Introduction to Statistical Methods

Bernoulli, Binomial, Poisson, and Normal Distributions

📅 Published: 2026-07-02

🎓 Level: postgraduate

👥 Audience: Postgraduate students in Statistics and Data Science

Prerequisite Knowledge

This lecture builds on the following concepts from earlier lectures. If any feel unfamiliar, review the linked notes before proceeding.

Previously Covered in This Subject

1.5.3 The Normal Distribution Preview — covered in Lecture 1
1.7 Preview — Probability Distributions — covered in Lecture 1
1.7.1 What is a Probability Distribution? — covered in Lecture 1
1.7.2 Named Distributions — covered in Lecture 1
1.7.3 Why the Normal Distribution Dominates — covered in Lecture 1
5.4.5 Analogy — Frequency Distribution to Probability Distribution — covered in Lecture 5
5.5 Introduction to Probability Distributions — covered in Lecture 5
5.5.2 Named Probability Distributions — covered in Lecture 5
6.2 Validating a Probability Distribution — covered in Lecture 6
6.4.3 Worked Example — Discrete Distribution (Finding Mean and Variance) — covered in Lecture 6
6.5 Cumulative Probability Distribution — covered in Lecture 6
6.6.5 Cumulative Distribution — covered in Lecture 6
6.8 Worked Example — Continuous Distribution (Exponential-Type) — covered in Lecture 6
6.9 Worked Example — Piecewise Continuous Distribution — covered in Lecture 6
6.10 Joint Probability Distributions — covered in Lecture 6
6.11 Marginal Probability Distributions — covered in Lecture 6
7.1.3 Joint Distributions — covered in Lecture 7
7.1.4 Marginal Distributions — covered in Lecture 7
7.4 Worked Example 1 — Discrete Joint Distribution Table — covered in Lecture 7
7.5 Worked Example 2 — Marginal Distributions from a Joint Table — covered in Lecture 7
7.7 Worked Example 4 — Continuous Joint Distribution — covered in Lecture 7
7.9 Named Probability Distributions — Overview — covered in Lecture 7
7.10 Bernoulli Distribution — covered in Lecture 7
7.10.2 The Bernoulli Table — covered in Lecture 7
7.10.4 When to Use Bernoulli — covered in Lecture 7
7.11 Binomial Distribution — covered in Lecture 7
7.11.4 Why "Binomial" — covered in Lecture 7
7.11.7 Binomial as a Generalization of Bernoulli — covered in Lecture 7
7.11.11 Worked Example — Given Mean and Variance, Find the Distribution — covered in Lecture 7
7.11.12 When Binomial Does NOT Apply — Two Diagnostic Examples — covered in Lecture 7
7.12 Poisson Distribution — covered in Lecture 7
7.12.3 Relationship to Binomial — The Rare Event Connection — covered in Lecture 7
7.12.4 Binomial vs Poisson — Quick Discrimination — covered in Lecture 7
7.12.5 Worked Example — Call Center (Poisson Identification) — covered in Lecture 7
7.13 Summary — Distribution Identification Cheat Sheet — covered in Lecture 7

Bernoulli, Binomial, Poisson, and Normal Distributions

8.1 Recap — Bernoulli and Binomial Distributions

Hook: If you flip a fair coin once, you know the chance of heads is 50%. But what if you flip it 5 times — what is the chance you get exactly 3 heads? A single coin flip is too simple to build a whole theory on, yet it is the atomic building block of the most widely used discrete probability model in statistics. The Bernoulli and binomial distributions take this one simple idea — success or failure — and scale it to answer questions about repeated trials everywhere, from clinical drug trials to factory quality control.

Intuition: Think of a Bernoulli trial as a single light switch — it can only be ON

s u ccess

or OFF

f ai l u r e

. A binomial distribution is like a row of

n

identical independent switches. You flip each one once and count how many turned ON. The binomial distribution tells you the probability of each possible count.

Imagine a basketball player shooting free throws. Each shot is a Bernoulli trial — make or miss. If the player takes 10 shots and their success rate is 70%, the binomial distribution answers: what is the probability they make exactly 7 out of 10? The key is that each shot is independent — making the first shot does not change the probability of making the second.

Where the analogy breaks: In real basketball, players get tired or gain confidence, so the success probability may drift. The binomial model assumes constant $p$ — a simplification that works well when the trials are truly independent draws.

8.1.1 Bernoulli Distribution — Definition and Properties

A Bernoulli distribution models a single trial with exactly two possible outcomes: success

1

or failure

0

The probability mass function $P M F$ of a Bernoulli random variable $X$ is:

$P (X = x) = {p q = 1 - p if x = 1 (success) if x = 0 (failure)$

Mean $e x p ec t e d v a l u e$ :

$μ = E [X] = 0 \cdot q + 1 \cdot p = p$

Variance:

$σ^{2} = E [(X - μ)^{2}] = (0 - p)^{2} q + (1 - p)^{2} p = p^{2} q + q^{2} p = pq (p + q) = pq$ where $q = 1 - p$ .

So a Bernoulli distribution is fully described by a single number $p$ . If you know $p = 0.3$ , you know everything: mean = 0.3, variance = 0.21.

8.1.2 Binomial Distribution — Definition and Conditions

A binomial distribution models the number of successes in

n

independent and identically distributed Bernoulli trials.

Probability mass function:

$P (X = x) = (x n) p^{x} q^{n - x}, x = 0, 1, 2, \dots, n$

where:

$(x n) = \frac{n !}{x ! ( n - x )!}$ is the binomial coefficient — the number of ways to choose $x$ successes from $n$ trials.
$p^{x} q^{n - x}$ is the probability of any single sequence with exactly $x$ successes.

Why the formula makes sense: To get exactly

x

successes in

n

trials, you need:

Choose which $x$ trials are successes — $(x n)$ ways.
Each of those $x$ trials succeeds with probability $p$ — contributes $p^{x}$ .
The remaining $n - x$ trials fail with probability $q$ — contributes $q^{n - x}$ .

Mean:

μ = n p

Variance: $σ^{2} = n pq$

Standard deviation: $σ = n pq$

Two critical conditions for a binomial distribution:

Independence: The outcome of any trial does not affect any other trial.
Constant probability: The probability of success $p$ stays the same for every trial.

Q: Does

p + q = 1

alone mean it is binomial?

A: No. You also need to know the value of $n$ , that $p$ stays constant throughout, and that the trials are independent. For example, getting a right answer on an exam question should not depend on whether you got the previous question right.

8.1.3 Worked Example — Multiple-Choice Guessing

Problem: Each of 5 questions on a multiple-choice exam has 4 choices, only one of which is correct. A student attempts to guess the answers. The random variable

X

is the number of questions answered correctly. Find:

$P (X = 3)$ — probability of exactly 3 correct
$P (X \leq 3)$ — probability of at most 3 correct
$P (X \geq 1)$ — probability of at least 1 correct
Mean and variance

Step 1 — Identify the distribution:

$n = 5$ $5 q u es t i o n s$
$p = \frac{1}{4} = 0.25$ $o n ecor r ec t c h o i ceo u t o f f o u r$
$q = 1 - p = \frac{3}{4} = 0.75$
Trials are independent and $p$ stays constant — this is binomial.

Step 2 — Compute $P (X = 3)$ :

$P (X = 3) = (3 5) (\frac{1}{4})^{3} (\frac{3}{4})^{5 - 3} = 10 \cdot \frac{1}{64} \cdot \frac{9}{16} = \frac{90}{1024} \approx 0.0879$

Step 3 — Compute $P (X \leq 3)$ :

Complement approach $e a s i er$ : $P (X \leq 3) = 1 - P (X = 4) - P (X = 5)$

$P (X = 4) = (4 5) (\frac{1}{4})^{4} (\frac{3}{4})^{1} = 5 \cdot \frac{1}{256} \cdot \frac{3}{4} = \frac{15}{1024} \approx 0.0146$

$P (X = 5) = (5 5) (\frac{1}{4})^{5} (\frac{3}{4})^{0} = 1 \cdot \frac{1}{1024} \cdot 1 = \frac{1}{1024} \approx 0.0010$

$P (X \leq 3) = 1 - 0.0146 - 0.0010 = 0.9844$

Step 4 — Compute $P (X \geq 1)$ :

$P (X \geq 1) = 1 - P (X = 0) = 1 - (0 5) (\frac{1}{4})^{0} (\frac{3}{4})^{5} = 1 - 1 \cdot 1 \cdot \frac{243}{1024} = 1 - 0.2373 = 0.7627$

Step 5 — Mean and variance:

$μ = n p = 5 \cdot 0.25 = 1.25$

$σ^{2} = n pq = 5 \cdot 0.25 \cdot 0.75 = \frac{15}{16} = 0.9375$

Sense-check: With 5 questions and a 25% chance each, expecting 1.25 correct answers makes sense. The student will most likely get 0, 1, or 2 right. This is consistent with the high probability $0.9844$ of getting at most 3 right.

