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2.4.2 Let W~Uniform[1 4]. Compute each of the following. (b) P(W >= 2) (c) P(W^2 <= 9)

For b, I was thinking P(2<=w<=4)= (4-2)/(4-1) = 2/3.

For c, I was thinking since w^2 <= 9, P(1<=w<=3)= (3-1)/(4-1) = 2/3.

However, I am not sure if these calculations are accurate or if I am missing something. Logically, it seems like P(W>=2) should be 3/4 since that could include 2, 3, and 4 from the included range [1,4] for part b. For part c, It would seem that the answer should be 3/4 as well to include (1, 2, 3) out of the 4 possible options in [1,4].

I am not sure which method to use. Please help and thank you in advance!

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    – sudeep5221
    Commented Jul 1 at 18:31
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    $\begingroup$ Are you referring to a continuous uniform random variable or a discrete one? $\endgroup$
    – Daniel5803
    Commented Jul 1 at 18:32
  • $\begingroup$ Regarding your question, $W$ is a continuous random variable, not a discrete random variable. All the calculations will need to be updated accordingly. $\endgroup$
    – sudeep5221
    Commented Jul 1 at 18:32
  • $\begingroup$ I'm not sure if it should be continuous or discrete. All of the info I have been provided is listed above. If it is a continuous random variable (I think I have seen this more in my textbook so I am thinking it is), how to I update the calculations? What formula should I be using? $\endgroup$
    – hsigmon
    Commented Jul 1 at 21:24
  • $\begingroup$ @hsigmon If it is continuous then the first calculation is correct and if it is discrete the second one is the correct one. Intuitively, the second calculation fails in the continuous case because you have to count intervals of length 1, of which there are 3, not 4, in [1, 4] (Namely [1,2], [2,3] and [3,4]) $\endgroup$
    – Daniel5803
    Commented Jul 1 at 21:33

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