Data Science Interview Questions: Statistical Foundations

The Binomial distribution models the count of successes in a fixed number n of independent Bernoulli trials, each with the same success probability p. Here n = 10 and p = 0.5, which matches counting heads in 10 fair coin flips. The Poisson distribution models event counts over a continuous interval, the Uniform distribution assigns equal probability to every outcome, and the Exponential distribution models waiting times between events, so none of those fit this fixed-trial counting scenario.

The empirical rule describes how a normal (Gaussian) distribution's probability mass is spread around its mean. Because the density is bell-shaped and symmetric, about 68% of values lie within ±1σ of the mean, about 95% within ±2σ, and about 99.7% within ±3σ. A complete answer must give all three percentages tied to the correct multiples of the standard deviation, which is what distinguishes it from a generic statement about spread.

The Binomial distribution counts successes in n independent Bernoulli trials that share a success probability p, so it fits any scenario with a fixed trial count such as coin flips or defects in a known sample size. The Poisson distribution models event counts over a time or space interval rather than a fixed number of trials, the Geometric distribution models how many trials are needed until the first success, and the Normal distribution is continuous, so it cannot model a count of successes.

The exponential distribution models the inter-arrival times of a Poisson process. Events in a Poisson process occur independently at a constant average rate, and this memoryless property of the waiting time between consecutive events is exactly what the exponential distribution describes. The answer is therefore the exponential distribution, not the Poisson distribution itself, which models the count of events in a fixed interval.

The normal (Gaussian) distribution is symmetric about its mean and is fully parameterized by its mean and variance (or standard deviation), so those two numbers determine the entire distribution. It is also the limiting distribution of standardized sums in the Central Limit Theorem, which is why it appears throughout statistical testing.

The Poisson distribution gives the probability of observing k events in a fixed interval when events occur independently at a constant average rate λ. It is the limiting case of the Binomial distribution as n → ∞ and p → 0 with np = λ. The question describes exactly these conditions (counts in a fixed interval, known constant rate, independent events), so the answer is the Poisson distribution.

A Bernoulli random variable X takes value 1 with probability p and 0 with probability 1−p, so its expected value is E[X] = 1·p + 0·(1−p) = p. The zero-valued outcome contributes nothing to the sum, so the mean equals the success probability.

A p-value is the probability of obtaining a test statistic at least as extreme as the observed one, assuming the null hypothesis is true. A p-value of 0.03 therefore means there is a 3% chance of seeing data this extreme or more extreme under the null hypothesis. It is not the probability that the null hypothesis is true, nor a measure of effect size. Because 0.03 is less than the chosen significance level of 0.05, the result is statistically significant and the null hypothesis is rejected.

A Type I error is a false positive: rejecting a null hypothesis that is actually true. Its probability equals the significance level α, which is the threshold chosen for the test. The opposite mistake, failing to reject a false null hypothesis, is a Type II error (false negative), so the statement as written correctly describes Type I error.

In the z-interval formula x̄ ± z_{α/2} · σ/√n, σ is the population standard deviation and n is the sample size. The ratio σ/√n is the standard error of the mean, and multiplying it by the critical z-value gives the margin of error.

Inferential statistics uses sample data to draw conclusions about a population; confidence intervals and hypothesis tests are core examples, since each estimates or tests a population parameter from a sample. Descriptive statistics, by contrast, only summarizes the data in hand (mean, median, standard deviation, histograms) without extending beyond the observed sample.

Descriptive statistics only summarize the data you actually have, using measures such as means, medians, and standard deviations. Drawing conclusions about a larger population that was not measured requires inferential statistics, which use sample data along with probability models to generalize and quantify uncertainty.

Computing the standard deviation of a sample summarizes the spread of the data you actually have, without drawing any conclusion about a larger population. That is the defining purpose of descriptive statistics, so the statement is true. Inferential statistics would only enter if the sample's value were used to estimate or test something about the population it came from.

A p-value is computed assuming the null hypothesis is true. It measures how surprising the observed data (or more extreme data) would be under that assumption, which is the probability of a test statistic at least as extreme as the one observed, given the null. It is not the probability that the null is true, not the probability that the alternative is true, and not the long-run false-positive rate of the test. Those are common misconceptions. The false-positive rate is the significance level alpha, which is a chosen threshold rather than a property of a particular dataset.

The Poisson distribution with rate λ has E[X] = λ and Var(X) = λ, so mean equals variance for every λ > 0. The Normal distribution has mean μ and variance σ², which are independent parameters and not necessarily equal. The Exponential distribution has mean 1/λ and variance 1/λ², which are equal only when λ = 1, so it fails in general. The Binomial(n, p) has mean np and variance np(1−p), which are equal only when p = 0, a degenerate case rather than a general property.

The Exponential distribution is the unique continuous distribution with the memoryless property. For X ~ Exp(λ), P(X > s + t) = e^{-λ(s+t)} = e^{-λs}·e^{-λt}, so P(X > s + t | X > s) = P(X > s + t)/P(X > s) = e^{-λt} = P(X > t). The Geometric distribution also satisfies this property, but it is discrete, not continuous. The Normal and Poisson distributions do not satisfy the memoryless property at all. Specifying 'continuous' in the prompt uniquely identifies the Exponential distribution.