Visual intuition: Picture a bar chart. The horizontal axis runs from $x = 0$ to $x = 5$ $t h e p oss ib l e n u mb er o f cor r ec t an s w er s$ . The vertical axis shows probability. The tallest bar sits at $x = 1$ — the most likely outcome is getting exactly one question right. The bars taper off toward both ends: $x = 0$ has moderate height, $x = 5$ is nearly invisible. The distribution is skewed right because $p = 0.25$ is below 0.5 — the longer tail stretches toward higher values. With $p = 0.5$ , the bars would be perfectly symmetric around the center.

Assumptions & Scope: The binomial distribution applies ONLY when:

The number of trials $n$ is fixed in advance.
Each trial has exactly two outcomes $s u ccess / f ai l u r e$ .
Trials are independent — the outcome of one trial gives no information about another.
The success probability $p$ is constant across all trials.

If the trials are not independent

e . g ., d r a w in g c a r d s w i t h o u t r e pl a ce m e n t f r o ma d ec k

, the binomial model gives wrong probabilities. When sampling without replacement from a finite population, use the hypergeometric distribution instead. If

p

drifts over time

e . g ., a l e a r nin g e f f ec t inan e x p er im e n t

, the binomial model is inappropriate.

Pitfalls — common beginner traps:

Confusing $p$ and $q$ : Always double-check which is which. If the problem says "30% are defective," then $p = 0.30$ is the probability of a defective $s u ccess in t ha t co n t e x t$ , and $q = 0.70$ is the probability of non-defective.

Forgetting the binomial coefficient: $P (X = 3)$ is NOT just $p^{3} q^{n - 3}$ . You must multiply by $(3 n)$ because there are many different sequences that yield exactly 3 successes.

Using the complement the hard way: For $P (X \leq 3)$ with $n = 5$ , computing $1 - P (X = 4) - P (X = 5)$ is faster than adding $P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)$ . Always check which tail has fewer terms.

Mixing up variance and standard deviation: The variance is $n pq$ . The standard deviation is $n pq$ . If a problem asks for $σ$ and you give $n pq$ , you lose the mark.

Q: Several students asked — what exactly does "independent" mean in the binomial context?

A: Independence means knowing the outcome of one trial tells you nothing about any other trial. If you flip a coin and get heads, the next flip is still 50-50. But if you draw cards from a deck without replacement, the draws are NOT independent — removing an ace changes the probability of the next card being an ace. That is why card-drawing problems often require the hypergeometric distribution, not the binomial.

Recap: Bernoulli models one yes/no trial. Binomial counts successes across

n

independent, identical Bernoulli trials. The formula

P (X = x) = (x n) p^{x} q^{n - x}

multiplies the number of ways

((x n))

by the probability of one way

(p^{x} q^{n - x})

Bridge: But what if $n$ is huge and $p$ is tiny — like defects in a factory producing thousands of units? Computing $(2 200)$ by hand is painful. The Poisson distribution, coming next, handles exactly that scenario by letting $n \to \infty$ and $p \to 0$ .

Real-world connection: Binomial models appear everywhere in quality control. A factory samples $n = 50$ units from each batch and counts how many are defective. Suppose the process is in control with $p = 0.02$ . The binomial distribution tells the quality engineer the probability of finding 3 or more defectives in a sample. If that probability is tiny and yet 3 defectives are observed, the process may be out of control. This is the foundation of statistical process control $S P C$ , used in manufacturing from automotive to pharmaceuticals.

In medicine, binomial models power clinical trial design. If a new drug has a true response rate $p$ , the binomial distribution predicts how many patients out of $n$ will respond. Researchers use this to calculate the sample size needed to detect a meaningful effect.

8.1.4 Symbol Registry — Binomial Distribution

Symbol	Meaning	LaTeX	Type / Domain
$n$	number of trials	$n$	integer, $n \geq 1$
$p$	probability of success per trial	$p$	scalar, $p \in [0, 1]$
$q$	probability of failure per trial, $q = 1 - p$	$q$	scalar, $q \in [0, 1]$
$X$	number of successes in $n$ trials	$X$	integer, $X \in {0, 1, \dots, n}$
$(x n)$	binomial coefficient, "n choose x"	$(x n)$	integer

8.2 Poisson Distribution

Hook: A call center receives about 10 calls per minute. What is the probability that exactly 3 calls arrive in the next minute? You cannot count the number of "non-calls" — there is no fixed

n

and no meaningful notion of a "failure" trial. Yet you still need a probability model for counting rare, random events. The Poisson distribution was invented exactly for this: modeling how many times something happens in a fixed window of time or space when events occur at a constant average rate.

Intuition: Imagine standing by a highway at night, counting cars that pass in the next hour. Each car arrival is random and independent of others. You do not know how many cars did not pass — only how many did. The Poisson distribution is the go-to model for these "count of random arrivals" problems.

Think of raindrops falling on a square tile. If rain falls uniformly at a constant average rate, the number of drops hitting a tile in one minute follows a Poisson distribution. The key insight: the Poisson emerges naturally when you take a binomial process with huge $n$ and tiny $p$ , and let $n \to \infty$ while keeping the expected count $λ = n p$ fixed.

Where the analogy breaks: The Poisson assumes events happen independently at a constant average rate. If rain comes in bursts $c l u s t er e d$ , or if cars travel in convoys, the independence assumption is violated — and the Poisson model underestimates the chance of extreme counts.

8.2.1 Definition and Formula

A Poisson distribution models the count of events occurring in a fixed interval of time or space, when events happen independently at a constant average rate.

Probability mass function:

$P (X = x) = \frac{e ^{- λ} λ ^{x}}{x !}, x = 0, 1, 2, \dots$

where:

$λ$ $l amb d a$ is the rate parameter — the average number of occurrences in the interval.
$e \approx 2.71828$ is Euler's number.
$x!$ $x f a c t or ia l$ is the product $1 \times 2 \times \dots \times x$ , with $0! = 1$ .

Key property — mean equals variance:

$μ = λ, σ^{2} = λ$

This is a distinctive fingerprint of the Poisson distribution. If you compute the sample mean and sample variance from data and they are far apart, the Poisson model is unlikely to be a good fit.

How Poisson emerges from binomial: Start with a binomial distribution $X \sim Binomial (n, p)$ . Let $n \to \infty$ and $p \to 0$ such that $λ = n p$ stays constant. Then:

$n \to \infty, p \to 0 lim (x n) p^{x} (1 - p)^{n - x} = \frac{e ^{- λ} λ ^{x}}{x !}$

The Poisson is the limiting case of the binomial when events are rare but opportunities are abundant.

8.2.2 When to Use Poisson Instead of Binomial

Two indicators point toward Poisson:

No natural upper bound: $X$ takes values $0, 1, 2, \dots$ without a finite cap $u n l ik e bin o mia l, w h er e (X \leq n)$ .
Large $n$ , small $p$ : The number of trials is large and the per-trial success probability is small.

Thumb rule: Use Poisson when

n \geq 100

and

n p \leq 10

. When these hold, binomial and Poisson probabilities are nearly identical — the Poisson saves you from computing enormous binomial coefficients.

Q: If we solve the same problem with binomial, do we get a different answer?

A: About the same. For large $n$ and small $p$ , binomial and Poisson give nearly identical results. The Poisson is an approximation that sacrifices a tiny bit of accuracy for enormous computational convenience. With $n = 200$ and $p = 0.03$ , the binomial requires computing $(x 200)$ for each term — the Poisson needs only powers and factorials.

8.2.3 Worked Example — Defective Units

Problem: 3% of electronic units manufactured by a company are defective. In a sample of 200 units, find the probability that less than 2 bulbs are defective.

Step 1 — Identify parameters:

$n = 200$ $l a r g e$
$p = 0.03$ $s ma l l$
$λ = n p = 200 \times 0.03 = 6$

Check thumb rule:

n = 200 \geq 100

✓ and

n p = 6 \leq 10

✓. Poisson is appropriate.

Step 2 — Compute $P (X < 2)$ :

$P (X < 2) = P (X = 0) + P (X = 1)$

$P (X = 0) = \frac{e ^{- 6} \cdot 6 ^{0}}{0 !} = e^{- 6}$

$P (X = 1) = \frac{e ^{- 6} \cdot 6 ^{1}}{1 !} = 6 e^{- 6}$

$P (X < 2) = e^{- 6} + 6 e^{- 6} = 7 e^{- 6}$

Using $e^{- 6} \approx 0.002479$ :

$P (X < 2) = 7 \times 0.002479 \approx 0.0174$

Sense-check: With an average of 6 defectives expected in a sample of 200, finding fewer than 2 is quite rare — about 1.7% chance. This makes sense: if the defect rate truly is 3%, seeing 0 or 1 defectives in 200 units would be an unusually clean batch.

