Question Bank

Units 1–2 — Descriptive Statistics

Q. Distinguish between descriptive and inferential statistics, with an example of each from business.

Descriptive statistics deals with collecting, summarising and simplifying data that would otherwise be unwieldy and voluminous. It is used when the population of interest is small enough to describe directly. Inferential statistics is the science of using a sample to make generalisations about the important aspects of a population, and is needed when the population is too large to measure.

A descriptive value computed on a population is called a parameter (μ, σ, p); one computed on a sample is called a statistic (x̄, s, p̂).

Business examples: a retailer computing last quarter's average basket size from every transaction is doing descriptive statistics. The same retailer surveying 400 shoppers to estimate the proportion of all customers who would use a subscription plan is doing inferential statistics — and the standard error tells them how far the estimate might be off. Deck slide 3

Q. Classify the types of statistical data and explain why the classification matters.

Data divides into two broad categories:

• Quantitative — measurable in definite units. Discrete: countable, such as customers per day, incoming flights, or defective items in a consignment. Continuous: measured on a scale, such as weight, length, height, thickness, velocity or temperature.
• Qualitative — a quality or attribute of the subject. Nominal: unordered categories, such as gender, skill level or education level. Rank: ordered, assigned as integers 1, 2, 3, …, n, such as positions in an interview or contest.

Why it matters: the data type determines every summarisation decision that follows. Qualitative data gets a frequency table and a bar graph with gaps between bars; quantitative data gets class intervals and a histogram whose bars touch. Using the wrong pairing misrepresents the data — a histogram implies a continuous scale that categorical data does not have. Deck slides 8–10

Q. Distinguish between primary and secondary data. What is the main limitation of secondary data?

Primary data does not already exist in any form and must be collected for the first time from the primary source, covering either the whole population or a sample drawn from it. Secondary data already exists in some form — published or unpublished — in an identifiable secondary source, and is generally available from published sources.

The main limitation of secondary data is that it is "not necessarily in the form actually required" Deck slide 7. It was collected for someone else's purpose, so its definitions, time periods, geographic coverage and categories may not match yours. It is cheap and fast; the price is that it may not answer your question. Primary data fits exactly but costs time and money.

Q. Why is the standard deviation preferred to the range as a measure of dispersion?

The range (highest − lowest) is fast and intuitive, but it uses only two observations out of the whole dataset and is destroyed by a single outlier. Everything between the extremes is ignored.

The standard deviation uses every observation, since it is built from each value's deviation from the mean. It is a "quantitative index of a distribution's spread or variability" and — critically — it is used to calculate the likelihood (probability) of an event occurring Deck slide 8.

That last property is what makes it indispensable: every z-score in the rest of the course is a distance measured in standard deviations. Chebyshev's theorem, the empirical rule, the normal table and the standard error are all expressed in units of σ. Nothing comparable can be built on the range.

Q. What is the coefficient of variation and when would you use it instead of the standard deviation?

The coefficient of variation is a relative measure of dispersion:

Coefficient of Variation

\[ \text{CV} = \left(\frac{\sigma}{\bar{x}}\right) \times 100\% \]

A standard deviation carries the units of the data, so an SD of ₹500 and an SD of 500 grams cannot be compared, and neither can two SDs measured in the same units if the means differ greatly. Dividing by the mean strips the units out and expresses variability relative to the average size of the thing being measured.

Use it whenever a question asks which of two things is "more consistent", "more stable" or "less risky" and the two have different means. Example: Branch A has mean sales ₹150 lakh with SD ₹12 lakh (CV 8.0%); Branch B has mean ₹450 lakh with SD ₹30 lakh (CV 6.67%). The raw SDs suggest B is far more variable; the CV shows B is actually the more consistent of the two. Class notes p. 1

Q. State Chebyshev's theorem and contrast it with the empirical rule.

Chebyshev's theorem states that no matter what the shape of the distribution, at least (1 − 1/k²) of the values lie within μ ± kσ. In particular, at least 75% lie within ±2σ and at least 89% within ±3σ Class notes p. 2.

The empirical rule applies only to normal distributions and states that around 68% of values lie within ±1σ, around 95% within ±2σ and around 99% within ±3σ.

	Chebyshev	Empirical rule
Assumption	None — any distribution	Distribution must be normal
Within ±2σ	At least 75%	About 95%
Nature of the bound	A guaranteed minimum	An approximate actual value

The trade-off: Chebyshev is always true but loose; the empirical rule is far more precise but must be earned by justifying normality. On the class's 200-patient hospital data, Chebyshev guaranteed at least 150 patients in the range while the normal assumption predicted 190 — and the actual count was 191. Assuming normality buys precision, when it is justified.

