There are four transport companies $A, B, C, D$, which transport passengers from Paris to Vienna.
Statistically $2000$ a month uses those companies' services and the probability that a person will choose company $A$ is $0,3$.
The company $A$ has a coach with $40$ places.
How many rides should company $A$ organize in order for the probability that a client will go away to another company is less than $0,01$.
Here is my approach:
Let $X(\omega) = X_1 + ... + X_n$ be the number of passengers transported monthly by $A$.
If we divide it by $40$, we will have the number of rides.
So the company $A$ needs to transport at least $0,3 \cdot 2000=600$ people with the probability at least $0,99$, that is $P(X \ge 600) \ge 0,99$.
Now we standardize the variable $X$ and get $Z = \frac{X - 0,3n}{\sqrt{0,3 \cdot 0,7 n}}$
And if I plug $X = \sqrt{0,21n}Z+0,3n$ I get $P(\sqrt{0,21n}Z+0,3n \ge 600) = P(Z \ge \frac{600-0,3n}{\sqrt{0,21n}}) = \Phi(\frac{0,3n-600}{\sqrt{0,21n}}) \ge 0,99$
So $\frac{600-0,3n}{\sqrt{0,21n}} \ge 2,33$
$0,3n + 1,06 \sqrt{n} -600 \le 0$
But the result isn't credible.
What am I doing wrong?
Could you help?
You have partly the right idea, but not quite. (Because I'm on the North American side of the Atlantic, I'll be using . instead of , in decimal fractions.)
First, just an intuitive approach. On average, you need to accommodate 0.3(2000) = 600 passengers a week. But to lose only a very small proportion to other companies you will have to have capacity for somewhat more than 600. The question is how many more.
Your weekly number of customers is a random number $X \sim Bin(n=2000, p=0.3),$ so you want to find $c$ such that $P\{X > c\} = 0.01$ or, equivalently, $P\{X \leq c\} = 0.99$. If you have statistical software available, you can find $c$ exactly and directly. For example, in R, 'qbinom(.99, 2000, 0.3)' returns 648. You would need 648/40 = 16.2 trips (busloads). If you round up to 17, you will take care of essentially 100% of the demand (lose no customers at all). If you round down to 16, you will have capacity for 16*40 = 640 people, which will be adequate 97.5% of the time. (In R, 'pbinom(640, 2000, 0.3)' returns 0.9754.)
If you don't have such software available, your idea to use the normal approximation will work: $$0.99 = P\{X \le c\} = P\{(X - 600)/20.5 \leq (c - 600)/20.5\} \approx P\{Z \leq 2.326\},$$ where $Z$ is a standard normal random variable, $\sqrt{2000*.3*.7} = 20.5,$ and solving $(c - 600)/20.5 = 2.326$ gives $c = 648.$
It seems your only mistake was to confuse the 600 customers on average with the $n = 2000$ potential customers when attempting the normal approximation to the binomial.