00:01
So for this problem, if we have that x is the result of the rand function, then x is going to be distributed as a uniform continuous random variable with minimum value zero, maximum value 1.
00:15
So generally, we have that if x is distributed as a uniform random variable between values a and b, we have the expected value of x, or pardon me, actually, i won't go for expect.
00:30
Value.
00:31
I'll go straight to finding the probability density function.
00:34
We'd have that the probability density function is going to be given by 1 over b minus a.
00:41
So in this case, 1 over b minus a is just 1 over 1 minus 0.
00:50
So f of x is equal to 1 for x in the interval from 0 up to 1.
01:01
Then moving on to part b, finding the probability of x being between 0 .25 and 0 .85, we can do that by finding the cumulative density function and evaluating it at 0 .85 minus the cumulative at 0 .25.
01:28
But we have, for our general uniform random variable, the cumulative density function is going to be given by x minus a x minus a divided by b minus a so for our specific case here that's just going to be x so we'd have the probability is going to be 0 .85 minus 0 .25 for a result of 0 .6 for part c probability of generating a random number with value less than or equal to 0 .3 so probability of x less than or equal to 0 .3 should simply be equal to f of 0 .3, which is itself 0 .3.
02:21
Then for part d, probability of x greater than or equal to, pardon me, not greater than or equal to, greater than 0 .7, it's going to be equal to 1 minus the probability of x less than or equal to 0 .7.
02:43
So that's then going to be 1 minus 0 .7 for a result of 0 .3...