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Question: How would you estimate the weight of Mount Everest? Give a 90% and 95% confidence interval.

I would define what Mount Everest is. Including its boundaries (length, width) and estimate the height. Then you estimate the density of a rock like composite such as granite to compute the weight, if you assume that such a mountain can be viewed as a pyramid. Now I am stuck on how to compute a 90% and 95% confidence interval for such questions.

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  • $\begingroup$ You are probably expected to give some measure of uncertainty around your estimates of length, width, height, density and then combine them suitably $\endgroup$
    – Henry
    Apr 16, 2015 at 19:24
  • $\begingroup$ @Henry Can you further clarify estimates? What would such an estimate be? Do you mean ranges? $\endgroup$
    – AG10
    Apr 16, 2015 at 19:28

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Defining what constitutes Mount Everest is very difficult because it sits on a high plateau. Do you just measure the material above the plateau? Do you measure all of Eurasia down to sea level?

To play along, I would prefer to do the big island of Hawaii. It roughly a cone sitting above the ocean floor, so at least we know what we are trying to estimate the weight of. I seem to recall it is a little over 30,000 feet (10,000 meters) high from the floor-looking it up seems against the rules here. Sea level is then halfway down the cone. I would guess the diameter of the island as 30 miles (50 kilometers) at sea level because you can drive across it in about an hour. Assuming the cone continues the same way to the ocean floor, we have a cone with diameter 100 km and altitude 10 km. The volume is $\frac 13 \pi 50^2\cdot 10=25,000 km^3=2.5\cdot 10^{13}m^3$ where I have used the convenient fact that $\pi=3$. Taking a density of $4 tonnes/km^3$ this gives $10^{14}$ tonnes as our estimate.

I enjoy problems like this. Asking people for $90\%$ and $95\%$ confidence bounds is a good way to get them mad at you because they always underestimate the uncertainties. I don't think there is any reasonable way to distinguish between the two. How good do I think this estimate is? I think the height and $\pi$ are very close, but the diameter at the base could easily be off a factor $5$-what goes on above water may be different from what goes on below. Thinking more about the density, I would change that to $3 tonnes/km^3$ with a factor $2$ possible error. So my final answer would be $8 \cdot 10^{13} tonnes$ and expect to be within a factor $50$ either way.

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  • $\begingroup$ Hi Ross can you further elaborate what off by a factor means? I am not sure I am fully understanding the confidence interval portion of the question. How would I derive the value for the factors? Is there an equation? $\endgroup$
    – AG10
    Apr 18, 2015 at 23:44
  • $\begingroup$ Off by a factor $5$ means it could be one fifth of my estimate or could be five times as large. My estimate of the diameter was 50 miles, so I am now claiming to be pretty sure it is somewhere in the range 10 miles to 250 miles. This is also a guess. Once you have made the guesses for each item that goes into the calculation, you combine them to get an estimate for the error bound on the result. In this case, as we multiply all the items, we can just multiply the error bounds to get the bound on the result. $\endgroup$ Apr 18, 2015 at 23:54
  • $\begingroup$ If I am understanding your explanation because you took 1/5 of the diameter, do I need to multiply every variable such as height and density by 1/5 then do the guesstimate calculation? And following that do the same and multiply by 5 to every variable parameter? I am having trouble following along and fully understanding. Thank you for your help. $\endgroup$
    – AG10
    Apr 19, 2015 at 2:08
  • $\begingroup$ How did you arrive at the 8 x 10^13 tones with factor 50? $\endgroup$
    – AG10
    Apr 19, 2015 at 2:11
  • $\begingroup$ No,look at the equation for the mass of a cone based on height, diameter, and density. I gave separate guesses for the uncertainty on height and density. I said the height was (to this level) exact and the density was a factor $2$. $\endgroup$ Apr 19, 2015 at 2:24
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These kinds of questions, with the requirement of a confidence interval are absolute nonsense. Your approach is correct, as this is more of an "order of magnitude" estimation problem, also known as a Fermi problem. You can certainly give some heuristic order-of-magnitude errors on your answer, but calling it a confidence interval is borderline nonsense.

On the other hand, you can explain to the interviewer (I've seen these come up during Job Interviews) that without even a sample of the height, the "confidence interval" part is complete nonsense. Equivalently, you could just say that with at least 95% confidence, the height is between 0 and 100km, the boundary between Earth and space. You can certainly "estimate" some distribution, say granite has some standard deviation of density that you pull out of a hat, etc. Maybe, you can call it a "hypothetical confidence interval."

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The phrase "Confidence intervals" assumes we know a lot about the distribution of the estimates, e.g. it is normal, with known mean and distribution. If we don't have that, then our estimates are not point samples from a population, but some other measure.

I would use interval arithmetic, the name "interval arithmetic" doesn't overstate the knowledge we have about the estimate, the intervals are just the highest and lowest we are comfortable with. The 90% and 95% would have to be thrown out of the analysis.

Fuzzy numbers come to mind as a option, but they're less known and not effortless to calculate, which is what this sort of fermi problem is trying to exploit (that some calculations are easy for humans make)

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