In the die example, since it's uniform distribution with a = 1 and b = 6, pdf is 1 / (b - a) = 1 / 5 for x in [1, 6] and 0 otherwise. E[X] could also be computed by (1/2) * (a + b) = 7 / 2 = 3.5
youtuyy
I think pdf should be 1/6 for [1,6] and 0 otherwise since [1,6] include 6 values. The expected number is also the average of the sum of the numbers.
In the die example, since it's uniform distribution with a = 1 and b = 6, pdf is 1 / (b - a) = 1 / 5 for x in [1, 6] and 0 otherwise. E[X] could also be computed by (1/2) * (a + b) = 7 / 2 = 3.5
I think pdf should be 1/6 for [1,6] and 0 otherwise since [1,6] include 6 values. The expected number is also the average of the sum of the numbers.