$$f(x) = \frac{1}{3\sqrt{2\pi}} \exp(-\frac{x^2}{18}), \quad x \in (-\infty, +\infty)$$
Which one of the following statements is correct about the random variable $X$ ?
Identify Normal Distribution from Probability Density Function
The given probability density function (PDF) for the random variable $X$ is:
$f(x) = \frac{1}{3\sqrt{2\pi}} \exp(-\frac{x^2}{18}), \quad x \in (-\infty, +\infty)$We need to determine which type of random variable $X$ represents.
Standard Normal Distribution PDF
The standard form of the probability density function for a normal (or Gaussian) distribution is:
$f(x; \mu, \sigma) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$Here, $\mu$ represents the mean and $\sigma$ is the standard deviation ($\sigma > 0$).
Comparing the Given PDF
Let's compare the given PDF with the standard normal PDF:
- The exponent term in the given PDF is $-\frac{x^2}{18}$. Comparing this with $-\frac{(x-\mu)^2}{2\sigma^2}$, we can infer that the mean $\mu = 0$.
- Equating the denominators of the exponent terms: $18 = 2\sigma^2$.
- Solving for $\sigma^2$: $\sigma^2 = \frac{18}{2} = 9$.
- Solving for $\sigma$: $\sigma = \sqrt{9} = 3$.
- Now, let's check the coefficient term. The coefficient in the given PDF is $\frac{1}{3\sqrt{2\pi}}$. Comparing this with $\frac{1}{\sigma\sqrt{2\pi}}$, we see that $\sigma = 3$.
Both comparisons yield $\mu = 0$ and $\sigma = 3$. This matches the parameters of a normal distribution.
Conclusion on Random Variable Type
Since the given probability density function perfectly matches the form of a normal distribution PDF with mean $\mu=0$ and standard deviation $\sigma=3$, the random variable $X$ is a normal random variable.
Other distributions mentioned have different PDF forms:
- Exponential: Defined for $x \ge 0$, involves $e^{-\lambda x}$.
- Poisson: A discrete distribution, not continuous.
- Uniform: Has a constant PDF over a finite interval.
Therefore, the correct statement is that $X$ is a normal random variable.
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. _______
\(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
