The Average Weight of American Women

I read an article in the newspaper recently and I read something I could not believe. The article was about Disneyland’s “It’s a Small World” ride being shut down to make modifications possibly because of the expanding waistlines of riders. That part was believeable. The unbelieveable part (at least to me) was that it said the average weight of American women (aged 20-74) is 164 pounds.

Yes, that’s right, 164 pounds. Do you believe it? I didn’t. I couldn’t think of any female friend who weighs that much. Admittedly, a large percentage of the female friends I have are Asian, so perhaps my sample is skewed. Still, I found it hard to believe. So I looked for and found the CDC report online. You can view it here:

It’s from 2004 and I couldn’t find anything wrong with it. I wondered if their sampling size was too small, or if it was unfairly biased (in scientific terms). But it looks like they’ve done a thorough job. My only complaint is that it (along with many other surveys I’ve seen) doesn’t include “Asian” as a race category. It only breaks it down into hispanic, non-hispanic white, and non-hispanic black. I’d be very interested to know what the average is for Asians, but I couldn’t find that information on the web.

So I guess I have to accept the results as fact. But I can still continue along on this note by asking the question: suppose I wanted to survey people and ask them for their weight. Assuming that many would be reluctant to tell the truth, can I get an accurate average?

This is similar to a problem solved in the book “Struck By Lightning” which I read a couple months ago. Basically it involves adding some randomness to the survey. We can extend that idea to this problem. (warning, math ahead)

Instead of simply asking for a person’s weight, we’ll do the following: Have the person roll a die, without letting us see the result. Tell them that if it comes up 1, tell us their exact weight. If it comes up 2, tell us their weight plus 10 pounds. If it comes up 3, tell us their weight plus 20 pounds, etc. This would help some. If someone then tells us their weight is 150 pounds, their actual weight is somewhere between 100 and 150 pounds, but we have no way of knowing what it is. But since we know the probability of each die roll is the same, the average difference is 25 pounds, so to get the average, we simply subtract 25 pounds from the average of all the weights we are told. Assuming our sampling size is large enough, we should have an accurate average.

The downside to this approach is that if someone rolls a 6 and has to tell us their weight plus 50 pounds, they may be reluctant to do so since we know they can’t weigh any less than 50 pounds less than the number they tell us.

One way to solve this problem would be to tell them that if they roll a 6, roll the die again. Keep rolling as long as they roll a 6, adding 50 pounds each time they roll a 6. This way, there’s no bound to how much they have to add, so there’s no conclusion we can draw if they tell us they weigh 995 pounds. Well, we can conclude that they got extremely unlucky with the die rolls, but it gives no clue as to their actual weight. The math gets a little more tricky, though. The adjustment to the average is no longer 25 pounds.

Let A = (1/6) * (0 + 10 + 20 + 30 + 40).

Then the adjustment is A + (1/6) * (50 + A + (1/6) * (50 + A + …))

I’m sure there’s some relatively simple mathematical method to figure out the exact answer (possibly related to continued fractions), but it’s 1:30am so I should probably go to sleep. I will say that I expanded it out a few terms and it looks to be about 30. I wouldn’t be surprised if it’s exactly 30 (maybe I’ll work out the details tomorrow).

Ok, I’m still awake. I’m a sucker for a math problem.

Let S = A + (1/6) * (50 + A + (1/6) * (50 + A + …))

Then:

A + (1/6) * (50 + S) = A + (1/6) * (50 + A + (1/6) * (50 + A + …))
A + (1/6) * (50 + S) = S
A + 50/6 + S/6 = S
A +50/6 = S * (5/6)
S = (A + 50/6) * (6/5)
S = (150 / 6) * (6/5)
S = 150/5 = 30

Aha! Now I can sleep.