Q. Explain skewness and kurtosis, and what each tells a manager about a service channel.

Skewness is departure from symmetry. Sk = 0 means perfect symmetry; Sk > 0 is a positive skew with a long right tail; Sk < 0 is a negative skew with a long left tail Class notes p. 1. It can be read from the three averages: Mean < Median ≤ Mode indicates negative skew.

Kurtosis is a property of the data describing peakedness. K > 3 is leptokurtic — tall and narrow, with most data piled at the centre. K < 3 is platykurtic — flat and wide. K = 3 is mesokurtic, the normal curve's own peakedness.

What a manager reads: in a satisfaction survey, negative skew means most customers are satisfied with a small dissatisfied tail — good news, despite the negative sign. Low SD combined with high kurtosis means consistent, predictable service; high SD with low kurtosis means the experience varies widely from customer to customer. Two channels can share a mean and still be completely different businesses to run.

Unit 3 — Theories of Probability

Q. Explain the three theories (methods) of assigning probability, with an example of each.

1. Classical method. Assigns probabilities on the assumption that all outcomes are equally likely: if an experiment has n possible outcomes, each gets 1/n. Example: rolling a fair die, S = {1,2,3,4,5,6}, each face has probability 1/6 Deck slide 12.

2. Relative frequency method. Assigns probabilities from experimentation or historical data, as frequency ÷ total frequency. Example: a rental firm's records show 2 polishers were rented on 18 of the last 40 days, so P(2 rented) = 18/40 = 0.45 Deck slide 14.

3. Subjective method. Assigns probabilities based on judgement, when neither symmetry nor data is available. Example: an analyst's own probabilities for the eight possible outcomes of two share investments Deck slide 15.

Whichever method is used, the assignments must satisfy the axioms: each probability lies between 0 and 1, and the probabilities of all sample points sum to 1.

Q. Distinguish between mutually exclusive events and independent events.

Two events are mutually exclusive if they have no sample points in common — when one occurs, the other cannot. Formally P(A ∩ B) = 0, so the addition law simplifies to P(A ∪ B) = P(A) + P(B) Deck slides 29–30.

Two events are independent if the occurrence of one does not change the probability of the other: P(A|B) = P(A), equivalently P(A ∩ B) = P(A)·P(B) Deck slide 37.

These are opposites, not synonyms. Mutual exclusivity is the strongest possible dependence: learning that A occurred tells you B definitely did not. Independence means learning A occurred tells you nothing about B. Two events with non-zero probabilities can never be both. Example: "the coin lands heads" and "the coin lands tails" are mutually exclusive but not independent; "the coin lands heads" and "the die shows a six" are independent but not mutually exclusive.

Q. State the law of total probability and Bayes' theorem, and prove Bayes' theorem.

Law of total probability. Let S₁, S₂, …, S_k be mutually exclusive and exhaustive events — one and only one must happen. Then for any event A:

Law of Total Probability

\[ P(A) = P(S_1)P(A \mid S_1) + P(S_2)P(A \mid S_2) + \dots + P(S_k)P(A \mid S_k) \]

Bayes' theorem. If those events have prior probabilities P(S₁)…P(S_k) and event A occurs, the posterior probability of S_i is:

Bayes' Theorem

\[ P(S_i \mid A) = \frac{P(S_i) \cdot P(A \mid S_i)}{\sum_{j=1}^k P(S_j) \cdot P(A \mid S_j)} \]

Proof. From the definition of conditional probability, P(A|Sᵢ) = P(A ∩ Sᵢ)/P(Sᵢ), so P(A ∩ Sᵢ) = P(Sᵢ)·P(A|Sᵢ). Also P(Sᵢ|A) = P(Sᵢ ∩ A)/P(A). Substituting the first result into the numerator and the law of total probability into the denominator gives the theorem. Class notes pp. 1–2

What it does: Bayes reverses a conditional probability. You are told P(evidence | cause) and want P(cause | evidence). The numerator is one branch; the denominator is every branch, which is exactly the P(A) built in the first step.

Q. A test is 99% accurate for a disease that affects 1 person in 1,000, but returns a positive for 2% of healthy people. Someone tests positive. Explain, with the calculation, why their chance of having the disease is under 5%.

Let A = has the disease, B = tests positive. Given P(A) = 0.001, P(A^c) = 0.999, P(B|A) = 0.99, P(B|A^c) = 0.02.