Visual intuition: Picture a bar chart for the Poisson distribution with $λ = 6$ . The horizontal axis shows $x = 0, 1, 2, \dots$ $n u mb er o f d e f ec t i v es$ . The vertical axis shows probability. The bars rise from near-zero at $x = 0$ , climb to a peak around $x = 5$ or $x = 6$ $s in ce (λ = 6) i s t h e m e an$ , then gradually taper off toward zero for larger $x$ . The distribution is slightly right-skewed for $λ = 6$ ; as $λ$ grows larger, the Poisson shape approaches a symmetric bell curve — previewing the normal approximation we will see later.

Assumptions & Scope: The Poisson model assumes:

Events occur independently — one occurrence does not make another more or less likely.
The average rate $λ$ is constant over the interval.
Two events cannot occur at exactly the same instant $r a r e in p r a c t i ce f or co n t in u o u s - t im e p r ocesses$ .

When independence fails

e . g ., co n t a g i o u s d i se a sec a sesc l u s t er

, the Poisson underestimates the probability of extreme counts. When the rate changes over time

e . g ., c a l l v o l u m e v a r i es b y h o u r

, you need a non-homogeneous Poisson process — the simple Poisson with constant

λ

is wrong.

Pitfalls — common beginner traps:

Forgetting $x!$ in the denominator: The Poisson formula has $x!$ below, not above. Writing $e ^{- λ} λ ^{x} \cdot x !$ is a common mistake.

Confusing the interval: If $λ = 5$ per hour and the question asks about a 30-minute window, you must adjust: $λ = 2.5$ for half an hour. The Poisson parameter scales linearly with the interval length.

Using Poisson when binomial is more accurate: If $n$ is small and $p$ is not extreme, use binomial directly. Poisson is an approximation — valuable when $n$ is large, but unnecessary for $n = 10$ .

Misreading "less than" vs. "less than or equal": In discrete distributions, these differ. $P (X < 2) = P (X = 0) + P (X = 1)$ . $P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2)$ . Always check the inequality sign.

Recap: The Poisson distribution models counts of rare, independent events in a fixed interval. It needs only one parameter

λ

— the mean, which also equals the variance. The formula

P (X = x) = \frac{e ^{- λ} λ ^{x}}{x !}

is the limiting case of the binomial when

n \to \infty

and

p \to 0

Bridge: Both binomial and Poisson are discrete distributions — they assign probability to individual integers. But many real-world quantities $h e i g h t s, t es t scor es, m e a s u r e m e n t er r or s$ are continuous. The normal distribution, coming next, handles continuous variables and is the single most important distribution in all of statistics.

Real-world connection: The Poisson distribution was originally developed by Siméon Denis Poisson in 1837 to model the number of wrongful convictions in French courts. Today, it powers queueing theory — call centers use it to predict incoming call volumes and staff appropriately. In traffic engineering, Poisson models estimate the number of cars arriving at an intersection per minute, which determines traffic light timing. In biology, the number of mutations in a strand of DNA per unit length follows a Poisson distribution. In finance, Poisson processes model the arrival of extreme market events $j u m p s$ in options pricing. Any field that counts random arrivals — customers, photons, earthquakes, website clicks — uses the Poisson distribution.

8.2.4 Symbol Registry — Poisson Distribution

Symbol	Meaning	LaTeX	Type / Domain
$λ$	Poisson parameter = mean = variance	$λ$	scalar, $λ > 0$
$e$	Euler's number ≈ 2.71828	$e$	constant
$X$	count of occurrences	$X$	integer, $X \in {0, 1, 2, \dots}$

8.3 Normal Distribution $G a u ss ian$

Hook: Measure the heights of 1000 randomly chosen adult men. Plot them as a histogram. The shape is unmistakable — a smooth, symmetric bell that peaks in the middle and tapers at the edges. This shape appears so often in nature that it earned the name "normal." But here is the surprising part: even when individual data are NOT bell-shaped, averages of data almost always are. This single fact — the Central Limit Theorem — makes the normal distribution the backbone of all statistical inference.

Intuition: Think of a skilled archer shooting arrows at a bullseye. Most arrows land near the center. A few drift slightly left or right. Very few miss by a lot. If you plot where the arrows hit along a horizontal line, the density forms a bell shape.

Imagine dropping a handful of sand onto a flat surface. The pile builds highest at the drop point and spreads out symmetrically in all directions. The cross-section of that pile is a bell curve. Many small independent influences — tiny gusts of wind, slight hand tremors, minute variations in arrow weight — add up to produce the bell shape. This "many small causes add up" mechanism is why the normal distribution appears everywhere: measurement errors, biological traits, test scores, and financial returns.

Where the analogy breaks: The sand pile is finite — it has edges. The normal curve extends infinitely in both directions, never quite touching zero. Real heights and weights cannot be negative, so the normal distribution is always an approximation for physical quantities, but an extraordinarily good one.

8.3.1 Definition

A normal distribution (also called Gaussian distribution) is a continuous probability distribution defined over all real numbers. A random variable

X

following a normal distribution is written:

$X \sim N (μ, σ^{2})$

Probability density function (PDF):

$f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}, x \in (- \infty, \infty)$

where:

$μ$ $m u$ is the mean — the center of the curve, where the peak sits.
$σ$ $s i g ma$ is the standard deviation — controls how spread out the curve is.
$π \approx 3.14159$ , $e \approx 2.71828$ .

Do not memorize the PDF. The formula looks imposing, but you will never need to plug numbers into it directly. The key insight: the entire infinite family of normal curves is determined by just two numbers —

μ

and

σ

. Change

μ

, and the curve slides left or right. Change

σ

, and the curve stretches wide or squeezes narrow.

The curve is called the normal curve, Gaussian curve, or most commonly, the bell curve. Different communities use different names: communication engineers say "Gaussian," management professionals say "bell curve."

8.3.2 Properties of the Normal Curve

Bell-shaped — symmetric, unimodal shape. One peak, no bumps.

Symmetric about the mean — the left half mirrors the right half exactly. For any symmetric distribution: mean = mode = median. All three measures of center coincide at the same point.

Total area under the curve equals 1 — because it is a probability distribution and total probability must equal 1. The area between any two x-values gives the probability of the random variable falling in that range.

Asymptotic — the curve approaches but never touches the x-axis. The tails extend to $- \infty$ and $+ \infty$ . In theory, a height of 10 meters or -2 meters has nonzero probability $v ani s hin g l y s ma l l, b u t n o t z er o$ .

The Empirical Rule $68 - 95 - 99.7$ :

- About 68.3% of values fall within

μ \pm σ

o n es t an d a r dd e v ia t i o n

. - About 95.4% of values fall within

μ \pm 2 σ

tw os t an d a r dd e v ia t i o n s

. - About 99.7% of values fall within

μ \pm 3 σ

t h r ees t an d a r dd e v ia t i o n s

This rule gives quick mental estimates without a table. If test scores have $μ = 75$ and $σ = 5$ , roughly 68% of students scored between 70 and 80, and roughly 95% scored between 65 and 85.

8.3.3 The Integration Problem

When

X

follows a normal distribution, finding probabilities requires integration:

$P (x_{1} < X \leq x_{2}) = \int_{x_{1}}^{x_{2}} \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}} d x$

This integral has no closed-form solution. The function $e^{- x^{2}}$ does not have an elementary antiderivative — you cannot solve it with basic calculus techniques $n o u - s u b s t i t u t i o n, n o in t e g r a t i o nb y p a r t s w or k s$ . This is a proven mathematical fact, not a shortcoming of your calculus skills. Numerical methods or pre-computed tables are the only practical way to evaluate normal probabilities.

Visual intuition: Picture the bell curve. The horizontal axis $(x)$ shows the values the random variable can take — test scores, heights, measurement errors. The vertical axis $(f (x))$ shows probability density — NOT probability itself, but how "concentrated" the probability is at each $x$ . The peak at $x = μ$ means values near the mean are most likely. The curve falls off symmetrically on both sides. The inflection points — where the curve changes from "scooping downward" to "leveling out" — sit exactly at $μ \pm σ$ . Mark $μ - σ$ , $μ$ , and $μ + σ$ on the axis: the middle 68% of probability lives between those two outer marks. The takeaway: the normal curve is the idealization of "most things cluster around average, with predictable scatter."

Recap: The normal distribution is the continuous, symmetric bell curve defined by two parameters — mean

μ

ce n t er

and standard deviation

σ

s p r e a d

. Its PDF looks scary, but you will never integrate it by hand. The empirical rule

68 - 95 - 99.7

gives quick probability estimates.

Bridge: Since we cannot integrate the normal PDF, how do we actually compute probabilities? The answer: we convert every normal problem to a single standardized version — the standard normal distribution — and use a pre-computed table. That is the topic of the next section.

