Probability that the cafe gets overcrowded using de Moivre-Laplace theorem

Question

1951 people live in the city. The probability of a single citizen eating breakfast in a cafe is 46%; eating lunch - 46%; eating dinner - 40%. In the city there are 4 cafes in total and a businessman plans to open one more to make it 5.

1. What is the least number of seats required in the new cafe so it gets overcrowded (all seats filled) no more than 24% of the evenings?
2. How often will the cafe be overcrowded with the same amount of seats at breakfast and at lunch? My solution:

(1) The probability of a single person eating at the new cafe in the evening is $$0.4/5=0.08,$$ since the probability of them eating the dinner is 40% and there will be 5 cafes in the city.

Since it cannot get overcrowded more than 24% of the all given evenings, the probability of the cafe not getting filled up is 0.24.

Now, using de Moivre-Laplace theorem, the equation I have to calculate is $$P(Sₙ≥x)=0.24,$$ searching for the highest possible x.

$$P(\frac{Sₙ−np}{\sqrt{np(1−p)}} ≥ \frac{x−np}{\sqrt{np(1−p)}})=0.24$$

Let's say that

$$\frac{x−np}{\sqrt{np(1−p)}} = y,$$

then

$$P(\frac{Sₙ−np}{\sqrt{np(1−p)}} ≥ y)=0.24 = 1 - Φ(y)$$

$$0.24 = 1 - Φ(y)$$

$$Φ(y) = 0.76$$

$$y = 0.706303$$

$$\frac{x−np}{\sqrt{np(1−p)}} = 0.706303$$

$$\frac{x−1951∗0.08}{\sqrt{1951∗0.08∗(1−0.08)}} = 0.706303$$

$$x = 164.54$$

Since the cafe cannot get overcrowded, the least number of seats required is rounded up to 165.

(2) At breakfast, 46% of people eat at 5 cafes, so the probability of a person eating at the new cafe is $$0.46/5=0.092,$$ since percentage is the same for lunch, I won't recalculate.

Now I'd use the formula

$$P(Sₙ=m)≈\frac{(np)ᵐ}{m!}∗e⁻ⁿᵖ.$$

Where n = 1951 (total amount of people), p = 0.092 (probability of them eating at the new cafe) and m = 165 (total amount of seats).

$$P(Sₙ=159)≈\frac{(1951∗0.092)¹⁶⁵}{165!}∗e⁻⁽¹⁹⁵¹∗⁰.⁰⁹²⁾.$$

However, all I get is some ridiculously complicated numbers, thus, I believe my solution is not right. Can you get me on the track and tell me whether the first part of the solution was actually correct?

Answer 1

About 1), if you don't use the Moivre-Laplace theorem and go with the Binomial (1951 trials and 0.08 success probability) cdf, you get that $P(x>164) = 1-F_1(164) = 0.2393<0.24$. You just need 164 seats.

Regarding 2), again with the corresponding Binomial cdf (1951 trials and 0.08 success probability) evaluated at 164 you get $P(x>164) = 1-F_2(164) = 0.8808$. Your restaurant would be overcrowded 88.08% of the time at breakfast and lunch with 164 seats.