Statistics for NDA — Mean, Variance, Standard Deviation and Probability Distributions

Concept

Statistics, in the context of NDA Mathematics, is the science of collecting, organising, and interpreting numerical data. The exam tests two closely linked ideas: descriptive statistics (how you summarise a dataset) and inferential probability (how you reason about uncertain outcomes). These appear together on the paper and you need to treat them as a unified chapter, not separate islands.

Here is the core mental model. Every dataset has a centre and a spread. The centre is captured by measures of central tendency — mean (arithmetic average), median (middle value), mode (most frequent value). The spread is captured by variance and standard deviation. Almost every NDA Statistics question asks you to compute one or both, sometimes after a transformation of the data.

Think of it this way: if you have the heights of 50 army recruits, the mean tells you the "typical" recruit, and the standard deviation tells you how much variation to expect. A tight cluster means low SD; a wide scatter means high SD.

The analogy that works best in the exam hall: mean is the balance point of a see-saw, variance is how wobbly the see-saw feels. When you shift every person on the see-saw by the same amount (add a constant), the balance point moves but the wobbliness does not. When you scale everyone up (multiply by a constant), both the balance point and the wobbliness scale — but variance scales by the square of that constant, SD scales linearly.

This single analogy unlocks almost every transformation question on NDA.

Beyond descriptive stats, the chapter bleeds directly into probability distributions — Binomial, Poisson, and the basics of expected value. NDA treats these as one chapter and so should you.

Deep Dive

Measures of Central Tendency

Arithmetic Mean for raw data: $\bar{x} = \dfrac{\sum x_i}{n}$

For grouped/frequency data: $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$

Median — middle value when data is sorted. For $n$ observations:

• If $n$ is odd: median = $\left(\dfrac{n+1}{2}\right)^{th}$ value
• If $n$ is even: median = average of $\left(\dfrac{n}{2}\right)^{th}$ and $\left(\dfrac{n}{2}+1\right)^{th}$ values

Mode — the most frequently occurring observation. For a frequency distribution, it is the observation with the highest frequency.

Relationship between the three for a moderately skewed distribution: $\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$. This is a useful check but is rarely tested directly.

Variance and Standard Deviation

Variance (population): $\sigma^2 = \dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2$

This is the version NDA uses. Remember it as: mean of squares minus square of mean.

Standard Deviation: $\sigma = \sqrt{\sigma^2}$

Coefficient of Variation (CV): $\text{CV} = \dfrac{\sigma}{\bar{x}} \times 100$

CV is a dimensionless measure used to compare variability across two datasets that have different units or different means. A higher CV means more relative variability.

Effect of Linear Transformation

This is the single most tested concept in NDA Statistics. If $Y = aX + b$:

| Measure | New Value | |---|---| | Mean | $a\bar{x} + b$ | | Variance | $a^2 \sigma^2$ | | Standard Deviation | $|a| \cdot \sigma$ | | CV | Changes (because both mean and SD change) |

Look — adding a constant $b$ does nothing to variance or SD. Only the scaling factor $a$ matters for spread.

Frequency Distribution and Variance Formula

For grouped data with frequencies $f_i$ and class midpoints $x_i$:

$\sigma^2 = \dfrac{\sum f_i x_i^2}{\sum f_i} - \left(\dfrac{\sum f_i x_i}{\sum f_i}\right)^2$

NDA sometimes gives you $\sum X$, $\sum X^2$, and $n$ directly (as in the 2025 PYQ). The formula then simplifies to:

$\sigma^2 = \dfrac{\sum X^2}{n} - \left(\dfrac{\sum X}{n}\right)^2$

Probability — Core Rules

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For independent events: $P(A \cap B) = P(A) \cdot P(B)$

So the union formula becomes: $P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$

Conditional Probability: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

Discrete Probability Distributions

Expected Value: $E(X) = \sum x_i \cdot P(X = x_i)$

E(X²): $E(X^2) = \sum x_i^2 \cdot P(X = x_i)$

Binomial Distribution $B(n, p)$: $P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$

Mean $= np$, Variance $= np(1-p)$, SD $= \sqrt{np(1-p)}$

Poisson Distribution with mean $\lambda$: $P(X = r) = \dfrac{e^{-\lambda} \lambda^r}{r!}$

Mean $= \lambda$, Variance $= \lambda$

Note: In a Poisson distribution, mean equals variance — this is a diagnostic fact that NDA tests directly.

Setting Up Probability Distributions

A common NDA pattern: you are given $P(X=r)$ in terms of $k$ for $r = 0, 1, 2, \ldots$ Use $\sum P(X=r) = 1$ to find $k$, then compute $E(X)$ or $E(X^2)$.

The trap here is forgetting that $E(X^2) \neq [E(X)]^2$. These are almost always different, and the question will have $[E(X)]^2$ as a wrong option to catch you.