With the rate parameterization, the density is λe^(−λx) for x ≥ 0. Integrating gives E[X] = ∫x·λe^(−λx)dx = 1/λ, and E[X²] = 2/λ², so Var(X) = E[X²] − (E[X])² = 2/λ² − 1/λ² = 1/λ². Substituting λ = 2 gives a mean of 1/2 = 0.5 and a variance of 1/4 = 0.25.

A Binomial(n, p) variable is a sum of n independent Bernoulli(p) trials, each with mean p, so its mean is np. An Exponential distribution with rate λ has mean 1/λ, the reciprocal of the rate. A Poisson(λ) distribution has mean λ, since λ is both its rate and its mean. The only pairing matching all three is Binomial → np, Exponential → 1/λ, Poisson → λ, which is the option listing those exact assignments.

The p-value is computed under the assumption that the null hypothesis is true: it is the probability of a test statistic at least as extreme as the observed one, counting both tails for a two-sided test. Here z = 1.96 gives a two-sided tail area of about 0.05 under the standard normal distribution. The correct reading is the statement that assumes H0 is true and describes the probability of data this extreme. The statement claiming a 5% probability that H0 is true confuses the p-value with a posterior probability, which frequentist testing does not provide. The statement equating the p-value with the Type II error rate confuses it with β, which is governed by power and the true alternative. The statement about the false-positive rate among rejections conflates the p-value with a conditional error rate that depends on the prior probability of H0 and on power.

In a Poisson process with rate λ, the waiting time between successive arrivals is exponential with mean 1/λ. With λ = 5 arrivals per hour, the mean gap is 1/5 hour = 0.2 hours. Converting to minutes, 0.2 × 60 = 12 minutes. The average gap is the reciprocal of the rate, so the answer is 12 minutes.

The Type I error rate of a test is set equal to α, so it drops from 0.05 to 0.01. Power is the probability that the test statistic lands in the rejection region when H₀ is false. Keeping the sample size and effect size fixed, a smaller α moves the critical value further into the tail, which shrinks the rejection region, so fewer true alternatives fall inside it and power decreases. The option claiming power stays the same is wrong because the test statistic's distribution may be unchanged while the critical value it is compared against is not. The option claiming power increases confuses the direction of the trade-off.

A distribution belongs to the regular one-parameter exponential family only if its density (or PMF) can be written in the form f(x | θ) = h(x) · exp(η(θ) T(x) − A(θ)), which requires, among other conditions, that the support of the distribution not depend on the unknown parameter θ. For a Uniform(0, θ) distribution the support is [0, θ]; since this interval changes with θ, the indicator function I(0 ≤ x ≤ θ) appears in the density and cannot be absorbed into h(x). Consequently, the density 1/θ · I(0 ≤ x ≤ θ) cannot be expressed in the standard exponential-family form, so Uniform(0, θ) is excluded from the regular exponential family. The other three distributions all have parameter-independent supports and admit standard exponential-family representations: Poisson(λ) has PMF exp(x log λ − λ − log x!), Bernoulli(p) has PMF exp(x log(p/(1−p)) + log(1−p)), and Exponential(λ) has density exp(−λx + log λ) on (0, ∞).

By the duality between two-sided tests and confidence intervals, a (1−α) confidence interval for μ excludes 0 if and only if the two-sided test of H₀: μ = 0 rejects at level α. With p = 0.06, at α = 0.05 (the 95% interval) we fail to reject, so 0 lies inside the 95% interval; at α = 0.10 (the 90% interval) we reject, so 0 lies outside the 90% interval. The claim that the probability H₀ is true equals 0.06 is wrong, since a p-value is not the posterior probability of the null, and the claim that power is 0.94 is wrong, since power depends on the true effect size, sample size, and α rather than on the observed p-value.

Inferential statistics use sample data to make probabilistic generalizations about a population parameter. The 95% confidence interval for the population proportion quantifies uncertainty about that parameter, and the p-value from the one-proportion z-test tests a hypothesis about the population proportion, so both are inferential. The sample proportion supporting Candidate A, the sample median age, and the sample standard deviation of ages merely summarize the observed data and are descriptive statistics.

The correct statement is that the test is conservative. Wilks' theorem needs the null value to lie in the interior of the parameter space, but σ²_b = 0 sits on the boundary of [0, ∞). Self and Liang (1987) showed that the asymptotic null distribution of 2 log Λ is a 50:50 mixture of a point mass at 0 (χ²(0)) and χ²(1). The point mass arises because the unconstrained MLE of σ²_b equals zero with probability about 1/2 under H₀, giving a statistic of exactly 0. Using the χ²(1) cutoff of 3.84 at α = 0.05 therefore yields an actual size of 0.5 × 0.05 = 0.025, so the test rejects too rarely. The nominal 0.05 level is recovered with the 0.10 quantile of χ²(1) (about 2.71) or by using the mixture directly. The statistic itself is well defined at the boundary and the test remains valid; only the reference distribution changes. The option claiming the test is exact is wrong because it ignores the boundary, the liberal option is wrong because the actual size is below 0.05, and the undefined-statistic option is wrong because the likelihood ratio is computable at zero.