Continuous Probability Distributions Questions and Answers

Q: 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 \(X\) ?

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.

Q: 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

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.

Q: The Chi-square distribution is :

The chi-square distribution is continuous.A chi-square variable is the sum of squares of independent standard normal variables, so it can take any non-negative real value, and its distribution is described by a continuous probability density on \((0, \infty)\), indexed by the degrees of freedom.It is not discrete. It is not symmetrical but positively skewed, becoming more symmetric only as the degrees of freedom grow, and it has a single mode (at \(k - 2\) for \(k\) above 2 degrees of freedom), so it is not multimodal.

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 \(X\) ?

A.\(X\) is an exponential random variable
B.\(X\) is a normal random variable
C.\(X\) is a Poisson random variable
D.\(X\) is a uniform random variable

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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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

A.\(f(y) =\) \(\begin{cases}\frac{1}{5}e^{-\frac{1}{5}(y-2)},&y\gt 2\\ 0,&y\le 2\end{cases}\)
B.\(f(y) =\) \(\begin{cases}\frac{2}{5}e^{-\frac{2}{5}(y-2)},&y\gt 2\\ 0,&y\le 2\end{cases}\)
C.\(f(y) =\) \(\begin{cases}\frac{3}{5}e^{-\frac{3}{5}(y-2)},&y\gt 2\\ 0,&y\le 2\end{cases}\)
D.\(f(y) =\) \(\begin{cases}\frac{4}{5}e^{-\frac{4}{5}(y-2)},&y\gt 2\\ 0,&y\le 2\end{cases}\)

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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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,

A.0.1353 and 0.4966
B.0.1827 and 0.379
C.0.2021 and 0.4493
D.0.2231 and 0.4066

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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The Chi-square distribution is :

A.Multimodal
B.Symmetrical
C.Continuous
D.Discrete

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The chi-square distribution is continuous.

A chi-square variable is the sum of squares of independent standard normal variables, so it can take any non-negative real value, and its distribution is described by a continuous probability density on \((0, \infty)\), indexed by the degrees of freedom.

It is not discrete. It is not symmetrical but positively skewed, becoming more symmetric only as the degrees of freedom grow, and it has a single mode (at \(k - 2\) for \(k\) above 2 degrees of freedom), so it is not multimodal.

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