Question

I just need an explanation as to why my answer is correct for the question below...

A large exhibition hall collected data about attendance at its exhibitors.Historical data shows that the average daily attendance for an event is 7850 people with a standard deviation of 367 people. If the capacity of the exhibition center is 8500 people, what is the probability that the center will reach full capacity?

These were my steps:
P(X <=8500) = 1- normalcdf (-1e99, 8500, 7850, 367)
= 0.0383.

The answer is correct, but I'm not sure why this method worked, as opposed to doing P(X = 8500)?

Answer 1

@SithsAndGiggles

Answer 2

Huh, I've never really thought about using Cumulative Distribution Function like that in this type of equation without having to use complimentary error function in the result.

Answer 3

is the complementary error function you were referring to as 1- normalcdf (-1e99, 8500, 7850, 367)? because I just did it as I could not get the answer any other way?

Answer 4

The attendance is normally distributed, so it's a continuous random variable. Points have zero probability of occurring with continuous random variables.

As for why you use \(X\le8500\), I'm not sure. I would have thought it would be \(X\ge8500\).

Answer 5

I used X<= 8500 because they wanted to fill the room to a max of 8500 people?

Answer 6

I tried X=8500 because I thought that made more sense as they wanted 8500, but the answer was wrong

Answer 7

Oh I see, I misread the question. I thought they wanted the probability that they'll exceed the total capacity. \(X\le8500\) makes sense in that case.