Poisson Distribution -- Rental of a number of television sets

[songoku]

Homework Statement

A dealer has a stock of 6 similar television sets which he rents out to customers on a monthly basis. It is known from past experience of the dealer that the monthly demand for the television sets have a Poisson distribution with mean 3.56
(i) Find the probability that in any month at least two sets are not rented out
(ii) Calculate the probability that in any month the demand is not fully met
(iii) Find the probability that exactly one set not rented out in exactly two out of four months.
(iv) If one of the television sets is temporarily withdrawn for repairs, find the expected number of television sets rented out in a month.

Homework Equations

Probability = e^-λ . λ^x / x!

Binomial distribution

The Attempt at a Solution

λ = 3.56 / month
(i) P (x ≤ 4) = P (x = 0) + P (x = 1) + P(x = 2) + P(x = 3) + P (x = 4)

(ii) P (x ≥ 7) = 1 - P (x ≤ 6)

(iii) I am not sure about the mean. Should we use mean per month or per 2 months or per 4 months?
I tried using per month so probability exactly one set not rented = P( x = 5) = p, then using binomial distribution the probability exactly two out of four months = 4C2 . p² . q²

(iv) expected value for Poisson distribution = λ
If no. of television = 6, mean = 3.56 / month
If no. of television = 5, mean = ... / month ---> what the question asks

How to find this one? Can simple ratio be applied for this question?Additional question: how to construct probability distribution table for Poisson distribution? I tried to find the probability from x = 0 until x = 6 (because there are 6 TV) but it doesn't add up to 1

Thanks

 

[StoneTemplePython]

I think $\text{i})$ and $\text{iii})$ are correct. To do $\text{ii})$ and $\text{iv})$ right, and to really understand the question, the first thing that needs to be addressed is:

Additional question: how to construct probability distribution table for Poisson distribution? I tried to find the probability from x = 0 until x = 6 (because there are 6 TV) but it doesn't add up to 1

Think about the physical interpretation of the problem. Ignore $iv$ for the moment. We have fixed supply of 6 televisions, but demand fluctuates and in any given month you have a Poisson distribution.

For a given month: Let's try enumerating the outcomes from the store owner's perspective:
If demand is zero, then store rents zero. If demand is 1, store rents 1. If demand is 2, store rents 2. If demand is 3, store rents 3. If demand is 4, store rents 4. if demand is 5, store rents 5. If demand is 6 store rents 6. If demand is 7, store rents 6. If demand is 8, store rents 6. If demand is 9, store rents 6, and if demand is 10, ...

How would you condense that into a probability distribution table that is meaningful to the store owner? (Hint: if you group things correctly, your table will have a finite number of entries and probabilities sum to one. )

After you have this table, re-visit the earlier questions. I really think drawing a picture, and in this case a table, should make things a lot clearer.

Key idea: find a way of representing this that makes sense to the dealer.

 

[songoku]

For a given month: Let's try enumerating the outcomes from the store owner's perspective:
If demand is zero, then store rents zero. If demand is 1, store rents 1. If demand is 2, store rents 2. If demand is 3, store rents 3. If demand is 4, store rents 4. if demand is 5, store rents 5. If demand is 6 store rents 6. If demand is 7, store rents 6. If demand is 8, store rents 6. If demand is 9, store rents 6, and if demand is 10, ...

How would you condense that into a probability distribution table that is meaningful to the store owner? (Hint: if you group things correctly, your table will have a finite number of entries and probabilities sum to one. )

Maybe the table will be like this:
x = 0 --> P = ...
x = 1 --> P = ...
x = 2 --> P = ...
x = 3 --> P = ...
x = 4 --> P = ...
x = 5 --> P = ...
x ≥ 6 --> P = 1 - P ( x ≤ 5) = ...

The total probability will add up to 1. Is this correct?

After you have this table, re-visit the earlier questions. I really think drawing a picture, and in this case a table, should make things a lot clearer.

Key idea: find a way of representing this that makes sense to the dealer.

For (ii), I am still stuck with the same idea. The demand won't be fully met if it is more than 6 TV
For (iv), I still don't have any idea what to do

Thanks

 

[Ray Vickson]

Maybe the table will be like this:
x = 0 --> P = ...
x = 1 --> P = ...
x = 2 --> P = ...
x = 3 --> P = ...
x = 4 --> P = ...
x = 5 --> P = ...
x ≥ 6 --> P = 1 - P ( x ≤ 5) = ...

