Question

Consider a new hotel deciding on cleaning staff hiring for the upcoming season. Cleaning times depend on whether it is a stay-over room or a check-out. Suppose that a guest will check-out on a given day with probability 40%. From your experience in similar hotels you estimate that a stay-over room cleaning time is well-described with normal distribution with average 15 minutes and standard deviation 1 minute. Check-out room cleaning time is also normal but with average 30 minutes and standard deviation 10 minutes.

i. Consider an occupied room (stay-over or check-out), what is the average cleaning time for such a room?

ii. Find the variance for the cleaning time for an occupied room.

iii. Suppose that the hotel has 200 rooms, and you estimate that on a given day a room will be occupied with probability 90%. Only occupied rooms need cleaning. Find the average total cleaning time for the hotel. iv. Find the variance of the total cleaning time for the hotel.

Hints: remember var(X) = EX^2 − (EX)^2 .

Answer 1

    Let S be the stayover room cleaning time                          
   S follows normal distribution with mean µs and standard deviation σs                          
   where    μs = 15 minutes       σs = 1 minute              
                              
   Let C be the check out room cleaning time                          
   C follows normal distribution with mean µy and standard deviation σy                          
   where    μc = 30 minutes       σc = 10 minutes              
                              
1)   Let O be the cleaning time for an occupied room                          
   Guests check out with 40% of the time                          
   Thus guests stay over 60% of the time                          
   Hence the cleaning time for occupied room is                          
   O = 0.4 * C + 0.6 * S                          
   Then O follows normal distribution with mean µ and standard deviation σ                          
   where                          
   μ = 0.4 * 30 + 0.6 * 15 = 21                          
   σ =                           
                              
   σ = 4.0448                          
   μ = 21                          
   Average cleaning time for an occupied room = 21 minutes                          
                              
2)   As calculated in (1), variance for an occupied room is                           
   σ² = (4.0448)² = 16.36                          
   Variance for an occupied room = 16.36 minutes                          
                              
3)   O is the cleaning time for the occupied rooms                          
   O follows normal distribution with mean µ and standard deviation σ                          
   where    μ = 21 minutes       σ = 4.0448 minutes              
   Since 90% of rooms are occupied,                          
   Average total cleaning time for 200 rooms = 0.9*200*21                          
   Average total cleaning time for 200 rooms = 3780 minutes                          
                              
                              
4)   Let T be the total cleaning time                          
   T = 0.9*200*O = 180*O                          
   Var(T) = (180)² * V(O)                          
                  = 32400 * 16.36                          
                  = 530064                          
   Variance of total cleaning time for 200 rooms = 530064 minutes