Thread: Uniform Distribution

I just want to see if I am doing this right. the width is 0 to b, with f(x)=1/2, first I try to find b first, since the area is 1=bXh ----> 1=bX(1/2) ---> b=2.

Now, I have to find P(x<1) not to be confused with []
so, in this case, P(0<x<1)= (1-0)*(1/2)= 1/2

P(1<x<4)

Since 4 is not in the function, it would be P(1<x<2), so, (2-1)*(1/2) = 1/2

Lastly, since the width goes from 0 to 2, the median of the distribution is 1.

I have a test in the morning and just want to get it right.

My side question is does it matter if it is greater or greater or equal to and vice-versa?

Thanks in advance.

Originally Posted by driver327

I just want to see if I am doing this right. the width is 0 to b, with f(x)=1/2, first I try to find b first, since the area is 1=bXh ----> 1=bX(1/2) ---> b=2.

Now, I have to find P(x<1) not to be confused with []
so, in this case, P(0<x<1)= (1-0)*(1/2)= 1/2

P(1<x<4)

Since 4 is not in the function, it would be P(1<x<2), so, (2-1)*(1/2) = 1/2

Lastly, since the width goes from 0 to 2, the median of the distribution is 1.

I have a test in the morning and just want to get it right.

My side question is does it matter if it is greater or greater or equal to and vice-versa?

Thanks in advance.

Your answers are correct.

No, it doesn't matter at this level.