I've got a poisson distribution problem that i'm having trouble with

IF on the average 0.3 customers arrive per minute at a cafeteria, what is the probability that exactly 3 customers will arrive during the five minute span?

 

P(X=5) = P(X<=5) - P(X<=4)

Use the distribution tables.

 

Or I believe it works this way too...

0.3 /min or 1.5 / 5 mins

Using the Poisson formula:

(1.5^3 * e ^ -1/5)/3! = 0.126

 

not that high :eek3:

 

those are some tiny customers

 

thanx man that helped out, i forgot how to do the customer/time conversion ratio.....do you have AIM? I may have a few other questions that I could use your help with.

 

my AIM is General Chang7

 

Post away in this thread... AIM doesn't love me anymore.

 

Please check this for me cause i'm missing a zero in my answer

If an average of 0.5 customers arrive per minute at a ticket counter, what is the probability that exactly 4 customers will arrive in a 5 minute span

a. 0.0534
b. 0.50
c. 0.68
d. 0.134

I got 0.534 which is just like a. w/o the 0 but rounded down would be b., I did it a couple times on my calculator and thats what i get

 

Answer is (d) kind sir

You have 0.5 arriving per minute... so 2.5 arrive every 5 minutes.

Doing P(X=4) we get (2.5^4 * e^-2.5)/4! = 0.134

The Poisson formula is basically:

P(X=n) = (mean^n * e^-mean)/n!