Bayesian Disease Test Calculation

\[ P(A \mid B) = \frac{P(A)P(B \mid A)}{P(A)P(B \mid A) + P(A^c)P(B \mid A^c)} = \frac{(0.001)(0.99)}{(0.001)(0.99) + (0.999)(0.02)} \approx 4.72\% \]

Why it feels wrong, and why it is right. Think of 100,000 people. About 100 have the disease and 99 of them test positive. The other 99,900 are healthy, and 2% of them — nearly 2,000 people — also test positive. So there are roughly 2,099 positives of which only 99 are genuine: about 4.7%.

The general principle: when the base rate is very low, most positives are false positives, no matter how accurate the test sounds. This is exactly the reasoning both papers' Q5 (a) is testing, restated as a credit-risk model or a churn model. Deck slides 43–44

Q. State the rule of at least one and explain why the complement is used.

For n independent trials with success probabilities p₁, p₂, …, pₙ:

Rule of At Least One

\[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - (1-p_1)(1-p_2)\dots(1-p_n) \]

Why the complement. "At least one" is the union of "exactly one", "exactly two", …, "exactly n" — computing it directly means n separate terms, each requiring its own combination count, and then adding them. Its complement is a single event, "none", whose probability is one product. For ten job applications, the direct route needs ten terms; the complement needs one line.

Never add the individual probabilities. For ten applications with success probabilities summing to 2.10, that would give an impossible answer above 1. The correct calculation for that case gives P(at least one offer) = 1 − 0.088 ≈ 0.91. Class notes p. 9

Q. What is the expected number of successes, and how does it differ from a probability?

The expected number of successes is the average count of successes over many repetitions:

Expected Successes

\[ E = \sum p_i \qquad (\text{or } E = n \cdot p \text{ for identical } p) \]

Distinguish it carefully from a probability. For ten job applications with probabilities 0.38, 0.35, …, 0.08, the expected number of offers is their sum, 2.10 offers, while the probability of getting at least one offer is 0.91. Both describe the same situation and both are correct. An expected count can exceed 1 — a probability cannot. The n·p form is also the mean of the binomial distribution, which is where Unit 4 picks it up.

Unit 4 — Probability Distributions

Q. State the conditions under which the binomial distribution applies, and give its mean and variance.

Four conditions Deck slide 23:

1. The data arise from a sequence of n independent trials.
2. At each trial there are only two possible outcomes — success and failure.
3. The probability of success, p, is the same in each trial.
4. The random variable of interest is the number of successes X in the n trials.

Binomial Formula

\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \qquad \mu = np, \quad \sigma^2 = npq, \quad \sigma = \sqrt{npq} \]

The deck adds a warning worth quoting: "The assumptions of independence and constant p are important. If they are invalid, so is the binomial distribution." More generally, "do not use a particular distribution unless you are satisfied that the assumptions which underlie it are at least approximately satisfied" Deck slide 22.

Q. Describe the Poisson distribution, its assumptions, and its distinguishing property.

The Poisson distribution is a discrete probability distribution for the count of events that occur randomly in a given interval of time or space Deck slide 28. Its two assumptions are that events occur at random (one occurrence does not change the probability of another) and at a constant rate.

Poisson Distribution

\[ P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \qquad \mu = \lambda, \quad \sigma^2 = \lambda, \quad \sigma = \sqrt{\lambda} \]

Its distinguishing property is that the mean and the variance are equal — both are λ. No other distribution in the syllabus has this.

Examples from the deck: tropical cyclones in a season, occasions when a river exceeds a level, accidents at a junction in a month, births in a hospital in 10 hours, car accidents at a site, requests for a document on a web server.

A Poisson variable can take any positive integer value, whereas the binomial always has a finite upper limit of n. The Poisson also approximates the binomial when n ≥ 20 and p ≤ 0.05, using λ = n·p Deck slide 29.

Q. State the properties of the normal distribution and explain why standardising is necessary.

Properties Deck slides 34–40:

• Bell shaped and symmetrical about the mean.
• Mean, median and mode are all equal and sit at the centre.
• The random variable has an infinite range — the curve approaches but never touches the horizontal axis.
• The total area under the curve is 1, and probability is area.
• Almost all the area lies between −3 and +3 standard deviations.
• There are infinitely many normal distributions, one for each pair (μ, σ).

