Question

The organizer of the Montreal International Art Exhibit is trying to determine its optimal operating hours for its next one-day exhibition. Studies have shown that the arrival times at any given exhibition form a normal distribution with the average time that visitors arrive being 2 hours and 56 minutes after doors open, with a standard deviation of 48 minutes.
a) If the organizer sets the opening of the exhibition at 10:00 a.m., at what time would they expect 95% of the visitors to have arrived?
b) If the organizer sets the opening of the exhibition at 9:00 a.m., at what time after the doors open will only 15% of the visitors have arrived?
c) At what time should the organizer open the exhibition if they would like 70% of the visitors to have arrived by 1:00 p.m. so that they can award the first door prize?

Answer 1

Answer on Question #60351 – Math – Statistics and Probability

Question

The organizer of the Montreal International Art Exhibit is trying to determine its optimal operating hours for its next one-day exhibition. Studies have shown that the arrival times at any given exhibition form a normal distribution with the average time that visitors arrive being 2 hours and 56 minutes after doors open, with a standard deviation of 48 minutes.

a) If the organizer sets the opening of the exhibition at 10:00 a.m., at what time would they expect 95% of the visitors to have arrived?

b) If the organizer sets the opening of the exhibition at 9:00 a.m., at what time after the doors open will only 15% of the visitors have arrived?

c) At what time should the organizer open the exhibition if they would like 70% of the visitors to have arrived by 1:00 p.m. so that they can award the first door prize?

Solution

2 h 56 min = 176 min. One should consider normal distribution with μ = 176 and σ = 48.

a) The fact “95% visitors arrived” means the cutoff score for the top 5% of arrival times, or P(x>X) = 0.05; P(x<X) = 0.95

The z-score associated with the given probability value can be obtained either from the standard normal table or by using the technology (NORM.S.INV() function of MS Excel).

For p = 0.95, z = 1.645

Converting z-score to the arrival time value:

X = μ + zσ = 176 + 1.645×48 = 255 min = 4 h 15 min

If the exhibition opens at 10 am, 95% of visitors will arrive by 2:15 pm

b) The fact “15% visitors arrived” means P(x<X) = 0.15

For p = 0.15, z = -1.036;

X = μ + zσ = 176 - 1.036×48 = 126 min = 2 h 6 min.

If the exhibition opens at 9 am, 15% of visitors will arrive by 11:06 am

c) P(x<X) = 0.7

For p = 0.7, z = 0.524,

X = μ + zσ = 176 + 0.524×48 = 201 min = 3 h 21 min.

If the organizer needs 70% arrival rate by 1 pm, he needs to open the exhibition at 9:39 am.

Answer: a) 2:15 pm; b) 11:06 am; c) 9:39 am.

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