1. ## continuous uniform distributions

for part 'a' I got: $1.6 . \frac{1}{20} = 0.08$

'b': mean =10
variance = $\frac{(20-0)^{2}}{12} = \frac{100}{3}$

for part 'c', I cant understand what the question is asking, is it that they want to find the probability of the 'dot appearing in a square, of side 4cm, in the middle of the screen? I know 'X' represents the distance of the dot from the left edge of the screen, and Y the distance from the bottom edge, so would it just be multiplying the both?

Also need help with part 'd', the question says 'within 2cm of the edge of the screen', should it not also tell me what 'edge' they are talking about?

Any help appreciated,
thank you.

2. Originally Posted by Tweety

for part 'a' I got: $1.6 . \frac{1}{20} = 0.08$

'b': mean =10
variance = $\frac{(20-0)^{2}}{12} = \frac{100}{3}$

for part 'c', I cant understand what the question is asking, is it that they want to find the probability of the 'dot appearing in a square, of side 4cm, in the middle of the screen? I know 'X' represents the distance of the dot from the left edge of the screen, and Y the distance from the bottom edge, so would it just be multiplying the both?

Also need help with part 'd', the question says 'within 2cm of the edge of the screen', should it not also tell me what 'edge' they are talking about?

Any help appreciated,
thank you.
Since the distribution of the dot in each direction is uniform and the directions are independent you can do it geometrically: Probability = (Area of required region)/(Total area).

By the way, it would be more helpful if you typed the question rather than attaching it (there is nothing in the question that makes that difficult to do).