
Could you give me a hint ?

:unsure:

 

[romsek]

a) 30 people per hour is 2 minutes per person. The wait time is exponentially distributed with this mean. Evaluate the probability that the wait time is less than 15 min given this distribution.

b) I'm not sure what the word "need" means here. If they mean find the probability the customer is served in 6-8 minutes it's just the CDF of the above mentioned distribution evaluated at 8 minutes same evaluated at 6 minutes.

 

[mathmari]

a) 30 people per hour is 2 minutes per person. The wait time is exponentially distributed with this mean. Evaluate the probability that the wait time is less than 15 min given this distribution.

b) I'm not sure what the word "need" means here. If they mean find the probability the customer is served in 6-8 minutes it's just the CDF of the above mentioned distribution evaluated at 8 minutes same evaluated at 6 minutes.

How do we know that we have an exponential distribution? :unsure:

 

[romsek]

How do we know that we have an exponential distribution? :unsure:

It's a property of the Poisson/Exponential distributions.

30 people an hr implies a Poisson distribution on the number of arrivals during a given period.

It's just a property that the time between Poisson arrivals has an exponential distribution.

 

[mathmari]

It's a property of the Poisson/Exponential distributions.

30 people an hr implies a Poisson distribution on the number of arrivals during a given period.

It's just a property that the time between Poisson arrivals has an exponential distribution.

Ah ok!

So do we have the following?

a) $P(X<15)=1-e^{-2\cdot 15}$

b) $P(6\leq X\leq 8)=P(X\leq 8)-P(X\leq 6)=(1-e^{-2\cdot 8})-(1-e^{-2\cdot 6})$

Is that correct? :unsure:

 

[romsek]

Is that correct? :unsure:

I believe so.