Finding a good lower bound for probability that sum of r.v.'s takes a value bigger than constant

Question

For motivation: In a town, there are $114$ busses. Each bus has a $10\%$ chance of being broken on any given day, independently of the other busses. I want to find a good estimate of the probability that on any day at least $99$ busses are able to drive.

Here is what I've done: The busses are i.i.d $X_i$'s with values in $\{0,1\}$ where $0$ means broken and $1$ not broken. We calculate $E(X)=0*0.1+1*0.9=0.9$, $Var(X)=E(X^2)-E(X)^2=0.9-0.81=0.09$ and $E(S):=E(\sum_{i=1}^{114}X_i)=\sum_{i=1}^{114}E(X_i)=114*0.9=102.6$, where we have used independence of the $X_i$. This means we get $$P[S\geq99]=P[S\geq E(S)-3.6]=P[S-E(S)\geq-3.6]$$$$=P[3.6\geq S-E(S)\geq -3.6]+P[S-E(S)>3.6]=P[|S-E(S)|\leq3.6]+P[S-E(S)>3.6]\geq P[|S-E(S)|\leq3.6]=^*1-P[|S-E(S)|>3.6]\geq\frac{Var(S)}{3.6^2}=1-\frac{114*Var(X)}{12.96}$$$$=1-\frac{0.09*114}{12.96}\approx0.21$$

where the second inequality is by Chebyshev.

Now this seems unnaturally low to me, I don't think this is a good estimate. Also, when looking up the solution to the exercise, I found that in the step I labeled with * above, the solution sets $1-P(|S-E(S)|>3.6)=1-P(|S-E(S)|\geq4.4)$, which I don't understand at all. They arrive at a $47\%$ estimate in the end instead of my $21\%$.

So a) Does somebody understand where the $4.4$ comes from or b) what other way could I arrive at the $47\%$?

Answer 1

You could indeed get a better result by using the central limit theorem.

Let $S_n= \sum_{i=1}^n X_i$ be the total number of none broken busses. We have $\mu = E(X_i)=p=0.9$ and $\sigma= \sqrt{var(X_i)} = \sqrt{p(1-p)}$ and hence have, by the central limit theorem (with $n=114$).

$$Z:= \frac{S_n-np}{\sqrt{n p(1-p) }} \stackrel{approx.}{\sim} N(0,1)$$ And hence: \begin{align*} P(S_n>99) & = 1-P(S_n\leq 99) \\ & = 1- P(\underbrace{\frac{S_n-114\cdot 0.9}{\sqrt{114\cdot 0.9\cdot 0.001}}}_{=Z \stackrel{approx.}{\sim}N(0,1)} \leq \frac{99-114\cdot 0.9}{\sqrt{114\cdot 0.9\cdot 0.1}})\\ & = 1- \Phi(\frac{99- 102.6}{\sqrt{114\cdot 0.9\cdot 0.1}})\\ & = 1- \Phi(-1.124) \\ & \approx 0.87. \end{align*}

Edit: I think that the 4.4 is just a mistake from them (which then leads to their result). I can not see anything wrong with your try (but I would understand "more than 99" as $S>99$ and not as $S\geq 99$).

Have another solution which leads to the same result as yours:

Let $\widetilde{S} = 114 -S$ be the total number of broken buses, with $E(\widetilde{S}) = 114-E(S) = 11.4$.
We then search (correcting for $>$ instead of $\geq $) \begin{align*} P(\widetilde{S}<15) & = P(\widetilde{S}-E(\widetilde{S})< 15-11.4 )\\ & = P(\widetilde{S}-E(\widetilde{S})< 3.6 ) \end{align*} While on the other hand, since $\widetilde{S}-E(\widetilde{S})\leq |\widetilde{S}-E(\widetilde{S})|$ we have $$P(|\widetilde{S}-E(\widetilde{S})|< 3.6 ) \leq P(\widetilde{S}-E(\widetilde{S})< 3.6 )$$ At the same time it follows from the Chebyshev's inequality that $$P(|\widetilde{S}-E(\widetilde{S})|< 3.6 ) \geq 1- \frac{10.26}{3.6^2} \approx 0.21$$ hence $$0.21 \leq P(\widetilde{S}-E(\widetilde{S})< 3.6 ) = P(\widetilde{S}<15).$$