Why standardise. The last property is the problem: to find areas we would need a separate table for every normal curve, and there are infinitely many. Standardising transforms every normal distribution into one particular distribution — the standard normal, with mean 0 and standard deviation 1 — for which a single table suffices Deck slide 37.

Standardizing Z Score

\[ z = \frac{x - \mu}{\sigma} \]

The z value is a distance measured in standard deviations: z = +2 means the value is two SDs above the mean.

Q. What is the continuity correction and why is it needed?

When a discrete binomial variable is approximated by a continuous normal curve, each integer count must be widened into an interval of width 1 centred on it. That adjustment is the continuity correction.

Continuity Correction Rules

\[ P(X = k) \to P(k - 0.5 \le X \le k + 0.5), \qquad P(X \ge k) \to P(X \ge k - 0.5) \]

Why it is needed. A continuous distribution assigns zero probability to any single point, so P(X = 55) computed directly on a normal curve is exactly 0 — which is obviously wrong for a binomial. Widening 55 into 54.5–55.5 recovers the area of the histogram bar the count actually represents.

How well it works: for 100 coin tosses, the exact binomial gives P(X = 55) = 0.0485 and the corrected normal approximation gives 0.0484. The approximation is only valid when np ≥ 5 and nq ≥ 5.

Q. How do you choose between the binomial, Poisson and normal distributions?

Distribution Decision Guide

Fixed \(n\), 2 outcomes, constant \(p\) \(\to\) Binomial
Average rate per interval \(\to\) Poisson
Continuous measurement \(\to\) Normal
Binomial with large \(n\) \(\to\) Normal (with continuity correction)
Binomial with large \(n\), small \(p\) \(\to\) Poisson (\(\lambda = np\))

The single most useful test is whether the question gives you an n. "10 bulbs, each 10% defective" has n = 10 and p = 0.1 → binomial. "8 customers arrive every 10 minutes" has no n at all, only a rate → Poisson. This choice, not the arithmetic, is what most often goes wrong in Q1 (b). Levin §5.7, p. 254

Units 5–6 — Sampling Theory

Q. What is a sampling distribution of the mean? Why is the concept necessary?

A sampling distribution of the mean is the probability distribution of the means of all the possible samples of a given size that could be drawn out of a given population Deck slide 3.

Why it is necessary. A sample mean x̄ is not a fixed number — draw a different sample and you get a different x̄. So x̄ is itself a random variable, and like any random variable it has a distribution, a mean and a standard deviation. Without that distribution there is no way to say how far a particular x̄ is likely to be from the true μ, and therefore no way to make any inference from a sample at all.

Example: from five employees with 3, 3, 7, 9 and 14 years of service, there are ⁵C₃ = 10 possible samples of size 3. Their means range from 4⅓ to 10, and the mean of those ten means is exactly 7.2 — the population mean. Class notes p. 1

Q. State the Central Limit Theorem and explain its significance.

The Central Limit Theorem states Deck slide 8:

1. The mean of the population is always equal to the mean of the sampling distribution of the mean: μ_x̄ = μ.
2. As the sample size increases, the sampling distribution of the mean approaches normality, regardless of the shape of the population distribution.

Significance. The deck calls it "perhaps the most important theorem in inferential statistics", because "without the knowledge of the frequency distribution of the population, it permits us to use the sample statistics to make inferences about the population parameters" Deck slide 9.

Practically: it is what lets you standardise a sample mean with a z and read an ordinary normal table even though you know nothing about the underlying population's shape. The conventional threshold is n ≥ 30; if the population is already normal, x̄ is exactly normal for any n. Every sampling-distribution question in both papers silently depends on it.

Q. Define the standard error of the mean. How does it relate to sample size, and what does that imply?

The standard error of a statistic is the standard deviation of the sampling distribution of that statistic Deck slide 10. For the mean:

Standard Error of Mean

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

Note carefully that it is not the standard deviation of the data. It measures how much the sample mean bounces around from sample to sample.

The chain of consequences Deck slide 19: as the standard error decreases, sample means cluster more tightly around the population mean; so any single sample mean is probably closer to μ; so the precision with which the sample mean estimates the population mean increases.

Because n sits under a square root, the returns diminish: quadrupling n only halves the error, and a hundredfold increase only cuts it by ten. This is the law of diminishing returns in sampling and the reason a national survey of 1.4 billion people stops at a few thousand respondents.

