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Continuous Probability Distributions Questions and Answers

1.

The probability density function \(f(x)\) of a random variable \(X\) which takes real values is

\(f(x) = \dfrac{1}{3\sqrt{2\pi}}\exp\left(-\dfrac{x^{2}}{18}\right)\), \(x \in (-\infty, +\infty)\)

Which one of the following statements is correct about the random variable \(X\) ?

Exam Year(s)

GATE CS - 2026,

Answer: B

X is a normal random variable.

The normal density with mean μ and standard deviation σ is

\[ \begin{aligned} f(x) &= \frac{1}{\sigma\sqrt{2\pi}} \\ &\quad \times \exp\left(-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right) \end{aligned} \]

With μ = 0 and σ = 3, the factor in front is 1/(3√(2π)) and the exponent is -x2/18, exactly the given function, defined on the whole real line.

An exponential density is zero for negative x and decays as e-λx, not e-x2. A Poisson variable takes only non-negative integer values and has a probability mass function, not a density. A uniform density is constant on a bounded interval.

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2.

A continuous random variable \(X\) has the probability density function

\[ f(x) = \begin{cases} \frac{3}{5} e^{-\frac{3}{5}x}, & x \gt 0 \\ 0, & x \le 0. \end{cases} \]

The probability density function of \(Y = 3X + 2\) is

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The density of \(Y\) is \(\frac{1}{5}e^{-\frac{1}{5}(y-2)}\) for \(y \gt 2\), and 0 otherwise.

\(X\) is exponential with rate \(3/5\). The map \(y = 3x + 2\) is increasing, with inverse \(x = (y - 2)/3\) and \(dx/dy = 1/3\). By the change of variables formula:

\[ \begin{aligned} f_{Y}(y) &= f_{X}\left(\frac{y-2}{3}\right)\cdot\frac{1}{3} \\ &= \frac{3}{5}e^{-\frac{3}{5}\cdot\frac{y-2}{3}}\cdot\frac{1}{3} \\ &= \frac{1}{5}e^{-\frac{1}{5}(y-2)},\quad y \gt 2. \end{aligned} \]

Since \(x \gt 0\) corresponds to \(y \gt 2\), the density is 0 for \(y \le 2\). As a check, scaling an exponential by 3 divides its rate by 3, from \(3/5\) to \(1/5\). The option keeping rate \(3/5\) forgets the Jacobian, and the rates \(2/5\) and \(4/5\) come from no consistent working.

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3.

The probability density function of the random variable \(X\) is

\[ f(x) = \begin{cases} \frac{1}{\lambda} e^{-x/\lambda}, & x \gt 0 \\ 0, & x \le 0, \end{cases} \]

where \(\lambda \gt 0\). For testing the hypothesis \(H_{0} : \lambda = 3\) against \(H_{A} : \lambda = 5\), a test is given as "Reject \(H_{0}\) if \(X \ge 4.5\)". The probability of type I error and power of this test are, respectively,

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The type I error is 0.2231 and the power is 0.4066, so the answer is 0.2231 and 0.4066.

For this exponential distribution with mean \(\lambda\), \(P(X \ge c) = e^{-c/\lambda}\).

\[ \begin{aligned} \alpha &= P(X \ge 4.5 \mid \lambda = 3) = e^{-1.5} = 0.2231 \\ 1 - \beta &= P(X \ge 4.5 \mid \lambda = 5) = e^{-0.9} = 0.4066 \end{aligned} \]

The type I error is the probability of rejecting \(H_{0}\) when it is true, and the power is the probability of rejecting it when \(H_{A}\) is true. 0.1353 is \(e^{-2}\) and 0.4493 is \(e^{-0.8}\); neither corresponds to the threshold 4.5 with means 3 and 5, and the pairs 0.1827/0.379 and 0.2021/0.4493 do not either.

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