Knowee
Questions
Features
Study Tools

Find k such that the function f defined byfX (x) =kx2, 0 < x < 1,0, otherwiseis a probability density function. Also, determine P (1/3 < X ≤ 1/2

Question

Find k such that the function f defined by

fX(x)={kx2,0<x<1,0,otherwise f_X(x) = \begin{cases} kx^2, & 0 < x < 1, \\ 0, & \text{otherwise} \end{cases}

is a probability density function. Also, determine ( P \left( \frac{1}{3} < X \leq \frac{1}{2} \right) .

🧐 Not the exact question you are looking for?Go ask a question

Solution

To find the value of k such that the function f is a probability density function, we need to ensure that the integral of f from -∞ to ∞ is equal to 1, because the total probability must be 1.

The function f is only non-zero between 0 and 1, so we only need to integrate from 0 to 1:

∫ from 0 to 1 of kx^2 dx = [kx^3 / 3] from 0 to 1 = k/3 - 0 = k/3.

Setting this equal to 1 gives k/3 = 1, so k = 3.

To find P(1/3 < X ≤ 1/2), we integrate the probability density function from 1/3 to 1/2:

∫ from 1/3 to 1/2 of 3x^2 dx = [x^3] from 1/3 to 1/2 = (1/2)^3 - (1/3)^3 = 1/8 - 1/27 = 0.04167.

This problem has been solved

Similar Questions

Let X be a random variable with density functionfX (x) = k1 + x2 , −∞ < x < ∞.Determine k and the distribution function

A random variable X has the distribution functionx: 0 1 2 3 4 5 6 7P(x): 0 k 2k 2k 3k2k22kkk 27Then evaluate)

Distribution function of a random variable X is given by F(x) = 1 - 1/(x ^ 2); 1 <= x < ∞ Then P(X <= 2) and respectively P(X > 3/2) are

Suppose the cumulative distribution function of a random variable X is given by :Calculate P(1 < X < 3)?

If one of the zeroes of the quadratic polynomial (k – 1)x2 + kx + 1 is –3, then the value of k isSelect an answerA4/3B– 4/3C2/3D– 2/3

1/1

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.