Random Squares

Probability Level 2

A factory manufactures wooden squares whose side length should be one meter. However, the side length of the squares is not always $\SI{1}{\meter}$ but varies uniformly between $\SI{1}{\meter}$ and $\SI{1.1}{\meter}.$ Note that the height and the breadth of each square piece remain equal to each other.

Two inspectors, Alice and Bob, want to calculate the average area of a square.

Alice claims that the average area is $\SI{1.1025}{\meter\squared}$ because the average side length is $\SI{1.05}{\meter}$ and $1.05 \times 1.05 = 1.1025.$
Bob claims that the average area is $\SI{1.1050}{\meter\squared}$ because the the area is between $(\SI{1}{\meter} \times \SI{1}{\meter})$ and $(\SI{1.1}{\meter} \times \SI{1.1}{\meter}),$ and $\frac{1 \times 1 + 1.1 \times 1.1}{2} = 1.105$ .

Who is correct?

Alice Bob Neither of them

7 solutions

Brian Charlesworth
Nov 12, 2017

With the side length $x$ varying uniformly from $100$ to $110$ cm the average area will be

$\displaystyle\dfrac{1}{110 -100}\int_{100}^{110} x^{2} dx = \dfrac{1}{10} \times \left(\dfrac{x^{3}}{3}\right)_{100}^{110} = \dfrac{1}{30}(110^{3} - 100^{3}) = \dfrac{331000}{30} = 11033.\bar{3}$ cm $^{2}$ .

Thus neither Bob nor Alice is correct.

How is that? What kind of average is it?

Malthe Christensen - 3 years, 6 months ago

The average, (or mean), of a function $f(x)$ over the interval $(a,b)$ is $\displaystyle \dfrac{1}{b - a} \int_{a}^{b} f(x) dx$ . In this case $f(x) = x^{2}, a = 100$ and $b = 110$ .

Brian Charlesworth - 3 years, 6 months ago

Why does f(x) = x^2?

Gabriel Hunt - 3 years, 6 months ago

@Gabriel Hunt – We're trying to find the average area of a square of side length $x$ , and the area of such a square is $x^{2}$ .

Brian Charlesworth - 3 years, 6 months ago

It's the average of a continuous function. The area can take on any real number value between 1.0 and 1.21 square meters. Alice only considers one possible value, Bob only considers two possible values. There are really infinite possible values.

Stephen Beck - 3 years, 6 months ago

This answer assumes that even though there is 10% error rate in the cut of one dimension, the height and width are always a perfect match. If we instead assume that the product is a rectangle with even distributions across the range, the 2-variable integration is more interesting, and the average is slightly less, but still does not match either of the wrong answers in the original question. If we allow the produced object to not have perfect right angles...then I can't do the math in my head. I might need a simplifying assumption.

Mel Nicholson - 3 years, 6 months ago

Good point. Since the shapes are described as "squares" it would seem that we can assume that the height and width vary in tandem, but this is not made explicit. If we employ your assumption, I get an answer of $11025$ cm $^{2}$ , which is indeed slightly less. If all four sides were to vary independently we would need to do 4-variable integration on Bretschneider's formula, (the general formula for the area of a concave quadrilateral), which would be extremely messy.

Brian Charlesworth - 3 years, 6 months ago

Agree. I assumed the errors would be independent (realistically), and thus came to 11025 cm2, Alice's answer. That needs to be more explicit in the problem statement.

Steve McMath - 3 years, 6 months ago

@Steve McMath – @Calvin Lin Given Mel's and Steve's comments, it might be worth adding a note stating that the side lengths of any specific object manufactured are all the same. This could be considered implicit due to the fact that the objects are referred to as "squares", but I can see how some ambiguity might arise. I could add the note myself but since the alternate interpretation leads to a different answer, (namely that Alice is correct), I know I should leave this decision up to you. Thanks. :)

Brian Charlesworth - 3 years, 6 months ago

@Brian Charlesworth – I've added a note to clarify. The problem did say "side length of the square" / "average area of the square", though I can see how the real world setting could lead to a lobsided quadrilateral which passes quality testing as a square.

Calvin Lin Staff - 3 years, 6 months ago

Whilst the answers I have read may give correct numerical answers, none explain the mistakes made by Alice and Bob.

Alice uses the average length but fails to recognise that the areas created with below average side lengths have a smaller variance in area than those created with above average side lengths. This is because the lengths are squared to give the area.

Bob's solution assumes a uniform distribution of areas, but if the sides are uniformity distributed, the areas will not be.

Garry Slocombe - 3 years, 6 months ago

How did you get 1/110-100?