Real-world connection: The normal distribution is the foundation of statistical inference. When pollsters report "margin of error ±3%," they rely on the fact that sample proportions are about normally distributed. When manufacturers set tolerance limits on parts, they use the normal distribution to predict what fraction will fall outside specs. In finance, the Black-Scholes options pricing model assumes stock returns are normally distributed. In machine learning, linear regression and many other models assume normally distributed errors. The normal distribution even underpins IQ scoring: IQ scores are deliberately normalized to have $μ = 100$ and $σ = 15$ , so exactly 68% of people score between 85 and 115.

8.4 Standard Normal Distribution

Hook: Every normal distribution has a different center and spread. If test scores have

μ = 75

and

σ = 5

, how do you find the probability of scoring above 82? The integral is unsolvable. The brilliant trick: convert your problem to a single universal scale where every normal curve becomes identical. Then use one pre-computed table for everything. This is the Z-transformation — the mathematical equivalent of converting all currencies to dollars so you can compare prices worldwide.

Intuition: Imagine you measure height in centimeters and your friend measures in inches. Comparing raw numbers is meaningless. But if you both report "how many standard deviations above or below average" you are, the numbers become directly comparable. A Z-score of

+ 1.5

always means "1.5 standard deviations above the mean" — regardless of the original units.

Think of standardization as putting every normal distribution through a universal translator. You feed in any $μ$ and $σ$ ; the Z-formula subtracts the mean $ce n t er in g$ and divides by the standard deviation $r esc a l in g$ . The output always has mean 0 and standard deviation 1. One table now serves all problems.

8.4.1 The Z-Transformation

To convert any normal random variable

X \sim N (μ, σ^{2})

to the standard normal

Z \sim N (0, 1)

, use:

$Z = \frac{X - μ}{σ}$

What this transformation does:

Centering: Subtracting $μ$ shifts the distribution so the new mean is 0.
Rescaling: Dividing by $σ$ compresses or stretches the spread so the new standard deviation is 1.

The standard normal PDF

a f t er t r an s f or ma t i o n

$f (z) = \frac{1}{2 π} e^{- \frac{z ^{2}}{2}}$

This is the normal PDF with $μ = 0$ and $σ = 1$ plugged in — simpler than the general form, but still not integrable by hand.

Why the transformation works — the calculus:

Start with the normal PDF: $f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}$

Substitute $Z = \frac{X - μ}{σ}$ , so $X - μ = σ Z$ and $d x = σ d z$ .

Inside the integral for probability:

$\int_{x_{1}}^{x_{2}} \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}} d x = \int_{z_{1}}^{z_{2}} \frac{1}{σ 2 π} e^{- \frac{z ^{2}}{2}} \cdot σ d z = \int_{z_{1}}^{z_{2}} \frac{1}{2 π} e^{- \frac{z ^{2}}{2}} d z$

The $σ$ from $d x$ cancels with the $\frac{1}{σ}$ outside the exponential. The result uses only the standard normal density — no $μ$ , no $σ$ .

Q: Where does the

\frac{1}{σ}

go during substitution?

A: When we substitute $X - μ = σ Z$ , the differential becomes $d x = σ d z$ . The $σ$ from the differential cancels with the $\frac{1}{σ}$ outside the exponential in the original PDF. Net result: the standard normal PDF has no $σ$ factor. This cancellation is why the standard normal table is universal — it works regardless of the original $μ$ and $σ$ .

8.4.2 Two Benefits of Standardization

Compute probabilities without integration. Values are read from a pre-computed standard normal table. No calculus needed.

Compare any datasets on the same scale. After standardizing, every variable has mean 0 and standard deviation 1. A Z-score of $+ 2.0$ means the same thing whether the original variable was height in cm, test scores, or income in dollars — it is always "two standard deviations above average."

Real-world application: In machine learning, this is called standard scaling or Z-score normalization. When preprocessing data for algorithms like support vector machines, K-means clustering, or neural networks, converting features to Z-scores ensures no feature dominates simply because of its scale. A feature measured in thousands $l ik e in co m e$ would otherwise overwhelm a feature measured in single digits $l ik e a g e$ . Standard scaling is used alongside min-max scaling $w hi c hma p se v er y t hin g t o ([0, 1])$ — the choice depends on the algorithm and whether outliers are a concern.

8.4.3 The Standard Normal Table

The standard normal table gives

F (z) = P (Z \leq z)

— the area under the standard normal curve from

- \infty

up to

z

. This is called the cumulative distribution function

C D F

How to read the table: The Z-value is split into two parts:

The row gives the first decimal place: the integer part and the tenths $e . g ., 1.7$ .
The column gives the second decimal place: the hundredths $e . g ., 0.05$ .

Example — finding $F (1.75)$ :

Find row 1.7.
Find column 0.05.
The intersection gives $F (1.75) = 0.9599$ .

This means the probability that

Z \leq 1.75

is 0.9599

ab o u t 95.99

. In plain words: about 96% of all values in any normal distribution fall less than 1.75 standard deviations above the mean.

The table is fixed and universal. Since the standard normal distribution always has $μ = 0$ and $σ = 1$ , the same table works for every problem. You only need to convert your $X$ to $Z$ first.

Q: Does the table change based on mean and variance?

A: No. It is a standard table that never changes. After standardizing, the mean is always 0 and the spread is always 1, so one fixed table works for everything. The table was computed once — by numerical integration — and printed in every statistics textbook since.

Visual intuition: Picture the standard normal bell curve centered at 0. The horizontal axis shows Z-values like $- 3, - 2, - 1, 0, 1, 2, 3$ . The vertical axis shows density. Now shade the area from $- \infty$ up to some Z-value, say $z = 1.75$ . The shaded area covers most of the left half plus a big chunk of the right half — that is the 0.9599 from the table. The unshaded right tail $t h e a r e ab ey o n d 1.75$ is $1 - 0.9599 = 0.0401$ — about 4%. The table is essentially a lookup of "how much area is to the left of this Z-value."

Pitfalls — common beginner traps:

Confusing $X$ and $Z$ : You MUST convert $X$ to $Z$ before using the table. Looking up $X = 40$ directly in the Z-table gives nonsense. Always compute $Z = \frac{X - μ}{σ}$ first.

Using variance instead of standard deviation: If the problem gives $σ^{2} = 25$ , then $σ = 5$ . Plugging 25 into the Z-formula as $σ$ is a common and costly mistake.

Reading the table backwards: The table gives $P (Z \leq z)$ . If you need $P (Z > z)$ , compute $1 - F (z)$ . Always check whether your answer should be more or less than 0.5 — if $z > 0$ and your answer is below 0.5, you probably forgot to subtract from 1.

Forgetting that the table typically only shows positive Z: For negative Z-values, use the symmetry property: $F (- z) = 1 - F (z)$ . Do not try to look up $- 1.5$ directly in a positive-only table.

Recap: The Z-transformation

Z = \frac{X - μ}{σ}

converts any normal distribution to the standard normal

N (0, 1)

. The standard normal table gives cumulative probabilities

F (z) = P (Z \leq z)

. Convert, look up, adjust — three steps that replace impossible integrals with simple arithmetic.

Bridge: Now that we have the table and the three fundamental cases $l ess t han, g r e a t er t han, b e tw ee n$ , let us work through a series of examples to build fluency. The next section is a problem-solving drill.

8.4.4 Symbol Registry — Standard Normal

Symbol	Meaning	LaTeX	Type / Domain
$X$	original random variable	$X$	continuous, $X \in (- \infty, \infty)$
$μ$	population mean	$μ$	scalar
$σ$	population standard deviation	$σ$	scalar, $σ > 0$
$Z$	standard normal variable	$Z$	continuous, $Z \in (- \infty, \infty)$
$F (z)$	cumulative probability $P (Z \leq z)$	$F (z)$	scalar, $F (z) \in [0, 1]$

8.5 Using the Standard Normal Table — Worked Examples

Hook: You have the Z-formula. You have the table. Now the real skill: looking at a word problem and instantly knowing which of three operations to apply. Every normal probability question — no matter how it is phrased — reduces to exactly one of: "less than," "greater than," or "between." Master these three patterns and you can solve any normal distribution problem in under a minute.

8.5.1 Three Fundamental Cases

When working with the standard normal table, every probability problem reduces to three cases:

Case	Expression	Rule	Mental picture
Less than $z$	$P (Z < z)$	$F (z)$ — read directly from table	Shade everything left of $z$
Greater than $z$	$P (Z > z)$	$1 - F (z)$	Shade everything right of $z$
Between $z_{1}$ and $z_{2}$	$P (z_{1} < Z < z_{2})$	$F (z_{2}) - F (z_{1})$	Shade the strip between them

The table always gives the left-tail area. Everything else is just subtraction from 1 or from another table value.

