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
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
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 _________
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs 500 is _______
The number of parameters in the univariate exponential and Gaussian distributions, respectively are
Find the value of λ such that the function f (x) is a valid probability density function. _______
\(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
