Given the probability distribution function $f(x) = \begin{cases} 0.25x & \text{for } 1 \le x \le 3 \\ 0 & \text{otherwise} \end{cases}$ The probability that the random variable x takes a value between 1 and $\sqrt{5}$ is __________.
Calculating Probability for Continuous Random Variable
The problem asks for the probability that a continuous random variable $X$, with a given probability density function (PDF), falls within a specific range. The PDF is defined as:
$f(x) = \begin{cases} 0.25x & \text{for } 1 \le x \le 3 \\ 0 & \text{otherwise} \end{cases}$
We need to find the probability $P(1 \le X \le \sqrt{5})$.
Probability Calculation using Integration
To find the probability for a continuous random variable, we integrate the PDF over the desired interval. The interval of interest is from $1$ to $\sqrt{5}$. Since $\sqrt{5}$ (approximately 2.236) is within the defined range of $f(x)$ (1 to 3), the calculation is straightforward:
$P(1 \le X \le \sqrt{5}) = \int_{1}^{\sqrt{5}} f(x) \, dx $
Substitute the function $f(x) = 0.25x$ into the integral:
$P(1 \le X \le \sqrt{5}) = \int_{1}^{\sqrt{5}} 0.25x \, dx $
Evaluating the Integral
Factor out the constant $0.25$ and integrate $x$ with respect to $x$:
$= 0.25 \int_{1}^{\sqrt{5}} x \, dx
$
$= 0.25 \left[ \frac{x^2}{2} \right]_{1}^{\sqrt{5}}
$
Now, apply the limits of integration (upper limit $\sqrt{5}$ and lower limit $1$):
$= 0.25 \left( \frac{(\sqrt{5})^2}{2} - \frac{1^2}{2} \right)
$
$= 0.25 \left( \frac{5}{2} - \frac{1}{2} \right)
$
$= 0.25 \left( \frac{4}{2} \right)
$
$= 0.25 \times 2
$
$= 0.5
$
The calculated probability is $0.5$. This value lies between $0.5$ and $0.5$, confirming the provided answer statement.
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