The total probability will add up to 1. Is this correct?For (ii), I am still stuck with the same idea. The demand won't be fully met if it is more than 6 TV
For (iv), I still don't have any idea what to do

Thanks

It is like the calculation of the mean number rented, when the supply is 5 instead of 6. In other words, if X is the demand and R is the number rented, we have R = X if X <= 5 and R = 5 if X > 5. You want the expected value of R.

 

[StoneTemplePython]

Maybe the table will be like this:
x = 0 --> P = ...
x = 1 --> P = ...
x = 2 --> P = ...
x = 3 --> P = ...
x = 4 --> P = ...
x = 5 --> P = ...
x ≥ 6 --> P = 1 - P ( x ≤ 5) = ...

The total probability will add up to 1. Is this correct?For (ii), I am still stuck with the same idea. The demand won't be fully met if it is more than 6 TV
For (iv), I still don't have any idea what to do

Thanks

Yes. Your ii) sounds good -- I was reading it as supply won't be met, last night-- you have it right. You might consider adding one more row to the table and having the bottom row instead be $x \geq 7$ though its your call.

iv.) should be easy, if you step back from the equations and remember that an expected value is just the sum of each probability times associated pay-off

 

[songoku]

Yes. Your ii) sounds good -- I was reading it as supply won't be met, last night-- you have it right. You might consider adding one more row to the table and having the bottom row instead be $x \geq 7$ though its your call.

Won't be met and won't be fully met are not the same? I will do the same for both because I think they are the same. The demand won't be met if it is more than 6

iv.) should be easy, if you step back from the equations and remember that an expected value is just the sum of each probability times associated pay-off

It is like the calculation of the mean number rented, when the supply is 5 instead of 6. In other words, if X is the demand and R is the number rented, we have R = X if X <= 5 and R = 5 if X > 5. You want the expected value of R.

Maybe I should make a new table consisting of x = 0 until x = 5 and x ≥ 6 then use formula expectation = ∑ x.p but how to find the probability of, let say, x = 1?
Will it be like this: P (x = 1) = e^-λ . λ^x / x! , where λ = 3.56 and x = 1?
I am not sure about the lambda (3.56) because it is for 6 TV and now we want to find new lambda for 5 TV.

Thanks

 

[Ray Vickson]

Won't be met and won't be fully met are not the same? I will do the same for both because I think they are the same. The demand won't be met if it is more than 6

Maybe I should make a new table consisting of x = 0 until x = 5 and x ≥ 6 then use formula expectation = ∑ x.p but how to find the probability of, let say, x = 1?
Will it be like this: P (x = 1) = e^-λ . λ^x / x! , where λ = 3.56 and x = 1?
I am not sure about the lambda (3.56) because it is for 6 TV and now we want to find new lambda for 5 TV.

Thanks

The 3.56 governs the demand; the 5 or 6 governs the supply. The demand distribution would be the same whether they had 6 TVs or 600 TVs.

 

[songoku]

The 3.56 governs the demand; the 5 or 6 governs the supply. The demand distribution would be the same whether they had 6 TVs or 600 TVs.

So let x be the random variable of the demand.
P( x = 0) = e^-3.56 . 3.56⁰ / 0! = 0.02848388
P (x = 1) = 0.1012422
P (x = 2) = 0.180211
P (x = 3) = 0.2138506
P (x = 4) = 0.190326997
P (x = 5) = 0.1355128
P (x ≥ 6) = 0.1504174

E(x) = 0 x 0.02848
388 + 1 x 0.1012422 + 2 x 0.180211 + 3 x 0.2138506 + 4 x 0.190326997 + 5 x 0.1355128 + ?

I am confused how to continue for P(x≥ 6). I think I need to understand severel other things:
1. If I want to find P(x = 6), is it e^-3.56 . 3.56⁶ / 6! or
e^-3.56 . 3.56⁵ / 5! Or both are wrong?

2. The meaning of P(x = 6) is the probability there are 6 demands in a month?

3. λ = 3.56 means there are, in average, 3.56 demands in a month? Independent of the number of TV? So if there are 100 TVs, the average demand per month still 3.56? This also means that the expectation value of demand is 3.56?

