Two railway companies respectively deploy one train (to get from city X to city Y). In total, $1000$ people randomly choose the train, respectively with probability $\frac{1}{2}$.

How many seats should make one railway company available in the train, such that the probability, that at least one of their passengers needs to stand, is less than $0.01$?

I think for these problem I need to use theorem of De Moivre Laplace.

I call total amount of people $n = 1000$

Probability for choose train is $p = \frac{1}{2}$

But formula for it is strange and I'm not sure how use it correct:

$$\lim_{n \rightarrow \infty}\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}}\leq x\right) = \Phi(x)$$

When insert everything in formula we have

$$\lim_{n \rightarrow \infty} \left(\frac{1000-0.5n}{\sqrt{0.5n(1-0.5)}}\right) = \lim_{n \rightarrow \infty} \left(\frac{1000-0.5n}{\sqrt{0.25n}}\right) = \lim_{n \rightarrow \infty}\left(\frac{1000}{\sqrt{0.25n}} - \sqrt{n}\right) = 0-\infty = -\infty$$

But I see from solution something went wrong :(

It was also hard find the correct formula on internet because in script there is strange thing. How solve this correct because I think similar question can asked in lesson and I want do it correct in test.

  • $\begingroup$ Since you should consider a fixed $n$ CLT resp. Moivre Laplace in this Bernoulli case is not the right choice. $\endgroup$ – Konstantin Jan 17 '18 at 20:56

$X\sim \text{Bin}(n,p)$. If $k$ are the available seats we are interested in the event $\{X \geq k+1 \}$ and we want $\Bbb P (X \geq k+1) \leq 0.01$

By $$\lim_{n \rightarrow \infty}\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}}\leq x\right) = \Phi(x)$$ we could say $n=1000$ is large enough and we can approximate $$\Bbb P (X \geq k+1) = 1-\Bbb P (X \leq k) =1- \Bbb P \left(\frac{X-1000 \cdot \frac 1 2}{\sqrt{1000 \cdot \frac 1 2(1-\frac 1 2)}}\leq \frac {k - 1000\cdot \frac 1 2}{\sqrt{1000 \cdot \frac 1 2(1-\frac 1 2)}} \right)\\ = 1-\Bbb P \left(\frac{X-np}{\sqrt{np(1-p)}}\leq \frac {k - 500}{\sqrt{250}} \right) \approx 1- \Phi\left(\frac {k - 500}{\sqrt{250}}\right) \overset{!}{\leq} 0.01 $$ This leads to $$\Phi\left(\frac {k - 500}{\sqrt{250}}\right) \geq 0.9 \Leftrightarrow \frac {k - 500}{\sqrt{250}}\geq 1.28155 \Leftrightarrow k \geq \sqrt{250}\cdot 1.28255 +500 = 520.278896065$$ The company should make 521 seats available.


Just a hint for starting the exercise:

Let's denote $X$ the number of people joining company A's railway. Furthermore use $s$ for the number of seats we search for.

You want to achieve: $0.01 > \mathbb{P}(X-s>0) = \mathbb{P}(X>s)$, where $X = \sum_{i_1}^n X_i$ and all $X_i$ admit the given Bernoulli distribution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.