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.

Q: The variance of a Chi-square distribution with n degrees of freedom is :

The variance is \(2n\).A chi-square variable with \(n\) degrees of freedom is the sum of the squares of \(n\) independent standard normal variables. Each \(Z^{2}\) has mean 1 and variance 2, because \(E(Z^{4}) = 3\):\[ \begin{aligned} \operatorname{Var}(Z^{2}) &= E(Z^{4}) - [E(Z^{2})]^{2} \\ &= 3 - 1 = 2 \\ \operatorname{Var}(\chi^{2}_{n}) &= n \times 2 = 2n \end{aligned} \]\(n\) is the mean of the distribution, not its variance, which is the usual trap. \(n^{2}\) and \(\sqrt{n}\) do not arise for chi-square; \(\sqrt{2n}\) would be its standard deviation.

Q: A continuous random variable X follows the probability law :\(f(x) = E.x^{2};\ 0 \le x \le 1\)What is the value of E ?

The value of E is 3.A probability density must integrate to 1 over its range:\[ \begin{aligned} \int_{0}^{1} E x^{2}\, dx &= 1 \\ E \left[ \frac{x^{3}}{3} \right]_{0}^{1} &= 1 \\ \frac{E}{3} &= 1 \Rightarrow E = 3 \end{aligned} \]E = 1 forgets the division by 3 from integrating \(x^{2}\), giving a total probability of only \(\frac{1}{3}\). E = 2 is the constant for \(f(x) = Ex\), and E = 6 would make the total probability 2.

Q: If x is normal variate with mean 1 and standard deviation 2 and y is another normal variate independent of x with mean 2 and standard deviation 3. What is the mean and variance of the normal variate z = x + 2y ?

The mean is 5 and the variance is 40, 5 and 40.For independent variables, means add linearly and variances add with the squares of the coefficients:\[ \begin{aligned} E(z) &= 1 + 2(2) = 5 \\ V(z) &= 2^{2} + 2^{2}(3^{2}) = 4 + 36 = 40 \end{aligned} \]5 and 36 forgets the variance of x, and 5 and 25 adds standard deviations instead of variances. 7 and 45 has the wrong mean.

Q: Which one of the following conditions would indicate that the dataset is not bell shaped ?

The condition that shows the data are not bell shaped is that the mean is much smaller than the median.In a bell-shaped (symmetric) distribution the mean and median are nearly equal. A mean far below the median means the data are skewed to the left, so the shape is not that of a bell. For a bell-shaped set the range is about five to six standard deviations, and the range is always at least as large as the interquartile range, so the statements in options B and C do not rule out a bell shape.

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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The variance of a Chi-square distribution with n degrees of freedom is :

A.2n
B.n
C.\(n^{2}\)
D.\(\sqrt{n}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The variance is \(2n\).

A chi-square variable with \(n\) degrees of freedom is the sum of the squares of \(n\) independent standard normal variables. Each \(Z^{2}\) has mean 1 and variance 2, because \(E(Z^{4}) = 3\):

\[ \begin{aligned} \operatorname{Var}(Z^{2}) &= E(Z^{4}) - [E(Z^{2})]^{2} \\ &= 3 - 1 = 2 \\ \operatorname{Var}(\chi^{2}_{n}) &= n \times 2 = 2n \end{aligned} \]

\(n\) is the mean of the distribution, not its variance, which is the usual trap. \(n^{2}\) and \(\sqrt{n}\) do not arise for chi-square; \(\sqrt{2n}\) would be its standard deviation.

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A continuous random variable X follows the probability law :

\(f(x) = E.x^{2};\ 0 \le x \le 1\)

What is the value of E ?

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The value of E is 3.

A probability density must integrate to 1 over its range:

\[ \begin{aligned} \int_{0}^{1} E x^{2}\, dx &= 1 \\ E \left[ \frac{x^{3}}{3} \right]_{0}^{1} &= 1 \\ \frac{E}{3} &= 1 \Rightarrow E = 3 \end{aligned} \]

E = 1 forgets the division by 3 from integrating \(x^{2}\), giving a total probability of only \(\frac{1}{3}\). E = 2 is the constant for \(f(x) = Ex\), and E = 6 would make the total probability 2.

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If x is normal variate with mean 1 and standard deviation 2 and y is another normal variate independent of x with mean 2 and standard deviation 3. What is the mean and variance of the normal variate z = x + 2y ?

A.5 and 25
B.5 and 36
C.5 and 40
D.7 and 45

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The mean is 5 and the variance is 40, 5 and 40.

For independent variables, means add linearly and variances add with the squares of the coefficients:

\[ \begin{aligned} E(z) &= 1 + 2(2) = 5 \\ V(z) &= 2^{2} + 2^{2}(3^{2}) = 4 + 36 = 40 \end{aligned} \]

5 and 36 forgets the variance of x, and 5 and 25 adds standard deviations instead of variances. 7 and 45 has the wrong mean.

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Which one of the following conditions would indicate that the dataset is not bell shaped ?

A.The mean is much smaller than median
B.The range is equal to five times the standard deviation.
C.The range is larger than interquartile range.
D.The range is twice the standard deviation.

Exam Year(s)

UGC NET - 2013,

Answer: A

The condition that shows the data are not bell shaped is that the mean is much smaller than the median.

In a bell-shaped (symmetric) distribution the mean and median are nearly equal. A mean far below the median means the data are skewed to the left, so the shape is not that of a bell. For a bell-shaped set the range is about five to six standard deviations, and the range is always at least as large as the interquartile range, so the statements in options B and C do not rule out a bell shape.

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For degrees of freedom (df) \(\gt\) 1, the mean (μ) of the t-distribution is

A.Zero
B.1
C.depends on df
D.2

Exam Year(s)

UGC NET - 2013,

Answer: A

For degrees of freedom above 1, the mean of the t-distribution is zero.

The t-distribution is symmetric about zero, like the normal distribution but with heavier tails, and its mean exists only when df is greater than 1. The mean does not depend on df (only the spread does), and it is not 1 or 2.

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