Thanks

 

[Ray Vickson]

So let x be the random variable of the demand.
P( x = 0) = e^-3.56 . 3.56⁰ / 0! = 0.02848388
P (x = 1) = 0.1012422
P (x = 2) = 0.180211
P (x = 3) = 0.2138506
P (x = 4) = 0.190326997
P (x = 5) = 0.1355128
P (x ≥ 6) = 0.1504174

E(x) = 0 x 0.02848
388 + 1 x 0.1012422 + 2 x 0.180211 + 3 x 0.2138506 + 4 x 0.190326997 + 5 x 0.1355128 + ?

I am confused how to continue for P(x≥ 6). I think I need to understand severel other things:
1. If I want to find P(x = 6), is it e^-3.56 . 3.56⁶ / 6! or
e^-3.56 . 3.56⁵ / 5! Or both are wrong?

2. The meaning of P(x = 6) is the probability there are 6 demands in a month?

3. λ = 3.56 means there are, in average, 3.56 demands in a month? Independent of the number of TV? So if there are 100 TVs, the average demand per month still 3.56? This also means that the expectation value of demand is 3.56?

Thanks

1. Why are you asking how to find P(X=6)? You already wrote the correct formula in your very first post: "Probability = e-λ . λx / x!" So, P(X=6) = 3.56⁶e^-3.56/6!.
2. Yes: that is the meaning you have been using all along, ever since post #1.
3. Yes: that is what I said. More importantly: do you understand why that is the case?

Finally: you are looking at the wrong random variable. The question asks for ER, the expected number rented; but you are trying to look at EX. the expected number demanded. So: look at R, not at X. In particular, fill in the following table of P(R=k) with appropriate numbers:
$$\begin{array}{c|c}
k & \Pr\{R=k\} \\ \hline
0 & ? \\
1 & ? \\
2 & ? \\
3 & ? \\
4 & ? \\
5 & ?
\end{array}
$$
Note that if only 5 TVs are available they cannot rent more than 5.

After filling out the table you ought to be able to compute the expected value of R.

 

[songoku]

Ahh I think I get it a little bit clearer now

1. Why are you asking how to find P(X=6)? You already wrote the correct formula in your very first post: "Probability = e-λ . λx / x!" So, P(X=6) = 3.56⁶e^-3.56/6!.

Because I keep mixing demand and supply

3. Yes: that is what I said. More importantly: do you understand why that is the case?

Because demand and supply have two separate independent distributions?

Finally: you are looking at the wrong random variable. The question asks for ER, the expected number rented; but you are trying to look at EX. the expected number demanded. So: look at R, not at X. In particular, fill in the following table of P(R=k) with appropriate numbers:
$$\begin{array}{c|c}
k & \Pr\{R=k\} \\ \hline
0 & ? \\
1 & ? \\
2 & ? \\
3 & ? \\
4 & ? \\
5 & ?
\end{array}
$$
Note that if only 5 TVs are available they cannot rent more than 5.

After filling out the table you ought to be able to compute the expected value of R.

Number of rented = 0 if the demand = 0 so P(R = 0) = P(x = 0) = 0.02848388 and it will the same until P(R = 4). For P(R = 5), it will be the same as P(x ≥ 5). Hence, the expected value is:
0 x 0.02848388 + 1 x 0.1012422 + 2 x 0.180211 + 3 x 0.2138506 + 4 x 0.190326997 + 5 x (0.1355128 + 0.1504174)

Is this correct? Thanks

 

[Ray Vickson]

Ahh I think I get it a little bit clearer now
Because I keep mixing demand and supply
Because demand and supply have two separate independent distributions?
Number of rented = 0 if the demand = 0 so P(R = 0) = P(x = 0) = 0.02848388 and it will the same until P(R = 4). For P(R = 5), it will be the same as P(x ≥ 5). Hence, the expected value is:
0 x 0.02848388 + 1 x 0.1012422 + 2 x 0.180211 + 3 x 0.2138506 + 4 x 0.190326997 + 5 x (0.1355128 + 0.1504174)

Is this correct? Thanks

Yes.

So, what is the final numerical answer?

 

[songoku]

So, what is the final numerical answer?

3.29

 

[Ray Vickson]

3.29

Right, and including that in your answer might get you more marks. It's these little things that can add up!

 

[songoku]

Thank you for the help Ray and StoneTemplePython