8.5.2 Example 1 — Less Than 1.75

Find $P (Z < 1.75)$ .

This is the "less than" case — read $F (1.75)$ directly from the table.

From the table: row 1.7, column 0.05 → $F (1.75) = 0.9599$ .

Sense-check: $Z = 1.75$ is positive, so more than half the area should be to its left. 0.9599 > 0.5 ✓.

8.5.3 Example 2 — Greater Than 2.06

Find $P (Z > 2.06)$ .

From the table: $F (2.06) = 0.9803$ .

The table gives the area from $- \infty$ to 2.06. We want the right tail beyond 2.06:

$P (Z > 2.06) = 1 - F (2.06) = 1 - 0.9803 = 0.0197$

Sense-check: $Z = 2.06$ is far to the right. The right tail should be small. 0.0197 ≈ 2% ✓.

8.5.4 Example 3 — Scores Less Than 40

Problem: Scores on a test are normally distributed with

μ = 30

and

σ = 4

. Find the probability of a student scoring less than 40.

Given: $X \sim N (30, 4^{2})$ , find $P (X < 40)$ .

Step 1 — Convert to Z: $Z = \frac{X - μ}{σ} = \frac{40 - 30}{4} = \frac{10}{4} = 2.5$

So $P (X < 40) = P (Z < 2.5)$ .

Step 2 — Read from table: Row 2.5, column 0.00 → $F (2.5) = 0.9938$ .

Answer: $P (X < 40) = 0.9938$ $ab o u t 99.38$ .

Sense-check: 40 is 2.5 standard deviations above the mean of 30. Almost everything falls below 2.5σ above the mean — 99.38% is consistent with the empirical rule $99.7$ .

8.5.5 Example 4 — Scores Greater Than 38

Problem: Same distribution,

μ = 30

σ = 4

. Find

P (X > 38)

Step 1 — Convert to Z: $Z = \frac{38 - 30}{4} = \frac{8}{4} = 2.0$

Step 2 — Greater than case: $P (X > 38) = P (Z > 2.0) = 1 - F (2.0)$ From the table: $F (2.0) = 0.9772$ .

Answer: $P (X > 38) = 1 - 0.9772 = 0.0228$ $ab o u t 2.28$ .

Sense-check: 38 is exactly 2σ above the mean. The empirical rule says about 95% of values fall within ±2σ, so about 5% fall outside. Half of that $2.5$ is above +2σ. Our answer 2.28% is very close ✓.

8.5.6 Example 5 — Score Between 70 and 80

Problem: Test scores are normally distributed with

μ = 75

and

σ = 4

. Find the probability a student scores between 70 and 80.

Step 1 — Convert both endpoints to Z: $Z_{lower} = \frac{70 - 75}{4} = \frac{- 5}{4} = - 1.25$ $Z_{upper} = \frac{80 - 75}{4} = \frac{5}{4} = 1.25$

So $P (70 < X < 80) = P (- 1.25 < Z < 1.25)$ .

Step 2 — Between case: $P (- 1.25 < Z < 1.25) = F (1.25) - F (- 1.25)$

Step 3 — Handle negative Z via symmetry: The standard normal table typically only has positive Z values. For negative Z, use:

$F (- z) = 1 - F (z)$

This works because the normal curve is symmetric. The area to the left of $- z$ equals the area to the right of $+ z$ .

So: $F (- 1.25) = 1 - F (1.25)$ .

$P (- 1.25 < Z < 1.25) = F (1.25) - [1 - F (1.25)$ = 2F $1.25$ - 1 \]

From the table: $F (1.25) = 0.8944$ .

Answer: $P (70 < X < 80) = 2 (0.8944) - 1 = 1.7888 - 1 = 0.7888$ $ab o u t 78.88$ .

Sense-check: 70 and 80 are ±1.25σ from the mean. The empirical rule says about 68% within ±1σ. Since ±1.25σ is wider, the answer should exceed 68%. 78.88% sits between 68% and 95% ✓.

8.5.7 Example 6 — Between 26 and 40 with $μ = 30, σ = 5$

Problem:

X

is normal with

μ = 30

and

σ = 5

. Find

P (26 \leq X \leq 40)

Caution: The problem gives $σ = 5$ directly. If it gave variance = 25, you would need $σ = 25 = 5$ . Always check whether you have $σ$ or $σ^{2}$ .

Step 1 — Convert to Z: $Z_{1} = \frac{26 - 30}{5} = \frac{- 4}{5} = - 0.8$ $Z_{2} = \frac{40 - 30}{5} = \frac{10}{5} = 2.0$

Step 2 — Between case with negative Z: $P (- 0.8 \leq Z \leq 2.0) = F (2.0) - F (- 0.8)$ $F (- 0.8) = 1 - F (0.8)$ From table: $F (2.0) = 0.9772$ , $F (0.8) = 0.7881$ .

$P = 0.9772 - (1 - 0.7881) = 0.9772 - 0.2119 = 0.7653$

Note on $<$ vs. $\leq$ : When the variable is continuous, strict inequality $(<)$ and non-strict inequality $(\leq)$ give the same probability. The probability at a single point is zero, so including or excluding the endpoint does not change the result.

Sense-check: The interval spans from 0.8σ below to 2.0σ above the mean. Most of the probability mass is captured — 76.53% is plausible ✓.

8.5.8 Example 7 — Greater Than 45 with $μ = 30, σ = 5$

Problem:

X \sim N (30, 5^{2})

. Find

P (X > 45)

Step 1: $Z = \frac{45 - 30}{5} = \frac{15}{5} = 3.0$

Step 2: From table: $F (3.0) = 0.9986$ .

$P (X > 45) = 1 - F (3.0) = 1 - 0.9986 = 0.0014$

Sense-check: 45 is 3σ above the mean. The empirical rule says 99.7% within ±3σ, so only 0.15% $0.0015$ lies above +3σ. Our answer 0.0014 matches ✓.

8.5.9 Example 8 — "At Least" and "Greater Than"

Problem: A distribution is normal. Find the probability that

X

is at least a certain value.

"At least" means $P (X \geq value)$ . Convert to Z, then use $1 - F (Z)$ — identical to "greater than."

In a continuous distribution, $P (X \geq k) = P (X > k)$ because $P (X = k) = 0$ . The distinction between $\geq$ and $>$ matters only for discrete distributions like binomial and Poisson — for normal, they are the same.

8.5.10 Student Question — Handling Negative Z

Q: When computing

P (- 0.8 \leq Z \leq 2.0)

, how do we handle the negative Z?

A: Two approaches, depending on your table:

If your table includes negative Z values: read $F (- 0.8)$ directly.
If your table only has positive values: use the symmetry formula $F (- z) = 1 - F (z)$ . So $F (- 0.8) = 1 - F (0.8) = 1 - 0.7881 = 0.2119$ .

The symmetry property works because the standard normal curve is symmetric about zero. Draw the bell curve: the area left of

- z

is the mirror image of the area right of

+ z

. Since the total area is 1, the area left of

- z

must equal

1 - F (z)

8.5.11 Summary of Standard Normal Table Rules

Problem type	Expression	Rule
Less than $z$	$P (Z < z)$	$F (z)$ — read directly from table
Greater than $z$	$P (Z > z)$	$1 - F (z)$
Between $z_{1}$ and $z_{2}$	$P (z_{1} < Z < z_{2})$	$F (z_{2}) - F (z_{1})$
Negative Z $p os i t i v e - o n l y t ab l e$	$F (- z)$	$1 - F (z)$

Bridge: The normal distribution is powerful on its own, but its real magic is that it can approximate other distributions. When

n

is large, the binomial distribution — which is discrete and requires summing many terms — can be replaced by a single normal table lookup. That is the normal approximation to the binomial, coming next.

Visual intuition for all cases: Draw the standard normal bell curve with 0 at center. For "less than $z = 1.5$ ": shade from the far left tail all the way to 1.5 — a big area, about 0.93. For "greater than $z = 1.5$ ": shade only the tiny right tail beyond 1.5 — about 0.07. For "between $- 1$ and $1$ ": shade the central stripe — about 0.68 by the empirical rule. The three cases correspond to left tail, right tail, and central stripe. The table always gives the left tail; everything else is subtraction.

8.6 Normal Approximation to the Binomial Distribution

Hook: You are asked: "A factory produces 500 units. Each has a 31% chance of being defective. What is the probability that fewer than 140 are defective?" Using the binomial formula directly means summing 140 terms, each with

(x 500)

— numbers so large they overflow most calculators. But the normal distribution can answer this in 30 seconds with a single table lookup. The bridge between discrete counting and continuous bell curves is the normal approximation to the binomial.

Intuition: Imagine a binomial bar chart for

n = 500

p = 0.31

. The bars form a shape. As

n

grows, that shape looks more and more like a bell curve. The normal approximation simply replaces the staircase of bars with a smooth curve.

