Let a continuous random variable \( X \) have probability density function (pdf): \[f(x) = \begin{cases} -0.75 \, x^2 + 1.5x & \text{for } 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\] Find the mode of \( X \).
1
The mode of a distribution is the value of the random variable that maximizes the probability density function (pdf). To find the mode of the given pdf, we first need to find the derivative of the pdf, set it to zero, and solve for x:
f'(x) = -1.5x + 1.5
Setting f'(x) = 0:
-1.5x + 1.5 = 0
x = 1.
The mode of the distribution is x = 1, since it maximizes the pdf in the range 0 < x < 2.
Important Questions from Mean and Variance of Random variables
A continuous random variable x has a probability density function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\) Then the variance of x is:
X is a non-negative integer valued random variable with
\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)
Then, mean and variance of X are respectively
A random variable X has the distribution law as given below:
X
1
2
3
P(X = x)
0.3
0.4
0.3
The variance of the distribution is:
