Please Help - Random Variables

[bobcat89]

You are given a straight stick of length 21.97 cm. You break the stick at a position chosen uniformly at random along its length. Each of the two stick portions you break in half and make a rectangle with the four bits of the stick. What is the expected area of the rectangle?

 

[bobcat89]

Any reason for 21.97; like, why not 21 or 22 ?

Hi Denis,

No particular reason. Its just the way the question is set.

 

[HallsofIvy]

Let x be one length of the two pieces. The other is 21.97- x. The two sides of the rectangle are, then x/2 and 10.985- x/2. The area of the rectangle is (x/2)(10.985- x/2)= 5.4925x- x^2/4. Since x is "uniformly chosen between 0 and 21.97" its probability density function is \(\displaystyle P(x)= \frac{1}{21.97}\) for \(\displaystyle 0\le x\le 21.97\). The expected value of f(x) is, of course, \(\displaystyle \int f(x)P(x)dx\).

 

[bobcat89]

Let x be one length of the two pieces. The other is 21.97- x. The two sides of the rectangle are, then x/2 and 10.985- x/2. The area of the rectangle is (x/2)(10.985- x/2)= 5.4925x- x^2/4. Since x is "uniformly chosen between 0 and 21.97" its probability density function is \(\displaystyle P(x)= \frac{1}{21.97}\) for \(\displaystyle 0\le x\le 21.97\). The expected value of f(x) is, of course, \(\displaystyle \int f(x)P(x)dx\).

Hi,

That makes sense. its also what I had figured but when posed with a different but similar question. I failed to make the logic work.

I am stuck at figuring out which the longer piece would be.
Question is below:
Suppose that a stick of length 8.67 cm is broken at random into two pieces, with the location of the break being uniformly distributed along the length of the stick. The longest of the two pieces is taken, and is itself broken into four equal-sized pieces from which a square is made.

What is the expected area of the square?