Chance error?

Question

On the average, hotel guests who take elevators weigh about 150 pounds with an SD of about 35 pounds. An engineer is designing a large elevator for a convention hotel, to lift 50 such people. if she does it to lift 4 tons, the change it will be overloaded by random group of 50 people is about _____. Explain.

Thanks!

Answer 1

You need the standard deviation of the wieght of samples of 50 people.

The mean weight of samples of 50 people is 150 * 50, or 7500 pounds. Four tons is 8000 pounds.

With such a large sample as 50, you would expect that your sample means will each be quite close to 150. In fact, the standard deviation of the means of your samples is

35 / sqrt(50) = 35 / 5sqrt(2) = 7 / sqrt(2) = 7sqrt(2) / 2

or around 5 pounds. That means that 2/3 of the time, the weight of 50 people will lie between 7495 and 7505 pounds, and 95% of the time, their weight will lie between 7490 and 7510 pounds.

The design difference between 8000 pounds and 7500 pounds represents 100 standard deviations of the sample means. That's off the charts. There is essentially "zero" probability that a sample as large as 50 will be so far from the mean as that.

Modulo does some calculations, but upon finding the result "enormous" abandons reason and uses some other (wrong) method that gives a more "reasonable" result. He was almost right the first time. The problem supposes an enormous design margin.

Answer 2

Thanks for Gary's comments.

Corrected from last night's posting: arithemetic error, I should have used 4 tons (8000 lbs)...

What you're saying is that you take a sample of 50 people out of your population. You know the population mean and SD. You want the probability that the weight of this sample exceeds 4 tons.

I haven't done these in awhile but I can suggest that the weight of the sample is the sum of 50 normals. I think that you want a Z value defined as

Z = (4 tons - 50*mean)/(sqrt(50)*sigma)


That would give a Z of about 500/(7*35) = about 2 which is about 2 SDs above the mean. That results in the probabiity being about 1%.

So, calculate your Z, find P(z 1-P(z