A Lego shop is open between 10-18.00.
question 'a': At the 15.th of December last year, 16 Star Wars lego was sold. What's the probability that between 10.00 and 12.00 more than five legos were sold(on that day)?
question b: Supposed, that this was an average day before Christmas, what's the probability that between 10.00 and 12.00 more than five legos were sold at any given day?
For question 'a', I tried to use Poisson distribution, with an expected value of E=2, based on the opening time and our two hour long time interval. How can I solve this example?
Thank you
In my opinion, without any further information, the Poisson distribution is definitely the right one to use here.
Since you're told that $\ 16\ $ items were sold over a period of $8$ hours on an "average day", you can assume that the number of sales over a period of length $\ T\ $ hours follows a Poisson distribution with parameter $\ 2T\ $, or $4$ (not $2$) for your given period of $2$ hours from $10.00$ to $12.00$.
For question $b$, you simply have to calculate $$ \mathrm{Pr}( S> 5 )=\sum_\limits{k=6}^\infty\mathrm{Pr}( S=k )\ ,$$ where $\ S\ $ is the number of sales in the two-hour period in question, and you simply have to replace $\ \mathrm{Pr}\ $ with the Poisson distribution with parameter $4$.
For qustion $a$ you know that $16$ was the actual, not merely the expected, number of items sold on December $15^\mathrm{th}$, so what you need to calculate is the conditional probability that more than $5$ items are sold between $10.00$ and $12.00$ given that a total of $16$ items were sold between $10.00$ and $18.00$—that is: \begin{eqnarray}\mathrm{Pr}\left( S_1 > 5\left\vert S_1+S_2=16\right. \right)&=& \frac{\mathrm{Pr}\left( \left\{S_1 > 5\right\}\cap\left\{ S_1+S_2=16\right\} \right)}{\mathrm{Pr}( S_1+S_2= 16 )}\\ &=&\frac{\sum_\limits{k=6}^{16}\mathrm{Pr}\left( \left\{S_1 = k\right\}\cap\left\{ S_2=16-k\right\} \right)}{\mathrm{Pr}( S_1+S_2= 16 )}\\ &=&\frac{\sum_\limits{k=6}^{16}\mathrm{Pr}\left( S_1 = k\right)\mathrm{Pr}\left( S_2=16-k \right)}{\mathrm{Pr}( S_1+S_2= 16 )}\ , \end{eqnarray} where $\ S_1\ $ is the number of items sold during the interval $10.00$ to $12.00$, and $S_2$ the number sold during the interval $12.00$ to $18.00$. You can typically assume that the numbers sold over disjoint intervals are independent, and you know that $\ S_1$ has a Poisson distribution with parameter $4$, $\ S_2\ $ a Poisson distribution with parameter $12$, and $\ S_1+S_2\ $ a Poisson distribution with parameter $16$.