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Probability Distributions & Random Variables (Statistics) Solved Questions & Notes (2026) - Apex Rankers

Statistics & Data Science > Statistics > Probability Distributions & Random Variables

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Probability Distributions & Random Variables

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Q. 1 Statistics
Difficulty: medium (1 Mark)
If a random variable $X$ follows a Poisson distribution with parameter $\lambda = 4$, what is the coefficient of skewness $\gamma_1 = \frac{\mu_3}{\sigma^3}$?
A
0.25
B
0.50
✓ Correct
C
1.00
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a Poisson distribution with parameter $\lambda$, the variance $\sigma^2 = \lambda$ and the third central moment $\mu_3 = \lambda$. Therefore, skewness $\gamma_1 = \frac{\lambda}{\lambda^{3/2}} = \frac{1}{\sqrt{\lambda}} = \frac{1}{\sqrt{4}} = 0.5$.
Q. 2 Statistics
Difficulty: easy (1 Mark)
For a Binomial distribution $B(n, p)$, under what condition does the distribution become perfectly symmetrical?
A
When $n \to \infty$ for any $p$
B
When $p = 0.5$
✓ Correct
C
When $n p = 1$
D
When $p = 0.25$
💡 Step-by-Step Explanation & Concept Rationale
The Binomial distribution is perfectly symmetrical (skewness $\gamma_1 = \frac{q-p}{\sqrt{npq}} = 0$) if and only if $p = q = 0.5$, regardless of the sample size $n$.
Q. 3 Statistics
Difficulty: medium (1 Mark)
What is the Moment Generating Function (MGF) $M_X(t)$ of an Exponential random variable $X \sim \text{Exp}(\lambda)$ with probability density function $f(x) = \lambda e^{-\lambda x}$ for $x > 0$ and $t < \lambda$?
A
$\frac{\lambda}{\lambda - t}$
✓ Correct
B
$\frac{\lambda}{\lambda + t}$
C
$e^{\lambda t}$
D
$\frac{1}{1 - \lambda t^2}$
💡 Step-by-Step Explanation & Concept Rationale
The MGF is $M_X(t) = \int_0^\infty e^{tx} \lambda e^{-\lambda x} dx = \lambda \int_0^\infty e^{-(\lambda - t)x} dx = \frac{\lambda}{\lambda - t}$ for $t < \lambda$.
Q. 4 Statistics
Difficulty: easy (1 Mark)
If $X$ and $Y$ are two independent standard normal random variables ($X, Y \sim N(0, 1)$), what is the probability distribution of $Z = X^2 + Y^2$?
A
Standard Normal distribution $N(0, 2)$
B
Chi-Square distribution with 2 degrees of freedom $\chi^2_{(2)}$ (or Exponential with mean 2)
✓ Correct
C
Student's t-distribution with 2 degrees of freedom
D
F-distribution with $(1, 1)$ degrees of freedom
💡 Step-by-Step Explanation & Concept Rationale
The sum of squares of $k$ independent standard normal variables follows a Chi-square distribution with $k$ degrees of freedom. For $k=2$, $\chi^2_{(2)}$ is identical to an Exponential distribution with parameter $\lambda = 1/2$ (mean 2).
Q. 5 Statistics
Difficulty: easy (1 Mark)
Which probability distribution possesses the unique 'Memoryless Property' among continuous random variables?
A
Normal distribution
B
Exponential distribution
✓ Correct
C
Uniform distribution
D
Weibull distribution with shape parameter $k = 2$
💡 Step-by-Step Explanation & Concept Rationale
The Exponential distribution satisfies $P(X > s + t \mid X > s) = P(X > t)$, meaning the past duration has no effect on future probability; it is the only continuous distribution with this property.
Q. 6 Statistics
Difficulty: easy (1 Mark)
Which discrete probability distribution possesses the memoryless property ($P(X > s + t \mid X > s) = P(X > t)$)?
A
Binomial distribution
B
Geometric distribution
✓ Correct
C
Poisson distribution
D
Hypergeometric distribution
💡 Step-by-Step Explanation & Concept Rationale
The Geometric distribution is the sole discrete distribution that is memoryless, modeling the number of Bernoulli trials until the first success.
Q. 7 Statistics
Difficulty: medium (1 Mark)
If a random variable $X$ has a standard Cauchy distribution ($f(x) = \frac{1}{\pi(1+x^2)}$), what is its mathematical expectation $E[X]$?
