I try to solve the following out of an old book on statistics:
Cardboard boxes are stacked. The boxes have an average height of 10 cm and the height is normally distributed with a standard deviation of 1 cm. The maximum height of the building is 106 cm.
If boxes are picked at random, determine the chance of:
- not having enough space to stack 10 boxes;
- having enough space to stack 11 boxes;
- having enough space to stack 10 boxes but not having any space (left) for a extra 11'th box.
My solution
Let $X\stackrel{d}{=} N(10,1)$ then $Y=10X \stackrel{d}{=}N(100,10)$ and $U=11X\stackrel{d}{=}N(110,11)$
The solution for the first problems is quickly found using the standard-normal distribution.
- $\mathcal{P}(Y\geqslant 106) = 0,0287$
- $\mathcal{P}(Z\leqslant 106) = 0,1131$
- The last 'problem' is bugging me. I tried the following:
$\mathcal{P}(Y\leqslant 106, Z\geqslant 106)$ $= \mathcal{P}(Y\leqslant 106)\cdot \mathcal{P}(Z\geqslant 106) = (1-0,0287)\cdot (1-0,1131) \approx 0,8614$
The book proposes the solution: 0,8582. (the difference isn't big, but I wonder what causes it?, the other (sub)answers align perfectly with the book)
I guess breaking the condition isn't allowed since $Y$ and $Z$ are not indepent? But I don't have any idea how to continue without it...
It should be clear that $A=\left\lbrace Z\leq 106\right\rbrace \subset \left\lbrace Y\leq 106\right\rbrace=B$ -- if $11$ fit then so do $10$. What you want is the probability of $\left\lbrace Y\leq 106\right\rbrace\setminus\left\lbrace Z\leq 106\right\rbrace=B\setminus A$. Since \begin{equation*} \mathbf{P}(B)=\mathbf{P}(B\setminus A)+\mathbf{P}(B\cap A), \end{equation*} we have \begin{align*} \mathbf{P}(B\setminus A)&=\mathbf{P}(B)-\mathbf{P}(B\cap A)\\ &=\mathbf{P}(B)-\mathbf{P}(A)\\ &=(1-0.0287)-(0.1131)\\ &=0.8582 \end{align*}