Stanton D. answered 01/23/20
Tutor to Pique Your Sciences Interest
So Madhumihta V.,
To convert this problem into a look-up, you need to scale the variance (given for one person's weight, i.e. standard deviation of single weight = √200 = 10√2 kg , or ~ 1.414 kg ) up for 10 adults. Personally, I prefer to work with the standard deviation. To obtain this, you scale up by (n)^0.5 ; so that value becomes SD = 10√2√10 or ~ 44.72 kg.
So, you are looking at average (¯x¯) +/- SD and comparing with the 800 kg limit.
So, you need first to compute how many times SD your limit is from the average passenger mass:
(800-700)/44.72 or about 2.236 standard deviations = Z (as it is called in these tables).
Then, you look up this value on a standard normal probability distribution table. You do not want a table showing "degrees of freedom" or anything like that. Make sure you get the "one-sided probability" value, or equivalent (it may indicate "area to the left" or "area to the right" but NOT "area outside both right and left"! -- since you are only interested in the probability of exceeding 800 kg, NOT the probability of being either <600 kg or >800 kg.
You should be able to interpolate a value of ~0.01268 .
So 0.01268 fraction of the time, the load will exceed 800 kg per the mathematical solution, and ~98.32% of the time the elevator will reach ground "safely".
Now let's get into the "nitty-gritty"! 800 kg is the stated rating for the elevator load. But there is a design factor for cables, called the safety factor, of from 6:1 to 10:1; hence, the actual failure of the cable might occur at 6*800 or ~ 4800 kg (and that's ignoring the weight of the cabin and counterweight). Assuming, that is, that the cable is in good condition. And, elevators may also include a run-away locking mechanism, which snaps the cabin into the shaft to immobilize it. But you don't want to go there, that abrupt halt would likely injure you anyway.
Perhaps more to the point -- when was the last time that you were on a crowded elevator (everyone is in a hurry and can't wait), but the passengers were conferring to total their mass and compare to the limit? I can think of several reasons why that process wouldn't occur, and even if it did, wouldn't be accurate!
"Mathematics brought down to ground"!
-- Cheers, -- Mr. d.
Stanton D.
Sorry, the second line above should read 14.14 kg, not 1.414 kg.01/23/20