Q. What is the finite population multiplier and when should it be used?

When sampling without replacement from a finite population, the standard error is smaller than the infinite-population formula suggests, and is corrected by the finite population multiplier:

Finite Population Multiplier

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \cdot \sqrt{\frac{N - n}{N - 1}} \]

The fraction n/N is called the sampling fraction. When N is very large relative to n the multiplier is close to 1 and changes nothing; the standard rule of thumb is that when the sampling fraction is less than 0.05 the multiplier need not be used Deck slide 21.

Why it works: each element measured is one fewer you are uncertain about. As n → N the multiplier → 0, which is exactly right — measure the whole population and there is no sampling error left. In an exam, if a question hands you N at all, apply the correction and state the sampling fraction that justified it.

Q. Describe the sampling distribution of a proportion and its standard error.

Used when the issue of interest is categorical — outcomes classified as occurrences or non-occurrences, and we estimate the proportion of occurrences Deck slide 13. Let x be the number of occurrences in n trials; x follows a binomial distribution with success probability p, and for large n that binomial is approximated by a normal distribution.

Sampling Proportion & SE

\[ \hat{p} = \frac{x}{n}, \qquad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}, \qquad z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \]

Before using the normal approximation, check n·p ≥ 5 and n·q ≥ 5 — the deck makes this an explicit first step Deck slide 18. With a finite population, multiply the standard error by the same √[(N−n)/(N−1)] factor.

Q. Explain the trade-off between precision and confidence in sampling.

Precision refers to the closeness of the estimate to the true population characteristic. It is measured by the standard error of the sampling distribution of the mean, and improved by narrowing the interval (a smaller E).

Confidence is the certainty that the estimate really holds for the population. It is stated as a confidence level — 90%, 95%, 99% — and improved by widening the interval (a larger z).

These pull in opposite directions: "other things being equal, the narrower the range, the lower the confidence. Hence a trade-off between precision and confidence is a must" Deck slide 23.

A concrete illustration from the class notes pp. 1–2: an estimate of mean height as 5'6″ ± 10″, i.e. 4'8″ to 6'4″, carries 95% confidence; widening it to 5'6″ ± 3′, i.e. 2'6″ to 8'6″, raises confidence to 99% but the interval is now almost certainly right and almost completely useless. "If the confidence level increases, z also increases, which causes the margin of error to increase as well." The only way to buy both at once is to increase n — and that costs money.

Q. What factors determine the required sample size? Is the population size one of them?

Three factors Deck slide 31, and they are exactly the three symbols in the formula:

1. Variance / heterogeneity (σ). "A heterogeneous population has more variance and will require a larger sample; a homogeneous population has less variance and permits a smaller sample."
2. Magnitude of error (E). "How precise must the estimate be?" n rises as 1/E², so halving the tolerance quadruples the sample.
3. Confidence level (z). "How much error will be tolerated?" n rises as z², so moving from 95% to 99% raises the requirement by about 73%.

Sample Size Formulas

\[ n = \left(\frac{z \cdot \sigma}{E}\right)^2 \quad (\text{Mean}), \qquad n = \frac{z^2 \cdot p \cdot q}{E^2} \quad (\text{Proportion}) \]

Population size is not on the list, and saying so explicitly is worth a mark. For large populations n depends only on σ, E and z. N enters only through the finite population correction, and only when the sample is a share of the population. That is why a poll of 1,000 people can describe a country of 1.4 billion as accurately as it describes a city of 100,000.

Q. Why is p = 0.5 used when no prior estimate of the population proportion is available?

The sample-size formula for a proportion, n = z²pq/E², needs a value of p — but p is precisely what the study is trying to estimate. The convention is to take p = 0.5, described by the deck as "a conservative estimate" Deck slide 38 and by the official answer key as assuming "maximum variability" Key p. 3.

The reason: the product p·q = p(1 − p) reaches its maximum at p = 0.5, where it equals 0.25. Any other value of p gives a smaller product and therefore a smaller required n. Choosing 0.5 therefore guarantees the sample is large enough whatever the true p turns out to be — it can never leave you under-sampled.

What a prior estimate is worth: at 95% confidence and E = 0.04, p = 0.5 requires n = 601; a known p = 0.25 requires only 451; and a known p = 0.05 requires just 73 Deck slide 41. The further the true p is from 0.5, the more a pilot study is worth. This is also the argument for sequential sampling — running a small pilot to estimate the parameters before drawing the full sample Deck slide 32.

Complete inventory of both past papers

Every question in the 2025-26 Final and Re-Examination, with its unit and whether it falls inside the Units 1–6 mid-term portion. Nothing is hidden.

In scope — worked in full on the Solved Papers page

Out of scope — listed so you know what was left out