Question

Choose numbers a and b that make the function f (x) = { 3/4
0
a valid continuous probability distribution.

...if x is between a and b everywhere else

Answer 1

The function is very unclear... is it
\[f(x)=\begin{cases}\dfrac{3}{4}&\text{for }a\le x\le b\\\\
0&\text{elsewhere}\end{cases}~~~~?\]
If it is, you must satisfy the following integral:
\[\int_a^bf(x)~dx=\int_a^b\frac{3}{4}~dx=1\]
Integrating, you have
\[\frac{3}{4}(b-a)=1\]
Now, since you have two unknowns, you must choose one as a starting point. The simplest might be \(a=0\), so that
\[\frac{3}{4}(b-0)=1~~\iff~~b=\frac{4}{3}\]
So,
\[f(x)=\begin{cases}\dfrac{3}{4}&\text{for }0\le x\le \dfrac{4}{3}\\\\
0&\text{elsewhere}\end{cases}\]
Of course, that's not the only possible answer. You could have picked \(a=1\), which would make \(b=\dfrac{7}{3}\). In general, if \(a=n\), then \(b=\dfrac{4}{3}+n\).

Answer 2

Thank you very much. I will work with what you have given me!

Answer 3

yw