Worked solution
Take fourth moments when an area is proportional to the square
S has distribution
| s | |
|---|---|
| 1 | 0.5 |
| 5 | 0.3 |
| 10 | 0.1 |
| 20 | 0.1 |
A circle has radius .
(i) Find . marks
(ii) Find and . marks
(iii) Hence find the mean and variance of the circle’s area A. marks
[2006/MJC/II/26]
CJC · Review problems · Question 7
Answer
(i)0.7; (ii), variance 33; (iii), variance .
(i) Retain the factorisation and both boundary scores
. The source’s next line prints , but its included score 10 and the original non-strict inequality require .
How did this part go?
(ii) Mean, second moment and variance
. . Therefore .
How did this part go?
(iii) Area and mean
. Hence .
How did this part go?
(iii) Retain the fourth moment and variance of the square
. Now . Thus .
How did this part go?
(iii) Apply the scale factor squared
.
How did this part go?
Common mistakes
Var(S²) is not [Var(S)]². The qualifying boundary score 10 must remain included.
Solution notes
All asks/marks and table entries retained; every source factorisation, moment/variance term and area transformation retained. Source strict-inequality slip corrected explicitly to include 10, native original preserved.