Prakhar Mathur - 3 years, 4 months ago

The average of a function $f(x)$ over an interval $(a,b)$ is $\displaystyle \dfrac{1}{b - a} \int_{a}^{b} f(x) dx$ .

Brian Charlesworth - 3 years, 4 months ago

@ Prakhur Mathur - You need grouping symbols, such as: 1/(110 - 100).

Linda Slovik - 2 years, 5 months ago

Arjen Vreugdenhil
Nov 27, 2017

Alice is wrong. She correctly identifies the median area, but generally, the median is not equal to the expectation value of a distribution. Generally, $\mathbb E f(X) \not= f(\mathbb E X)$ : the expectation value of a function of a stochastic variable is not equal to the function's value at the expectation of its distribution.

Bob's mistake is to assume that the square of a uniform distribution is distributed uniformly. For the actual distribution of $X^2$ , see below.

For the correct answer, we must take the average (by integration) of all possible areas, weighted according to the distribution of the side $X$ : $\mathbb E(X^2) = \int x^2\:d\mathbb P_X = \int_1^{1.1}\:x^2\:10dx = \left[\frac{10}3x^3\right]_1^{1.1} = \frac{10}3(1.1^3 - 1^3) \approx 1.103.$

Alternatively, consider that $\text{Var}\ X = \mathbb E X^2 - (\mathbb E X)^2$ , and that the variance on a uniform distribution between $a$ and $b$ is equal to $\tfrac1{12}(b-a)^2$ . Then $\mathbb E X^2 = \text{Var}\ X + (\mathbb E X)^2 = \tfrac1{12}(0.1)^2 + (1.05)^2 \approx 0.000833 + 1.10250 = 1.10333.$ Thus, the variance of the distribution is precisely the correction necessary for Alice's answer.

The distribution of $X^2$ is But if $x$ is uniformly distributed between $0 < a < b$ , then $p_X(x) = \begin{cases} \dfrac 1{b-a} & a < x < b \\ 0 & \text{otherwise} \end{cases}$ then $p_{X^2}(y) = \begin{cases} \dfrac 1 {2(b - a)\sqrt y} & a^2 < y < b^2 \\ 0 & \text{otherwise} \end{cases}$ which is obviously not uniform: smaller values are more likely than larger values.

See my comment on Brian's answer. Is it fair to assume that even though they can miss the length by 10%, but will always have a perfect match between height and width?

Mel Nicholson - 3 years, 6 months ago

Since the problem speaks of the "side length of the square", rather than the "dimensions of the rectangle", this is indeed what appears to be implied. A clarification would have been nice.

In case both the width and the length vary uniformly and independently, we have $\mathbb E(XY) = \mathbb EX\cdot \mathbb EY = 1.05\cdot 1.05 \approx 1.103,$ and Alice would be correct.

Both interpretations are covered by the equation $\mathbb E(XY) = \mathbb EX\cdot \mathbb EY + \text{Cov}\ (X,Y).$ If we know that the board is always square, $X = Y$ and $\text{Cov}\ (X,Y) = \text{Cov}\ (X,X) = \text{Var}\ X$ .

If we know that length and width are independent, $\text{Cov}\ (X,Y) = 0$ .

Arjen Vreugdenhil - 3 years, 6 months ago

Be X one of the lenght and be Y the other lenght. X and Y are independent random variables uniformly distributed. The expected value E() of the area is E(XY) = E(X)E(Y). For a uniform distribution, the expected value is (a+b)/2 being a and b the extreme value of the distribution. Therefore E(X) = E(Y) = (1+1.1)/2 = 1.05.... Then, E(XY) = 1.05^2 = 1.1025.

Elvis Zavatti - 3 years, 6 months ago

If the board always turns out to be a square (as I read the problem), $X$ and $Y$ are not independent so that $\mathbb E(XY) \not= \mathbb E X\cdot \mathbb E Y$ .

Arjen Vreugdenhil - 3 years, 6 months ago

Yes but: "She correctly identifies the median area" is untrue. if you graph the curve xy = 1.1025 on the square regions 1< x,y < 1.1 then the area above the curve is smaller than the area below. We have a curve that touches the top left to bottom right diagonal at (1.05, 1.05) but is above it elsewhere. The median value would come from the curve xy=a which splits the square's area into two equal areas.

