uniform distrubution

[maths_arghh234]

hopefully this will be one of the final times i and a question on here, i think i've been using it too much, it's just such a great website.
okay here's the question

the isosceles right-angled triangle ABC in which AB=BC=Xcm where Xcm is a continuous random variable uniformly distributed on the interval (4,6)
a) find the probability that

i) the length of AC exceeds 8cm ii) the area of the triangle is less that 10cm^2

thanks for reading

 

[galactus]

A uniform distribution forms a rectangle.

Since the triangle is isosceles and the range of their lengths is from 4 to 6, the hypoteneuse of the triangle ranges from

\(\displaystyle \sqrt{4^{2}+4^{2}}=4\sqrt{2}\approx 5.657\) to \(\displaystyle \sqrt{6^{2}+6^{2}}=6\sqrt{2}\approx 8.485\)

What is the probability it is greater than 8?. Look at the graph. It is only a matter of 'pro-rating'.

The area of the triangle ranges from 8 to 18. So, what is the probability it is less than 10?.

 

[maths_arghh234]

thanks for replying

okay, so for a. i) do you work out the probability of X being between 8 and 8.485??
and for ii) work out the probability between 10 and 18 ??

 

[galactus]

Yes, find that area of the shaded regions as a percentage of thre whole area. since it is a uniform distribution, this represents the probability.

 

[maths_arghh234]

thanks

so i've got a formula in my book that says, probability the X is between c and d , is worked out by, (d-c)/ (b-a)
so if that's right is the answers for a) i) (8.485 -8 )/ (8.485-5.657) = 0.1715??
and for ii) (10-8)/ (18-8) = 0.2 ??