00:01
Hi, i'm david and i'm here to help you answer your question.
00:03
Now let me pick up your question here.
00:07
In this question, we're given the probability density function under random variable x.
00:13
And the first step we need to find will have to find the case for that the fx will be a valid pdf.
00:22
Remind you that the integral of the fx dx from minus infinity to infinity must equal to 1.
00:31
So to do the part a, we will go to do integral now.
00:35
We see x between 0 and the 1.
00:38
So we have a k -1 minus x, the x.
00:42
K gives the constant i can bring outside.
00:45
Untie derivative of 1 equals to x.
00:48
Untie derivative x equals the x squared over 2 e -valuate from 0 to 1.
00:54
So we should get k.
00:56
Now input the 1 in 7, we have 1 .5.
00:58
0 in second and 0 therefore we have the k out of 2 and it means that the 1 equal to the k out of 2 it means that k must equal to 2 therefore the pdf it has the form of the 2 1 minus x between the interval zone 1 0 to 1 now to do the other part before we do that i will find the cumulative distribution function f capital x by definition change equal to the probability x smaller equitaphethical equal to integral starting from the zero or no action here we have the two reduce the x by the variable t d t so we have the two untie derivative that 1 equal to t untid derivative the t equal to the t square over 2 e value from 0 to the x therefore we have the x minus x square over 2 and then find the x between the interval from 0 to 1 with that in mind we will be able to find the rest now now for the b we want to find the proper building of x and the 0 .2 notice that we can write this down to the proper b to x more than 0 .12 minus proper being x small is equal to 0 from 1 now for the continuous random equal or not equal will be the same thing so we can insert the equal for the smaller equal to 0 .2 the reason why we do that because now this one will be the community distribution and the 0 1 so if we're black pin we got equal to the 2 0 .0 2 minus 0 1 square over 2 minus 2 0 1 minus 2 0 1 minus 0 1 minus 0 1 minus 0 1 minus 0 2 1 square over 2 and if we compute this one get equal to the 0 .2 minus 0 .2 square divided by 2 and then we will have times 2 minus 2 times 2 times 0 .1 minus 1 square divided by 2 equal to the 0 .1 1 equal to the 0 .1 1 square divided by 2 equal to the 0 .1 square.
03:38
7.
03:39
Next we want to find the probability of x by common this one will equal to 1 minus problem by beta beta x more equal to 0 .5 f capital under 0 .5 we'll get equal to 1 minus 2 times 0 .5 minus 0 .5 square over 2 it will compute this one we have 0 .5 minus 0 .5 square divided by 2 times 2 times 2.
04:11
1 minus the answer equal to 0 .15.
04:18
Then 1 to 5, we did the probability that the x in list 3.
04:26
So the next one, 1 to 5 ,000 0 .3.
04:35
So this one will be the same as probability x more equal to 0 .3.
04:40
Then we get equal to the n...