homework check

Question:

A production line produces washers of circular shape with a diameter that varies randomly. If the designed diameter of a washer is 16mm and the observed values of the diameter after a long production run were 15, 15.2, 15.5, 16.2 mm, what is the expected error for the area of a washer. (Error = absolute value of the difference between the designed value and the observed value)

... based on the question, i need to find the expected error for the area of the washer, so i need to use the area formula of a circle, (pi)r^2.
... and, for the observed value 16.2mm, the error is +0.20mm, because the error supposed to be in absolute value.

so, i made a table..

X p(X=x)
0 (pi)(0.5)^2
1 (pi)(0.4)^2
2 (pi)(0.25)^2
3 (pi)(0.1)^2

E(x) = (sigma)(X)(p(X=x)
= 0.9896.

i'm really sorry, i don't know how to put the symbols etc.
hope you can understand my "writings".

do correct me if i'm totally wrong.
thank you.

Exact Value -- \(\displaystyle \pi r^{2}\)

A little error -- \(\displaystyle \pi (r+\Delta)^{2} \; = \; \pi (r^{2} + 2r\Delta + \Delta^{2})\)

The error portion, then -- \(\displaystyle \pi (2r\Delta + \Delta^{2})\;=\;\pi \cdot \Delta \cdot (2r + \Delta)\)

Oftentimes, it is thought that squared errors are small enough to ignore amd we are left with \(\displaystyle \pi \cdot \Delta \cdot 2r\)

You didn't say if you were allowed to utilize the calculus. With it, we could get there a little more directly with a derivative (differential variation).
If \(\displaystyle Area(t)\;=\;\pi \cdot r(t)^{2}\) then \(\displaystyle dArea\;=\;2\pi r \cdot dr\)

Care to rethink your original response?

okay, i understand now.
thanks a lot!