00:01
In the given question, by having x be a discrete random variable, variable and here probability of getting x is equal to small x.
00:26
Here for 0 .1, 0 .2, 0 .2, 0 .3, and 0 .2 and 0 .3, 0 .0 .0.
00:40
So the values are for x is 0 .2.
00:46
Again, x is equal to 0 .4 here.
00:50
X is 0 .5 and x is equal to 0 .8.
00:56
Then x is equal 1 and for 0 it is otherwise.
01:02
So first of all, in this question we need to find the cda.
01:07
In the first part, a part, we need to find the cdia and plot it.
01:13
So for plotting cdf we have to make a table first.
01:18
So here is the table.
01:20
So here we can see this is our tabular data or tabular representation of the given information.
01:28
This is a table.
01:29
Here the values of x are is certainly 0 .2, 0 .4, 0 .5, 0 .8 and 1.
01:37
So px and fx frequency and probability are computed here in these two rows.
01:44
We can see and with the help of these values of tx and fx i have plotted this graph here so we can study all the values from the graph here we are having our frequency and this is our probability so this is the solution for part a of the question that we have find it and plotted the cdf of the given question now let's move on to the b part of the question here in the part we need to compute a value for the probability of x is less than or equal to 0 .5.
02:26
So we can calculate this value as probability for x is less than or equal to 0 .5 will be equal to probability of x is equal to 0 .2, 1 value is 0 .2 which is less than 0 .5 and then added by probability of x is equal to 0 .4.
02:49
So for these we have our values in the table we can see here.
02:54
Only for 0 .2 is 0 .1 and for 0 .4 we have 0 .2 so we have to add these two values here 0 .1 added by 0 .2 so together it is equal to 0 .3.
03:11
So here we have our answer for part b now that probability of getting x is less than or equal to 0 .5 is equal to 0 .3.
03:23
So this is a second solution.
03:26
Now let's move on to the third part c...