Think of pixel art: from far away, a detailed pixel image looks smooth. The binomial is the pixel art — discrete bars, each an integer wide. The normal curve is the smooth photograph you see from a distance. The continuity correction $t h e \pm 0.5 a d j u s t m e n t$ accounts for the fact that each "pixel" $in t e g er$ spans from $x - 0.5$ to $x + 0.5$ in the continuous world.

8.6.1 Motivation — Why Approximate?

Consider a binomial problem:

n = 500

p = 0.31

. To find

P (X < 250)

using the binomial formula directly:

$P (X < 250) = x = 0 \sum 249 (x 500) (0.31)^{x} (0.69)^{500 - x}$

This requires computing 250 terms — each with a binomial coefficient involving factorials of 500. This is nearly impossible by hand and numerically unstable even for computers without careful algorithms.

By contrast, using the normal distribution: convert to $Z = \frac{250 - μ}{σ}$ and read one value from the table. The same amount of work regardless of whether $n = 50$ or $n = 5000$ .

If someone allows us to use the normal distribution even when the problem is binomial, it is a huge simplification.

8.6.2 How Normal Approximation Works

Even though the original problem is discrete and binomial, we can use the continuous normal distribution by:

Computing the binomial mean: $μ = n p$
Computing the binomial standard deviation: $σ = n pq$
Converting to Z: $Z = \frac{X - n p}{n pq}$
Using the standard normal table

This bridges discrete distributions

bin o mia l

and continuous distributions

n or ma l

8.6.3 When to Apply Normal Approximation

Condition $p r o f esso r^{'} sr u l e$ : Use normal approximation when BOTH:

$n p \geq 15$
$n q \geq 15$

If either condition fails, the probabilities from the normal approximation will differ significantly from the exact binomial probabilities — do not use it.

Note on thresholds: Some textbooks $e . g ., A n d er so n e t a l .$ use a less strict threshold of $n p \geq 5$ and $n q \geq 5$ . The professor uses $n p \geq 15$ and $n q \geq 15$ , which is more conservative and gives more accurate results. Use the professor's threshold $(n p \geq 15, n q \geq 15)$ for this course's exams.

Rationale: Research has shown that once both $n p$ and $n q$ exceed 15, the normal and binomial probabilities are close enough that the approximation is reliable. The stricter threshold ensures better accuracy, especially in the tails.

8.6.4 Worked Example — Validating the Conditions

Problem 1:

n = 50

p = 0.31

. Can we use normal approximation?

$n p = 50 \times 0.31 = 15.5 \geq 15$ ✓
$n q = 50 \times 0.69 = 34.5 \geq 15$ ✓

Answer: Yes, normal approximation can be used.

Problem 2: 36% probability, $n = 25$ . Can we use normal approximation?

$n p = 25 \times 0.36 = 9 < 15$ ✗
$n q = 25 \times 0.64 = 16 \geq 15$ ✓

Answer: No — one condition fails. Better not to use the approximation; the probabilities would differ too much.

8.6.5 The Continuity Correction — Why 13.5 Instead of 14?

When we convert a discrete binomial problem to a continuous normal one, a subtle issue arises.

The problem: In a discrete distribution, $X = 14$ is a single point — it has probability mass. In a continuous distribution, a single point has probability zero. Continuous distributions always need an interval.

Values like 13.5, 13.6, 13.7, ..., up to 14.4 — when rounded to the nearest integer — all become 14. So in the continuous world, $X \geq 14$ actually means $X \geq 13.5$ .

This adjustment is called the continuity correction.

The rules:

Discrete statement	Continuous equivalent
$P (X < 14)$	$P (X < 13.5)$
$P (X \leq 14)$	$P (X < 14.5)$
$P (X = 14)$	$P (13.5 < X < 14.5)$
$P (X > 14)$	$P (X > 14.5)$
$P (X \geq 14)$	$P (X > 13.5)$

The pattern: "less than" subtracts 0.5; "greater than" adds 0.5. "Equals" spans ±0.5 around the integer.

Q: Why exactly 13.5 instead of 14?

A: In discrete math, the integer 14 corresponds to the continuous interval \[13.5, 14.5\). If you only use $X > 14$ in the continuous integral, you miss the values from 13.5 to 14 that all round to 14. The continuity correction includes them. The difference in probability is usually small, but the correction makes the result more accurate.

Q: For greater than 14, do we need the correction?

A: Yes — for $P (X > 14)$ you would use $X > 14.5$ in the continuous formula, because 14.1, 14.2, ... all round to 14 $w hi c hi s N O T g r e a t er t han 14$ , and "greater than 14" means starting from 15, which means 14.5 and above.

Q $se v er a l s t u d e n t s$ : Does $P (X = 14)$ make sense in a continuous distribution?

A: Exactly — and this is the one case where the continuity correction is NOT optional. $P (X = 14)$ cannot be computed directly from a continuous distribution because a single point has zero probability. The continuity correction converts it to $P (13.5 < X < 14.5)$ , which has a meaningful area. This is the most important use of the correction.

8.6.6 Full Worked Example — Normal Approximation

Problem:

n = 50

p = 0.31

. Find

P (X < 14)

Step 0 — Validate conditions:

$n p = 15.5 \geq 15$ ✓
$n q = 34.5 \geq 15$ ✓

→ Normal approximation is valid.

Step 1 — Compute $μ$ and $σ$ : $μ = n p = 50 \times 0.31 = 15.5$ $σ = n pq = 50 \times 0.31 \times 0.69 = 10.695 \approx 3.27$

Step 2 — Apply continuity correction and convert to Z: $P (X < 14)$ → use $X < 13.5$ $s in ce " l ess t han 14" m e an s u pt o 13, w h ose u pp er b o u n d i s 13.5$ .

$Z = \frac{13.5 - 15.5}{3.27} = \frac{- 2}{3.27} \approx - 0.61$

Step 3 — Use the table: $P (Z < - 0.61) = 1 - F (0.61) = 1 - 0.7291 = 0.2709$

Without continuity correction $u s in g (X = 14) d i r ec tl y$ : $Z = \frac{14 - 15.5}{3.27} = - 0.46$ , giving a different $l ess a cc u r a t e$ probability.

Sense-check: The mean is 15.5 and we want the probability of being below 14. Since 14 is below the mean, the probability should be less than 0.5 but not tiny $i t i so n l y ab o u t 0.46 σ b e l o w$ . 0.2709 is plausible ✓.

8.6.7 Summary — Normal Approximation Procedure

Check conditions: $n p \geq 15$ AND $n q \geq 15$ .
Compute $μ = n p$ and $σ = n pq$ .
Apply continuity correction to the X value $a d j u s t b y \pm 0.5$ .
Convert to Z: $Z = \frac{X _{corrected} - μ}{σ}$ .
Use the standard normal table.

Bridge: We have now covered the four core discrete and continuous distributions: Bernoulli, Binomial, Poisson, and Normal — plus the normal approximation that bridges the discrete and continuous worlds. The professor offers a brief glimpse of two more distributions

c hi - s q u a r e an d t

that will be needed later for hypothesis testing.

Visual intuition: Draw a binomial bar chart for $n = 50$ , $p = 0.31$ — bars at integer positions from roughly 8 to 24, peaking near 15.5. Now overlay a smooth bell curve with the same mean and standard deviation. The curve passes through the tops of the bars. For $P (X < 14)$ , the binomial sums the areas of bars at 0 through 13 — a staircase. The normal approximation integrates the smooth curve from $- \infty$ to 13.5 — a smooth area that closely matches the staircase sum. The continuity correction aligns the staircase edge $a t 13.5$ with the integral boundary.

Pitfalls — common beginner traps:

Forgetting the continuity correction: Computing $Z = \frac{14 - μ}{σ}$ instead of $Z = \frac{13.5 - μ}{σ}$ for $P (X < 14)$ . The error is usually small but can matter in borderline cases.

Mixing up the ±0.5 direction: For $P (X < 14)$ use 13.5 $s u b t r a c t 0.5$ . For $P (X \leq 14)$ use 14.5 $a dd 0.5$ . For $P (X > 14)$ use 14.5. For $P (X \geq 14)$ use 13.5. Write down the discrete-to-continuous conversion explicitly before computing Z.

Skipping the condition check: Always validate $n p \geq 15$ and $n q \geq 15$ first. If the problem asks for the approximation and one condition fails, state that the approximation is unreliable and explain why.

Using $σ^{2}$ instead of $σ$ : The Z-formula needs $σ = n pq$ . Using $n pq$ directly in the denominator gives a completely wrong Z-value.

