Understanding Continuous Random Variables
A continuous random variable, like one following a normal distribution, can take any value within a given range. Unlike discrete variables, the probability of a continuous variable equaling a *specific* exact value is always zero.
Probability for a Specific Value
For any continuous random variable $X$, the probability density function (PDF), denoted $f(x)$, describes the likelihood. However, the probability of $X$ being exactly equal to a specific value $c$ is calculated by integrating the PDF from $c$ to $c$.
Mathematically, this is represented as:
$ P[X = c] = \int_{c}^{c} f(x) \, dx $
The integral of any function over an interval of zero width is zero. Therefore:
$ P[X = c] = 0 $
Applying to the Question
In this specific question, we are asked for the probability $P[X = 5]$. Since $X$ is a continuous random variable (following a normal distribution), the probability of it taking the exact value of 5 is:
$ P[X = 5] = 0 $
The information about the sample size ($n=25$), sample mean ($\bar{x}=5$), and sample standard deviation ($s=1.5$) is extra information not needed to determine the probability of $X$ equaling a single specific value for a continuous distribution.
Conclusion
The probability $P[X = 5]$ for a continuous random variable $X$ is 0.
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
