Question

Please help me do this question​

Answer 1

Answer: This is a problem about a continuous random variable X with a probability density function (pdf) given by:

f(x) = kx(1-x), 0 < x < 1

f(x) = 0, otherwise

The problem asks you to determine the value of k, the mean of X, and the variance of X.

To find the value of k, we need to use the fact that the pdf must integrate to 1 over the entire range of X. That is:

∫−∞∞​f(x)dx=1

Since f(x) is zero outside the interval (0, 1), we only need to integrate over that interval. That is:

∫01​kx(1−x)dx=1

Expanding and integrating term by term, we get:

k[2x2​−3x3​]01​=1

Evaluating at the limits and simplifying, we get:

k(21​−31​)=1

Solving for k, we get:

k=2−36​=6

So, the value of k is 6.

To find the mean of X, we need to use the formula:

E(X)=∫−∞∞​xf(x)dx

Again, since f(x) is zero outside the interval (0, 1), we only need to integrate over that interval. That is:

E(X)=∫01​xkx(1−x)dx

Plugging in the value of k and expanding, we get:

E(X)=∫01​6x2(1−x)dx

Integrating term by term, we get:

E(X)=6[3x3​−4x4​]01​

Evaluating at the limits and simplifying, we get:

E(X)=6(31​−41​)=21​

So, the mean of X is 0.5.

To find the variance of X, we need to use the formula:

Var(X)=E(X2)−[E(X)]2

We already know that E(X) is 0.5, so we only need to find E(X^2). We can use a similar approach as before, but with x^2 instead of x. That is:

E(X2)=∫−∞∞​x2f(x)dx

Again, since f(x) is zero outside the interval (0, 1), we only need to integrate over that interval. That is:

E(X2)=∫01​x2kx(1−x)dx

Plugging in the value of k and expanding, we get:

E(X2)=∫01​6x3(1−x)dx

Integrating term by term, we get:

E(X2)=6[4x4​−5x5​]01​

Evaluating at the limits and simplifying, we get:

E(X2)=6(41​−51​)=103​

So, E(X^2) is 0.3.

Now, we can plug in these values into the formula for variance and simplify. We get:

Var(X)=E(X2)−[E(X)]2=0.3−(0.5)2=0.05

So, the variance of X is 0.05.