Uniform Problem - Expected Value

Question

A fraternity is throwing a party for its members. The cost of booking a band to play at the party, the amount that the supplier is going to charge, the cost of renting some place and some other costs are random uniform variables over the intervals (1300,1800),(1800,2000),(800,1200) and (400,700) respectively. If the number of people invited is a random integer in the interval (150,200). What is the minimum amount, in average, the fraternity will have to charge each person to not lose money?

What I did was: The sum of all the costs and then the average, that gave me 5000, then the average of the attendees, 175. I divided them. But I don't know how to continue.

Answer 1

Here is a simulation in R of total costs for the party and cost per ticket.

set.seed(2021)
m=10^5
x1 = runif(m, 1300, 1800)
x2 = runif(m, 1800, 2000)
x3 = runif(m, 800, 1200)
x4 = runif(m, 400, 700)
t = x1 + x2 + x3 + x4
n = sample(150:200, m, rep=T)

summary(t);  sd(t)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   4333    4850    5000    5000    5150    5673 
[1] 212.0116

You can sum the means of the four uniform distributions to get the mean of the total cost (about $\$5000).$ Also, you can sum the variances of these four distributions (provided they're independent) to get the variance of the total cost; the take the square root to get the standard deviation (about $\$212.)$ The distribution is not exactly normal but nearly normal.

hist(t, prob=T, col="skyblue2", 
     main="Simulated Dist'n of Cost")

summary(t/n)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  21.88   26.62   28.55   28.77   30.81   37.40 
q = quantile(t/n, .95);  q
     95% 
33.45204 
hist(t/n, prob=T, col="skyblue2", 
     main="Approx. Dist'n per Ticket")
 abline(v = q, col="red", lwd=2)

If the probability models for costs and attendance are correct, the fraternity is unlikely to lose money charging $\$33.50$ per ticket. [If the ticket price were set at $\$28.55$ (median of distribution) there would be a 50:50 chance of losing money.]

Answer 2

The question in the exercise points IMO to an exercise where a so called "fair price" is to be calculated, which means that the expected costs should be equal to the expected revenue from the price.

You almost did the job already:

Expected costs: $5000$

Expected number of attendees: $175$

Fair price $p$: $5000 = 175p \Rightarrow p = \frac{200}7 \approx 28.57143$.

If you want to be picky, you may round it up to $28.58$ in order not to loose money "in average" (as the exercise text requires).