A
0
B
Undefined (does not exist)
✓ Correct
C
1
D
$\pi$
💡 Step-by-Step Explanation & Concept Rationale
The integral $\int_{-\infty}^\infty \frac{x}{\pi(1+x^2)} dx$ is not absolutely convergent (diverges to $\infty - \infty$). Hence, the mean and all higher moments of the Cauchy distribution do not exist.
Q. 8 Statistics
Difficulty: easy (1 Mark)
If $X_1, X_2, \dots, X_n$ are independent random variables from a Poisson distribution with parameters $\lambda_1, \lambda_2, \dots, \lambda_n$, what is the distribution of their sum $S_n = \sum_{i=1}^n X_i$?
A
Normal distribution with mean $\sum \lambda_i$
B
Poisson distribution with parameter $\lambda = \sum_{i=1}^n \lambda_i$
✓ Correct
C
Binomial distribution with $p = \sum \lambda_i / n$
D
Negative Binomial distribution
💡 Step-by-Step Explanation & Concept Rationale
By the additive (reproductive) property of the Poisson distribution, the sum of independent Poisson variables is itself Poisson distributed with parameter equal to the sum of their individual means.
Q. 9 Statistics
Difficulty: easy (1 Mark)
What is the variance of a Continuous Uniform distribution $U(a, b)$ on the interval $[a, b]$?
A
$\frac{(b - a)^2}{12}$
✓ Correct
B
$\frac{(b - a)^2}{6}$
C
$\frac{b - a}{2}$
D
$\frac{(b + a)^2}{12}$
💡 Step-by-Step Explanation & Concept Rationale
For a uniform distribution $U(a, b)$, the mean is $\mu = \frac{a+b}{2}$ and the variance is $\sigma^2 = \frac{(b-a)^2}{12}$.
Q. 10 Statistics
Difficulty: medium (1 Mark)
If $X \sim N(\mu, \sigma^2)$, what are the values of the third and fourth central moments $\mu_3$ and $\mu_4$?
A
$\mu_3 = 0$ and $\mu_4 = 3\sigma^4$
✓ Correct
B
$\mu_3 = \sigma^3$ and $\mu_4 = \sigma^4$
C
$\mu_3 = 0$ and $\mu_4 = 0$
D
$\mu_3 = 3\sigma^3$ and $\mu_4 = 3\sigma^4$
💡 Step-by-Step Explanation & Concept Rationale
Due to symmetry around $\mu$, all odd central moments of the normal distribution are zero ($\mu_3 = 0$). The fourth central moment is $\mu_4 = 3\sigma^4$, giving a kurtosis coefficient $\beta_2 = \mu_4 / \sigma^4 = 3$ (mesokurtic).
Q. 11 Statistics
Difficulty: medium (1 Mark)
What is the relationship between the Hypergeometric distribution and the Binomial distribution?
A
The Binomial distribution models sampling without replacement, whereas Hypergeometric models sampling with replacement
B
The Hypergeometric distribution converges to the Binomial distribution as the population size $N \to \infty$ with a fixed sampling fraction $\frac{n}{N} \to 0$
✓ Correct
C
The Hypergeometric distribution always has a higher variance than the Binomial distribution
D
They are mathematically identical for all population sizes
💡 Step-by-Step Explanation & Concept Rationale
The Hypergeometric distribution models sampling without replacement from a finite population. As population size $N \to \infty$, sampling without replacement becomes effectively equivalent to sampling with replacement (Binomial).
Q. 12 Statistics
Difficulty: medium (1 Mark)
In a Negative Binomial distribution $NB(r, p)$, what does the random variable $X$ represent?
A
The number of trials required to achieve exactly $r$ successes
B
The number of failures before the $r$-th success is observed
✓ Correct
C
The number of successes in $r$ independent trials
D
The total time elapsed between $r$ independent Poisson events
💡 Step-by-Step Explanation & Concept Rationale
The Negative Binomial distribution models the number of failures $k$ (or total trials $n$) preceding the $r$-th success in independent Bernoulli trials with success probability $p$.
Q. 13 Statistics
Difficulty: medium (1 Mark)
If a continuous random variable $X$ has a Gamma distribution $\text{Gamma}(\alpha, \beta)$ with PDF $f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x}$ for $x > 0$, what are its mean and variance?
