continuous probability q (1 Viewer)

[yashbb]

q 5b

 

[LennonTheLemon]

I think you have to justify it by thinking about how y=1/c is a horizontal line. If c=2 and you were finding the area from 0 to 2 under y=1/2, then your area would be 2/2=1 according to your E(x). Now this makes sense, as the probability from your lowest to highest bounds for a PDF should always be equal to 1.

 

[yashbb]

I think you have to justify it by thinking about how y=1/c is a horizontal line. If c=2 and you were finding the area from 0 to 2 under y=1/2, then your area would be 2/2=1 according to your E(x). Now this makes sense, as the probability from your lowest to highest bounds for a PDF should always be equal to 1.

oh sorry i meant part c my b so sorry

 

[LennonTheLemon]

oh sorry i meant part c my b so sorry

Right. So if you remember, the formula for Variance is , and is just the mean squared. In part b you found that the mean is , so the square of that is .

Now to find , integrate from 0 to c of the original PDF function but multiplied by . So find .

Once you have those two components, just plug them into the Variance formula above and you're done.

 

[Trebla]

Right. So if you remember, the formula for Variance is , and is just the mean squared. In part b you found that the mean is , so the square of that is .

Now to find , integrate from 0 to c of the original PDF function but SQUARED. So find .

Once you have those two components, just plug them into the Variance formula above and you're done.

FYI

 

[LennonTheLemon]

FYI

oh crap, yes! thx i will edit my reply