Question

for this problem
f the probability density of a random variable is given by:
f(x)= Kx^2 for 0 11 years ago

Answer 1

For any probability distribution \(f(x)\) over support \(D\), you must have that
\[\int_Df(x)\,dx=1\]
In this case, you want to find \(k\) such that
\[\int_{-\infty}^\infty f(x)\,dx=1\]
Since \(f\) is zero everywhere but over the interval \((0,1)\), the integral reduces to
\[\int_0^1kx^2\,dx=1\]
From here you can integrate and solve for \(k\).

Answer 2

The probabilities can be determined by finding the distribution function \(F(x)\), where
\[F(x)=\int_{-\infty}^xf(t)\,dt=\begin{cases}0&\text{for }x<0\\\displaystyle\int_0^xf(t)\,dt&\text{for }0<x<1\\1&\text{for }x>1\end{cases}\]

Answer 3

To actually compute the probabilities above, you can determine the values of \(P\left(\dfrac{1}{4}<x<\dfrac{3}{4}\right)\) for part (a) and \(P\left(x>\dfrac{2}{3}\right)\) for part (b), where
\[P(X<c)=F(c)=\int_{-\infty}^cf(t)\,dt\\
P(X>c)=1-P(X<c)\\
P(c<X<d)=P(X<d)-P(X<c)\]