A
Mean $= \frac{\alpha}{\beta}$, Variance $= \frac{\alpha}{\beta^2}$
✓ Correct
B
Mean $= \alpha \beta$, Variance $= \alpha \beta^2$
C
Mean $= \frac{\alpha}{\beta^2}$, Variance $= \frac{\alpha}{\beta}$
D
Mean $= \frac{1}{\alpha \beta}$, Variance $= \frac{1}{\alpha \beta^2}$
💡 Step-by-Step Explanation & Concept Rationale
For the rate parameterization of the Gamma distribution $\text{Gamma}(\alpha, \beta)$, $E[X] = \frac{\alpha}{\beta}$ and $\text{Var}(X) = \frac{\alpha}{\beta^2}$.
Q. 14 Statistics
Difficulty: easy (1 Mark)
What distribution results from the ratio of two independent Chi-square random variables, each divided by their respective degrees of freedom: $F = \frac{U / d_1}{V / d_2}$?
A
Student's t-distribution
B
Snedecor's F-distribution with $(d_1, d_2)$ degrees of freedom
✓ Correct
C
Beta distribution of the first kind
D
Gamma distribution with parameters $(d_1, d_2)$
💡 Step-by-Step Explanation & Concept Rationale
By definition, Snedecor's F-statistic is the ratio of two independent Chi-square variables scaled by their degrees of freedom: $F(d_1, d_2) = \frac{\chi^2_{(d_1)}/d_1}{\chi^2_{(d_2)}/d_2}$.
Q. 15 Statistics
Difficulty: easy (1 Mark)
If $Z \sim N(0, 1)$ is a standard normal variable and $V \sim \chi^2_{(k)}$ is an independent Chi-square variable with $k$ degrees of freedom, what is the distribution of $T = \frac{Z}{\sqrt{V / k}}$?
A
Standard Normal distribution
B
Student's t-distribution with $k$ degrees of freedom ($t_{(k)}$)
✓ Correct
C
F-distribution with $(1, k)$ degrees of freedom
D
Cauchy distribution with parameter $k$
💡 Step-by-Step Explanation & Concept Rationale
Gosset defined Student's t-distribution as the ratio of a standard normal deviate to the square root of an independent Chi-square deviate divided by its degrees of freedom.
Q. 16 Statistics
Difficulty: medium (1 Mark)
What is the relationship between the Student's t-distribution with $k$ degrees of freedom and the F-distribution?
A
$T_{(k)}^2 \sim F(1, k)$
✓ Correct
B
$T_{(k)} \sim F(k, 1)$
C
$T_{(k)}^2 \sim F(k, k)$
D
$\sqrt{T_{(k)}} \sim F(1, k)$
💡 Step-by-Step Explanation & Concept Rationale
Squaring a t-distributed variable with $k$ degrees of freedom yields $T^2 = \frac{Z^2}{V/k} = \frac{\chi^2_{(1)}/1}{\chi^2_{(k)}/k} \sim F(1, k)$.
Q. 17 Statistics
Difficulty: easy (1 Mark)
As the degrees of freedom $k \to \infty$, what does Student's t-distribution converge to?
A
Standard Uniform distribution $U(-1, 1)$
B
Standard Normal distribution $N(0, 1)$
✓ Correct
C
Chi-square distribution $\chi^2_{(1)}$
D
Cauchy distribution
💡 Step-by-Step Explanation & Concept Rationale
As degrees of freedom approach infinity, the heavier tails of the t-distribution diminish, and the density function converges asymptotically to the standard normal distribution $N(0, 1)$.
Q. 18 Statistics
Difficulty: medium (1 Mark)
What is the value of the integral $\int_{-\infty}^{\infty} e^{-x^2} dx$ (the Gaussian Gaussian integral)?
A
$\pi$
B
$\sqrt{\pi}$
✓ Correct
C
$\frac{\sqrt{\pi}}{2}$
D
$2\pi$
💡 Step-by-Step Explanation & Concept Rationale
The standard Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$. (This is derived using double integration in polar coordinates).
Q. 19 Statistics
Difficulty: medium (1 Mark)
For a Weibull distribution with PDF $f(x) = \frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k}$ ($x \ge 0$), what happens when the shape parameter $k = 1$?
A
It becomes a Standard Normal distribution
B
It reduces exactly to the Exponential distribution with rate $\frac{1}{\lambda}$
✓ Correct
C
It becomes a Rayleigh distribution
D
It becomes a Lognormal distribution
💡 Step-by-Step Explanation & Concept Rationale
When shape parameter $k = 1$, the Weibull PDF simplifies to $f(x) = \frac{1}{\lambda} e^{-x/\lambda}$, which is the Exponential distribution with scale parameter $\lambda$ (rate $\beta = 1/\lambda$).