Robert Creamer - 3 years, 6 months ago

By definition, the median of a random variable $X$ is $m$ such that $\mathbb P(X < m) = \tfrac12$ . And $\mathbb P(X^2 < 1.1025) = \mathbb P(X < \sqrt{1.1025}) = \mathbb P(X < 1.05) = \tfrac12.$

Note that I interpreted the problem as saying that all boards are exact squares. If the length and width vary independently between 1.0 and 1.1 m, so that the most boards are non-square rectangles, then Alice's answer is actually correct. To find the median in that case, we consider (for $m < 1.21$ ) $\mathbb P(XY < m) = \int p_X(x)\:\mathbb P(Y < m/x)\:dx = 10 \int_1^{1.1}\:\mathbb P(Y < m/x)\:dx$ $= \begin{cases} 0 & m \leq 1 \\ 100 \int_1^m (m/x - 1)\:dx = 100 m\ln m-100m+100 & 1 \leq m \leq 1.1 \\ 10 (m/1.1 - 1) + 100 \int_{m/1.1}^{1.1} (m/x-1)\:dx = 100 m\ln \frac{1.21}m + 100m - 120 & 1.1 \leq m \leq 1.21 \\ 1 & m \leq 1 \end{cases}$ A quick calculation shows that $\mathbb P(XY < 1.1) \approx 0.48412,$ so that the median $m > 1.1$ . We solve numerically $100m\ln\frac{1.21}m + 100m - 120 = \tfrac12$ and find $m \approx 1.10168$ for the median in this situation. Admittedly, a lot more complicated!

Arjen Vreugdenhil - 3 years, 6 months ago

Right - got it now. I missed the bit about length and breadth being equal.
Rather an unnatural condition mind, that the factory can't manufacture the width of wooden squared to a tolerance of better than 10%, but they can manufacture the lengths to be exactly equal to the widths!.. But still it was part of the question.

Robert Creamer - 3 years, 6 months ago

David Fairer
Dec 2, 2017

Whilst the answers given are correct, I think that I must respectfully say that I don't care for them. I think that the answer can be seen easiest by drawing a graph of the length of the square on the x axis, and the area of the square (y = x^2) on the y axis. And then the average from x = 1 to x = 1.1 is a constant line from x = 1 to x = 1.1 with the value AVG (which will of course be between 1 and 1.1^2 = 1.21. So area of rectangle with length 0.1 and height AVG = Integration between 1 and 1.1 of x^2 (I don't know how to generate the integration sign). So 0.1 x Avg = [(x^3)/3] between 1 and 1.1. So 0.1 x AVG = (1 - 1.1^3) / 3. So AVG = 0.331 x 10 / 3. So AVG = 3.31 / 3. Which is 331 / 300. Which is 1.1 + 1 / 300, a rather nasty value! Regards, David Ps All the values of the area are in meters squared of course.

The question is incorrectly framed. The length of the sides of the wooden squares - or of anything - cannot be one square meter. It should read "A factory manufactures wooden squares whose side length should be one meter."

&y Edwards - 3 years, 6 months ago

Mihaly Hanics
Dec 1, 2017

Could you solve this with only a good diagram? My idea was to let x be between 1 and 1.1, and graph the points between (x,1 x) and (x,1.1 x). If you draw these points, they will form a trapezoid (with vertices at (1;1), (1;1.1 1), (1.1;1.1 1.1), (1.1,1.1)). Those points are the possible values for the area (their y coordinate), and maybe somehow you could count their average. Notice how you can cut the trapezoid into two pieces, the two right triangles, and work on them differently, then count their weighted average.

Well it varies uniformly. Alice used 2 decimal points. I think we would have to know the frequency of variation otherwise.

Jane Stewart - 3 years, 6 months ago

Michel Latil
Dec 1, 2017

Without doing any calculation or verification, none is correct because the initial data is given inthe problem statmen is wrong and thus not computable. The problem states: “whose side length should be one square meter“. This is wrong because a length cannot be measured in square meters because that is a surface measure. So, since the data of the problem is not correct, no computation using it can yield a correct answer. Again: thinking out of the box

Daniel Langstaff
Nov 30, 2017

This question was difficult to understand because of word choice. I'm guessing that the factory is making cubes and not squares. Next I guessed that the front and rear faces of the cube are always squares. This leaves only the side length to vary between 1 and 1.1m in length. Assuming equal amounts of uniform statistical distribution between those numbers then the average works out to be {1x1.1+1x1}/2=1.05

I do agree with you the question is ill posted since you they give you two dimension and subsequently you imagine is a board of one meter width by one meter high. In my more than five decade as engineer I never have seen such a cubic wooden block of one meter edge, nor I can imagine the purpose of making it ..unless to be registered in the Guinness

Mariano PerezdelaCruz - 3 years, 6 months ago

Edwin Gray
Nov 27, 2017

the average would be (1/.1)*(integral of (1 + x)^2 from 1 to 1.1 = .11033333/.1 = 1.10333333 m^2 Ed Gray

Random Squares

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