Memory Tricks and Shortcuts

Fast-Solving Framework

When you see a Statistics/Probability question in the NDA paper, run this decision tree in the first 10 seconds:

Step 1 — Identify the type:

• Given $\sum X$, $\sum X^2$, $n$ → Variance formula directly
• Given CV and SD → Mean = SD × 100 / CV
• Linear transformation ($Y = aX + b$) → New mean = $a\bar{x} + b$, New SD = $|a|\sigma$
• Probability distribution with unknowns → Sum of probabilities = 1 first, then compute $E(X)$ or $E(X^2)$
• Conditional probability → List the sample space of the conditioning event, count favourable outcomes
• Independence given → Use $P(A \cap B) = P(A) \cdot P(B)$ in the union formula

Step 2 — One formula, one substitution. NDA Statistics questions almost never require more than two lines of arithmetic once you identify the right formula. If you find yourself doing more than three lines, you have the wrong formula.

Step 3 — Sanity check the answer. Variance is always non-negative. SD $\leq$ range. CV is a percentage, typically between 0% and 100% for well-behaved data. Probability is always between 0 and 1.

Solved PYQs

Why this question: This 2025 NDA PYQ tests the variance formula directly with $\sum X$ and $\sum X^2$ given — the cleanest possible test of whether you know the "mean of squares minus square of mean" pattern.

Solving path: Variance(X) $= \frac{900}{50} - \left(\frac{200}{50}\right)^2 = 18 - 16 = 2$. Variance(Y) $= \frac{1400}{50} - \left(\frac{250}{50}\right)^2 = 28 - 25 = 3$. Since $2 < 3$, Variance(X) $<$ Variance(Y). Done in under 30 seconds.

Why this question: A classic discrete probability distribution setup. The trap is computing $E(X^2)$ directly rather than $[E(X)]^2$.

Solving path: Sum all probabilities: $k + 2k + 3k + 4k = 10k = 1$, so $k = 1/10$. Then $E(X^2) = 0^2(1/10) + 1^2(2/10) + 2^2(3/10) + 3^2(4/10) = 0 + 2/10 + 12/10 + 36/10 = 50/10 = 5$.

Why this question: Conditional probability on dice — tests whether you correctly identify the reduced sample space.

Solving path: Condition is sum = 8. List all pairs: (2,6), (3,5), (4,4), (5,3), (6,2) — exactly 5 outcomes. Of these, first die = 5 is only (5,3) — 1 outcome. So $P = 1/5$.

Why this question: Independence + union formula. This is a direct drill on the formula $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ when independence is stated.

Solving path: $0.6 = 0.3 + P(B) - 0.3 \cdot P(B) = 0.3 + 0.7 \cdot P(B)$. So $0.7 \cdot P(B) = 0.3$, giving $P(B) = 3/7$.

Why this question: Tests the transformation rule for mean and SD — the single most important NDA transformation fact.

Solving path: $Y = 3X + 10$. New mean $= 3(50) + 10 = 160$. New SD $= 3(5) = 15$ (adding 10 does not affect SD). Answer: 160 and 15.

Why this question: Coefficient of Variation reverse-calculation — tests whether you can invert the CV formula to find the mean.

Solving path: Mean$_A = (21 \times 100)/60 = 35$. Mean$_B = (35 \times 100)/70 = 50$. Means are 35 and 50.

Why this question: Poisson condition $P(1) = P(2)$ — tests the pattern-based cancellation trick.

Solving path: $e^{-\lambda}\lambda = e^{-\lambda}\lambda^2/2$. Cancel $e^{-\lambda}$: $\lambda = \lambda^2/2$, so $\lambda = 2$.

Why this question: At-least Binomial probability — the classic $n = 6$ coin toss question that appears in multiple NDA papers.

Solving path: $P(X \geq 4) = \dfrac{\binom{6}{4} + \binom{6}{5} + \binom{6}{6}}{2^6} = \dfrac{15 + 6 + 1}{64} = \dfrac{22}{64} = \dfrac{11}{32}$.

Common Mistakes

• Forgetting that variance scales by $a^2$, not $a$. When $Y = aX + b$, variance multiplies by $a^2$. A common error is to write new variance $= a \cdot \sigma^2$. You will see this trap as an option in the paper.

• Using $E(X^2) = [E(X)]^2$. These are equal only in the degenerate case where all probability is at a single point. In virtually every NDA problem, $E(X^2) > [E(X)]^2$. Compute them separately.

• Not listing the conditional sample space. In conditional probability questions, you must restrict to outcomes where the conditioning event has already occurred. A common error is computing $P(A \cap B) / P(\text{total})$ instead of $P(A \cap B) / P(B)$.

• Misapplying the independence condition. Independence means $P(A \cap B) = P(A) \cdot P(B)$. Mutually exclusive events with non-zero probabilities are NOT independent — they are the opposite. These are distinct concepts that NDA options deliberately conflate.

• Poisson mean-variance confusion. In a Poisson distribution, mean $=$ variance $= \lambda$. If a question gives you the variance and asks for the mean (or vice versa), they are the same number. This is a giveaway if you know it, a trap if you do not.

• Wrong denominator in variance formula. NDA uses population variance $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$, not the sample variance with $n-1$ in the denominator. Dividing by $n-1$ will give you a slightly different number that will match a wrong option in the choices.

Related Topics

• probability
• binomial distribution
• poisson distribution
• measures of central tendency
• frequency distribution
• normal distribution

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