Q. 20 Statistics
Difficulty: medium (1 Mark)
What is the Rayleigh distribution in terms of the Weibull distribution?
A
Weibull distribution with shape parameter $k = 0.5$
B
Weibull distribution with shape parameter $k = 2$
✓ Correct
C
Weibull distribution with shape parameter $k = 3.6$
D
Weibull distribution with shape parameter $k \to \infty$
💡 Step-by-Step Explanation & Concept Rationale
The Rayleigh distribution is a special case of the Weibull distribution when the shape parameter $k = 2$. It models wind speed distributions and wave heights.
Q. 21 Statistics
Difficulty: easy (1 Mark)
If $X \sim \text{Lognormal}(\mu, \sigma^2)$, what is the distribution of $Y = \ln(X)$?
A
Exponential distribution with mean $\mu$
B
Normal distribution $N(\mu, \sigma^2)$
✓ Correct
C
Standard Uniform distribution
D
Gamma distribution
💡 Step-by-Step Explanation & Concept Rationale
By definition, a positive random variable $X$ is lognormally distributed if its natural logarithm $\ln(X)$ is normally distributed with mean $\mu$ and variance $\sigma^2$.
Q. 22 Statistics
Difficulty: medium (1 Mark)
What is the mean of a Lognormal random variable $X \sim \text{Lognormal}(\mu, \sigma^2)$?
A
$e^{\mu}$
B
$e^{\mu + \frac{\sigma^2}{2}}$
✓ Correct
C
$\mu + \frac{\sigma^2}{2}$
D
$e^{\mu + \sigma}$
💡 Step-by-Step Explanation & Concept Rationale
The expectation of a lognormal variable is $E[X] = e^{\mu + \frac{\sigma^2}{2}}$, which is always strictly greater than the geometric mean $e^\mu$ for $\sigma > 0$.
Q. 23 Statistics
Difficulty: medium (1 Mark)
What is Chebyshev's Inequality for any random variable $X$ with finite mean $\mu$ and variance $\sigma^2$, for any $k > 1$?
A
$P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}$
✓ Correct
B
$P(|X - \mu| \ge k\sigma) \ge 1 - \frac{1}{k^2}$
C
$P(|X - \mu| < k\sigma) \le \frac{1}{k^2}$
D
$P(|X - \mu| \ge k\sigma) \le \frac{\sigma^2}{k}$
💡 Step-by-Step Explanation & Concept Rationale
Chebyshev's inequality guarantees that for any distribution, the probability of observing a value beyond $k$ standard deviations from the mean is at most $\frac{1}{k^2}$, i.e., $P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}$.
Q. 24 Statistics
Difficulty: easy (1 Mark)
According to Markov's Inequality, if $X$ is a non-negative random variable ($X \ge 0$) with expectation $E[X]$, what is the upper bound on $P(X \ge a)$ for any $a > 0$?
A
$P(X \ge a) \le \frac{E[X]}{a}$
✓ Correct
B
$P(X \ge a) \le \frac{E[X^2]}{a^2}$
C
$P(X \ge a) \ge 1 - \frac{E[X]}{a}$
D
$P(X \ge a) \le \frac{a}{E[X]}$
💡 Step-by-Step Explanation & Concept Rationale
Markov's inequality states that for any non-negative random variable $X$ and constant $a > 0$, $P(X \ge a) \le \frac{E[X]}{a}$.
Q. 25 Statistics
Difficulty: medium (1 Mark)
What does the Weak Law of Large Numbers (WLLN) state regarding the sample mean $\bar{X}_n$ of $n$ independent and identically distributed (i.i.d.) random variables with finite mean $\mu$?
A
$\bar{X}_n$ converges in distribution to a standard normal variable
B
$\bar{X}_n$ converges in probability to the population mean $\mu$ as $n \to \infty$ (i.e., $\lim_{n \to \infty} P(|\bar{X}_n - \mu| > \epsilon) = 0$)
✓ Correct
C
The variance of the population decreases to zero as $n \to \infty$
D
$\bar{X}_n$ equals $\mu$ with probability 1 for any sample size $n$
💡 Step-by-Step Explanation & Concept Rationale
The Weak Law of Large Numbers (Khinchin's theorem) guarantees convergence in probability: as sample size grows arbitrarily large, the sample mean converges in probability to the expected value $\mu$.
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