# Math Help - poisson distribution

1. ## poisson distribution

can anyone show me how do you do part (i) and (ii) , this involve, i dont know how to start . thanks in advance..

2. ## Re: poisson distribution

Since tea and coffee sales are independent the joint distribution of their sold amounts in a minute is the product of their individual distributions.

Coffee sales are Poisson w/mean $\lambda_c=1.5$

Tea sales are Poisson w/mean $\lambda_t=0.5$

So the Joint PMF is given by

$p_{C,T}(c,t)=\left(\dfrac{(1.5)^c}{c!}e^{-1.5}\right)\left(\dfrac{(0.5)^t}{t!}e^{-0.5}\right)$

for (i) simply plug (1,1) in for (c,t)

for (ii) just recast things to be over the course of 3 minutes rather than 1.
Coffee sales are now Poisson with $\lambda_c=3*1.5=4.5$
Similarly Tea sales are now Poisson with $\lambda_t=1.5$

You should be able to finish from here.

3. ## Re: poisson distribution

Originally Posted by romsek
Since tea and coffee sales are independent the joint distribution of their sold amounts in a minute is the product of their individual distributions.

Coffee sales are Poisson w/mean $\lambda_c=1.5$

Tea sales are Poisson w/mean $\lambda_t=0.5$

So the Joint PMF is given by

$p_{C,T}(c,t)=\left(\dfrac{(1.5)^c}{c!}e^{-1.5}\right)\left(\dfrac{(0.5)^t}{t!}e^{-0.5}\right)$

for (i) simply plug (1,1) in for (c,t)

for (ii) just recast things to be over the course of 3 minutes rather than 1.
Coffee sales are now Poisson with $\lambda_c=3*1.5=4.5$
Similarly Tea sales are now Poisson with $\lambda_t=1.5$

You should be able to finish from here.

hi , i am now having problem with part II
less than 5 cups drink sold= 4 cup + 3 cup + 2 cup +1 cup + 0 cup, ( I let coffee = X , tea = Y) ,
so i have ( 3X , 1Y ) + ( 3Y , 1X ) +( 2X , 1Y) +( 2Y , 1X) + ( 1X , 1Y) +(2X , 0Y) + (2Y , 0X) + (1X) + (1Y) = 0.877 , please refer to the photo below for the working.

4. ## Re: poisson distribution

Originally Posted by delso

hi , i am now having problem with part II
less than 5 cups drink sold= 4 cup + 3 cup + 2 cup +1 cup + 0 cup, ( I let coffee = X , tea = Y) ,
so i have ( 3X , 1Y ) + ( 3Y , 1X ) +( 2X , 1Y) +( 2Y , 1X) + ( 1X , 1Y) +(2X , 0Y) + (2Y , 0X) + (1X) + (1Y) = 0.877 , please refer to the photo below for the working.
you're missing a bunch of cases and you seem to have your PDF wrong.

Valid cases of (c,t) are

(0,0)

(1,0) (0,1)

(2,0) (1,1) (0,2)

(3,0) (2,1) (1,2) (0,3)

(4,0) (3,1) (2,2) (1,3) (0,4)

5. ## Re: poisson distribution

Originally Posted by romsek
you're missing a bunch of cases and you seem to have your PDF wrong.

Valid cases of (c,t) are

(0,0)

(1,0) (0,1)

(2,0) (1,1) (0,2)

(3,0) (2,1) (1,2) (0,3)

(4,0) (3,1) (2,2) (1,3) (0,4)

then what's the correct PDF?

6. ## Re: poisson distribution

Originally Posted by delso
then what's the correct PDF?
ok looking at your worksheet you do have the right PDF. You must have some calculator errors or something.