Real-world connection: The normal approximation to the binomial is the theoretical foundation of political polling. A pollster surveys $n = 1000$ voters and finds 52% support for a candidate. The margin of error $\pm 3$ comes from treating the sample proportion as about normal with $μ = p$ and $σ = p (1 - p) / n$ . Quality control charts in manufacturing $p - c ha r t s$ use this same approximation to set control limits. In A/B testing for websites, the normal approximation determines whether a difference in conversion rates between two page designs is statistically significant or just random noise.

8.7 A Glimpse — Chi-Square and T-Distributions

Hook: The normal distribution is the undisputed king of statistics. But it has two close cousins that appear whenever you work with small samples or test relationships between categories. The t-distribution handles the uncertainty of estimating

σ

from a small sample. The chi-square distribution tests whether observed counts match expected counts. These distributions are not on the mid-semester exam, but they will dominate the second half of the course.

8.7.1 Chi-Square Distribution

A chi-square $(χ^{2})$ distribution is a continuous distribution that takes only non-negative values

s t a r t s a t z er o

. The shape depends on a parameter called degrees of freedom

df

At low df $1 or 2$ : the curve starts high near zero and drops sharply — heavily right-skewed.
At moderate df $5-10$ : the curve rises from zero to a peak, then gradually declines — like a wave.
At high df $30 +$ : the curve approaches a symmetric bell shape — resembling the normal distribution.

Why it matters: The chi-square distribution arises naturally when you square and sum independent standard normal variables: if

Z_{1}, Z_{2}, \dots, Z_{k}

are independent

N (0, 1)

, then

Z_{1}^{2} + Z_{2}^{2} + \dots + Z_{k}^{2} \sim χ^{2} (k)

. This makes it the go-to distribution for testing variances and for goodness-of-fit tests

c h ec k in g i f o b ser v e dd a t ama t c h e x p ec t e d p a tt er n s

The formula involves the gamma function — do not worry about it. Chi-square tables exist, similar to the standard normal table, and will be provided when needed for hypothesis testing.

8.7.2 T-Distribution

The t-distribution

or S t u d e n t^{'} s t - d i s t r ib u t i o n

is a continuous, symmetric, bell-shaped distribution that looks similar to the normal but with heavier tails.

Why it exists: You estimate the population standard deviation $σ$ from a small sample using $s$ . The extra uncertainty from that estimate makes the distribution of $\frac{X ˉ - μ}{s / n}$ wider in the tails than the normal. The t-distribution accounts for this extra uncertainty. As the sample size grows, $s$ becomes a better estimate of $σ$ , and the t-distribution converges to the normal.

Comparing normal and t-distributions:

Property	Normal $N (0, 1)$	t-distribution
Shape	Symmetric, bell-shaped	Symmetric, bell-shaped
Tails	Thin — probability drops fast	Heavier — more probability in extremes
Center	Peaked at 0	Flatter — less probability near center
Parameters	None $f i x e d s ha p e$	Degrees of freedom $df$
As df $\to \infty$	—	Approaches $N (0, 1)$

Analogy $f r o m t h e p r o f essor$ : Think of building a sand hill on a beach. The normal distribution is the perfect, undisturbed hill — rock-solid and fixed. The t-distribution is what you get when removing some sand from the center and sprinkling it onto the edges — the top flattens and the tails spread outward. At small degrees of freedom, the t-distribution is like a delicate, shifting hill that changes shape noticeably with each grain of sand. As degrees of freedom grow

a dd in g s an d ba c k

, it firms up into the fixed normal shape. At infinity degrees of freedom, the t-distribution is exactly the normal distribution — the sand hill is fully restored.

Where the analogy guides intuition: When df is small $s a y, (n = 5) o b ser v a t i o n s$ , the t-distribution is noticeably wider than normal — you need a bigger margin for error. When df is large $(n > 30)$ , the difference is negligible — you can use the normal table. This is why many statistical procedures switch from t to z at large sample sizes.

Recap: The chi-square distribution handles sums of squared normals and will be used for variance tests and categorical data. The t-distribution is a "small-sample normal" with heavier tails that converges to the normal as sample size grows. Both are previews of the second half of the course.

Bridge: The next section collects all the exam-relevant guidance the professor shared during this lecture — what is tested, what is not, and the key traps to avoid.

Real-world connection: The t-distribution was discovered by William Gosset, a chemist working at the Guinness brewery in Dublin. He published under the pseudonym "Student" because Guinness forbade employees from publishing. His problem was practical: testing the quality of small batches of beer — where sample sizes were tiny $3-5 ba r r e l s$ and the normal distribution gave overconfident results. The t-distribution solved this by widening the confidence intervals for small samples. Today, every clinical trial, every A/B test, and every scientific experiment with fewer than 30 observations relies on the t-distribution for valid inference.

8.8 Exam Guidance

8.8.1 Syllabus Scope

Exam note: The mid-semester exam covers up to and including the normal distribution. The normal approximation to the binomial is part of the normal distribution topic and is included.

NOT on the exam: Chi-square distribution, t-distribution, hypothesis testing, sampling — these come after the mid-semester.

8.8.2 Expected Question Types

Based on the professor's discussion, the question paper may include topics from the entire first half of the course:

Basic statistics — mean, median, mode; symmetry and skewness; detecting non-symmetric distributions; outliers.
Basic probability — simple probability problems, sample spaces, events.
Conditional probability and Bayes' theorem — problems involving conditional and inverse probabilities.
Random variables — discrete and continuous; expect at least one numerical problem $P M F, P D F, e x p ec t e d v a l u e, v a r ian ce$ .
Distributions — 1 or 2 problems expected; could be binomial, Poisson, or normal. The hardest part is identifying which distribution a problem belongs to. A problem may have sub-parts $a, b, c$ worth 3–6 marks each.

8.8.3 Key Exam Tips

Exam note — critical traps to avoid:

Identification is the hardest part: The problem will not always state "this follows a binomial distribution." You must recognize it from context — look for fixed $n$ , constant $p$ , independent trials. For Poisson: large $n$ , small $p$ , counting events in an interval. For normal: continuous data, bell-shaped context, or problems that provide $μ$ and $σ$ .

Variance vs. standard deviation trap: If the problem says "variance = 5," then $σ = 5$ . If it says "standard deviation = 5," use $σ = 5$ directly. Always double-check which one is given before plugging into $Z = \frac{X - μ}{σ}$ .

Normal table will be provided on the exam. If you cannot finish a calculation, write the Z-score and the appropriate expression in terms of $F (z)$ — this earns partial credit.

Less than and greater than are the same with or without the equal sign in continuous distributions. $P (X \leq 40) = P (X < 40)$ for normal — the probability of hitting exactly 40 is zero.

Write your steps in full. Write $Z = \frac{X - μ}{σ}$ , substitute the numbers, read $F (z)$ from the table, and state the final probability. Even if the final number is wrong, clear steps earn most of the marks.

8.8.4 Tomorrow — Problem-Solving Session

The next session is a dedicated problem-solving and revision class covering all topics from the first half of the course. The plan:

Solve 1–2 problems per topic: basic statistics, probability, conditional probability/Bayes, discrete/continuous random variables, and each distribution $bin o mia l, P o i sso n, n or ma l, n or ma l a pp r o x ima t i o n$ .
Students should actively participate — have a notebook ready and solve alongside the class.

8.9 Key Industry Applications and Real-World Connections

Z-score normalization $s t an d a r d sc a l in g$ is used extensively in machine learning data preprocessing. Converting features to mean 0 and standard deviation 1 ensures all features contribute proportionally during model training — otherwise, a feature measured in thousands $l ik e ann u a l in co m e$ would dominate a feature measured in single digits $l ik e a g e$ . This is used alongside min-max scaling, with the choice depending on whether the algorithm assumes normally distributed inputs.

The bell curve terminology is widely used in management contexts. Performance reviews often use "grading on a curve" — forcing employee ratings into a normal distribution where most people are "average," a few are "exceptional," and a few are "needs improvement." The same idea appears in college grading curves and standardized test scoring $S A T, GR E, I Q$ .

Gaussian distribution is the preferred term in communications engineering and signal processing. Channel noise in wireless communications is modeled as additive white Gaussian noise $A W GN$ — the foundational assumption behind Shannon's channel capacity theorem and all modern digital communication systems.

Normal distribution tables have been a standard tool since before calculators. They remain the most efficient way to compute normal probabilities without software, and every statistics exam provides one. The ability to read a Z-table is a fundamental skill tested across all distributions covered in this lecture.

Poisson processes model everything from call center staffing $h o w man y a g e n t s a r e n ee d e df or a g i v e n c a l l v o l u m e ?$ to website server capacity planning $h o w man y r e q u es t s p er seco n d c an t h eser v er han d l e b e f or e q u e u in g ?$ . Insurance companies use the Poisson distribution to model the number of claims per policy period and set premiums accordingly.

Binomial models power acceptance sampling in manufacturing. If a batch has a defect rate $p$ , the binomial distribution tells quality engineers the probability that a sample of $n$ units contains $x$ or more defectives. That probability can trigger a batch rejection if it is too low under the "acceptable quality" assumption.

ISM Lecture 8 notes · Bernoulli, Binomial, Poisson, and Normal Distributions

Introduction to Statistical Methods· postgraduate· 2026-07-02

Sections Breakdown

1Recap — Bernoulli and Binomial Distributions

Definition, properties, PMF, mean, and variance of Bernoulli and binomial distributions with worked multiple-choice guessing example.

2Poisson Distribution

Definition, rate parameter lambda, mean equals variance property, when to use Poisson vs binomial, worked defective units example.

3Normal Distribution (Gaussian)

Definition, PDF, properties (bell-shaped, symmetric, asymptotic), empirical rule (68-95-99.7), and the integration problem.

4Standard Normal Distribution

Z-transformation, centering and rescaling, standard normal table, reading cumulative probabilities, handling negative Z values.

5Using the Standard Normal Table — Worked Examples

Three fundamental cases (less than, greater than, between) with fully worked examples for each pattern.

6Normal Approximation to the Binomial

Motivation, conditions (np >= 15, nq >= 15), continuity correction, full worked example with sense-check.

7A Glimpse — Chi-Square and T-Distributions

Preview of chi-square and t-distributions with comparisons to the normal distribution.

8Exam Guidance

Syllabus scope, expected question types, key exam tips, and problem-solving session overview.

9Key Industry Applications and Real-World Connections

Real-world applications including Z-score normalization, bell curve in management, Gaussian in communications, Poisson in queueing theory, and binomial in quality control.

Postgraduate students in Statistics and Data Science

Exam Revision Notes

Below is the distilled, exam-ready core of this lecture. Every entry is built from the full textbook notes above. Use this section for rapid review — but if something doesn't make sense, go back to the full explanation in the main content.

Bernoulli Distribution

Must-know: A Bernoulli distribution models a single trial with two outcomes. The parameter p fully defines it: mean = p, variance = p(1-p).

$P (X = x) = {p 1 - p x = 1 x = 0, μ = p, σ^{2} = p (1 - p)$

Top pitfall: Confusing p and q (1-p). Always identify which outcome is 'success' before computing.

Self-check: If a Bernoulli trial has p = 0.3, what are the mean and variance?

Connects to: Binomial Distribution

Binomial Distribution

Must-know: Models the number of successes in n independent, identically distributed Bernoulli trials. The PMF combines the binomial coefficient with success and failure probabilities.

$P (X = x) = (x n) p^{x} q^{n - x}, μ = n p, σ^{2} = n pq$

Top pitfall: Forgetting the binomial coefficient — P(X = x) is NOT just p^x q^{n-x}. Also, mixing up variance (npq) with standard deviation (sqrt(npq)).

Self-check: With n = 10 and p = 0.5, what is P(X = 3) and why must you include binom(10, 3)?

Connects to: Bernoulli Distribution, Poisson Distribution, Normal Approximation

Poisson Distribution

Must-know: Models count of rare, independent events in a fixed interval. Mean equals variance, both equal to lambda. Emerges as the limit of binomial when n is large and p is small.

$P (X = x) = \frac{e ^{- λ} λ ^{x}}{x !}, μ = λ, σ^{2} = λ$

Top pitfall: Forgetting x! in the denominator. Also, failing to adjust lambda when the interval changes (e.g., lambda = 5 per hour becomes 2.5 for 30 minutes).

Self-check: If a call center receives 10 calls per minute on average, what is the probability of exactly 3 calls in the next minute?

Connects to: Binomial Distribution

Normal Distribution (Gaussian)

Must-know: The continuous, symmetric bell curve defined by mean mu (center) and standard deviation sigma (spread). The empirical rule (68-95-99.7) gives quick probability estimates. The PDF integral has no closed form — always use Z-transformation and tables.

$X \sim N (μ, σ^{2}), f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}$

Top pitfall: Trying to integrate the PDF by hand. Always convert to Z and use the table. Also, using variance instead of standard deviation in the Z formula.

Self-check: Scores are normally distributed with mu = 75 and sigma = 5. What percentage of students score between 70 and 80?

Connects to: Standard Normal Distribution, Normal Approximation to Binomial

Standard Normal Distribution and Z-Transformation

Must-know: The Z-transformation Z = (X - mu) / sigma converts any normal distribution to N(0, 1). The standard normal table gives F(z) = P(Z <= z). Three fundamental cases: less than (read directly), greater than (1 - F(z)), between (F(z2) - F(z1)).

$Z = \frac{X - μ}{σ}, Z \sim N (0, 1), F (z) = P (Z \leq z)$

Top pitfall: Looking up X directly in the Z-table without converting. Also, forgetting to use symmetry F(-z) = 1 - F(z) for negative Z values with positive-only tables.

Self-check: If Z = 1.75, what is P(Z < 1.75)? If Z = -0.8 and your table only has positive values, how do you find F(-0.8)?

Connects to: Normal Distribution

Normal Approximation to the Binomial

Must-know: When np >= 15 and nq >= 15, the binomial distribution can be approximated by the normal. Use continuity correction (+/- 0.5) because binomial is discrete and normal is continuous.

$μ = n p, σ = n pq, Z = \frac{X _{corrected} - μ}{σ}$

Top pitfall: Forgetting the continuity correction (±0.5) and mixing up the direction of the correction. Always check np >= 15 AND nq >= 15 first.

Self-check: For n = 50, p = 0.31, find P(X < 14) using normal approximation. What is the continuity-corrected Z value?

Connects to: Binomial Distribution, Standard Normal Distribution

Previous LectureLecture 07 · Joint, Marginal, and Conditional Distributions — Discrete and Named Distributions

Next LectureLecture 09 · Descriptive Statistics, Probability, and Random Variables

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Prerequisite Knowledge

Previously Covered in This Subject

Bernoulli, Binomial, Poisson, and Normal Distributions

8.1 Recap — Bernoulli and Binomial Distributions

8.1.1 Bernoulli Distribution — Definition and Properties

8.1.2 Binomial Distribution — Definition and Conditions

8.1.3 Worked Example — Multiple-Choice Guessing

8.1.4 Symbol Registry — Binomial Distribution

8.2 Poisson Distribution

8.2.1 Definition and Formula

8.2.2 When to Use Poisson Instead of Binomial

8.2.3 Worked Example — Defective Units

8.2.4 Symbol Registry — Poisson Distribution

8.3 Normal Distribution Gaussian

8.3.1 Definition

8.3.2 Properties of the Normal Curve

8.3.3 The Integration Problem

8.4 Standard Normal Distribution

8.4.1 The Z-Transformation

8.4.2 Two Benefits of Standardization

8.4.3 The Standard Normal Table

8.4.4 Symbol Registry — Standard Normal

8.5 Using the Standard Normal Table — Worked Examples

8.5.1 Three Fundamental Cases

8.5.2 Example 1 — Less Than 1.75

8.5.3 Example 2 — Greater Than 2.06

8.5.4 Example 3 — Scores Less Than 40

8.5.5 Example 4 — Scores Greater Than 38

8.5.6 Example 5 — Score Between 70 and 80

8.5.7 Example 6 — Between 26 and 40 with μ=30,σ=5

8.5.8 Example 7 — Greater Than 45 with μ=30,σ=5

8.5.9 Example 8 — "At Least" and "Greater Than"

8.5.10 Student Question — Handling Negative Z

8.5.11 Summary of Standard Normal Table Rules

8.6 Normal Approximation to the Binomial Distribution

8.6.1 Motivation — Why Approximate?

8.6.2 How Normal Approximation Works

8.6.3 When to Apply Normal Approximation

8.6.4 Worked Example — Validating the Conditions

8.6.5 The Continuity Correction — Why 13.5 Instead of 14?

8.6.6 Full Worked Example — Normal Approximation

8.6.7 Summary — Normal Approximation Procedure

8.7 A Glimpse — Chi-Square and T-Distributions

8.7.1 Chi-Square Distribution

8.7.2 T-Distribution

8.8 Exam Guidance

8.8.1 Syllabus Scope

8.8.2 Expected Question Types

8.8.3 Key Exam Tips

8.8.4 Tomorrow — Problem-Solving Session

8.9 Key Industry Applications and Real-World Connections

ISM Lecture 8 notes · Bernoulli, Binomial, Poisson, and Normal Distributions

Sections Breakdown

Exam Revision Notes

Bernoulli Distribution

Binomial Distribution

Poisson Distribution

Normal Distribution (Gaussian)

Standard Normal Distribution and Z-Transformation

Normal Approximation to the Binomial

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8.3 Normal Distribution $G a u ss ian$

8.5.7 Example 6 — Between 26 and 40 with $μ = 30, σ = 5$

8.5.8 Example 7 — Greater Than 45 with $μ = 30, σ = 5$