What is a probability distribution in the context of IB DP Mathematics?

In IB DP Mathematics, a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. It is a fundamental concept in probability theory, used to describe the likelihood of various outcomes. Probability distributions can be discrete or continuous, depending on whether they deal with discrete or continuous random variables.

How do discrete and continuous probability distributions differ?

Discrete probability distributions apply to scenarios where the set of possible outcomes is countable, such as rolling a die or flipping a coin. Examples include the binomial and Poisson distributions. Continuous probability distributions, on the other hand, apply to scenarios where outcomes are not countable and can take any value within a range, such as measuring height or temperature. The normal distribution is a common example of a continuous distribution.

What is the significance of the normal distribution in probability?

The normal distribution, also known as the Gaussian distribution, is significant in probability because it describes how values of a variable are distributed. It is characterized by its bell-shaped curve and is defined by its mean and standard deviation. The normal distribution is important in the IB DP curriculum because it models many natural phenomena and is used in various statistical methods, including hypothesis testing and confidence intervals.

How is the probability mass function (PMF) used in probability distributions?

The probability mass function (PMF) is used in discrete probability distributions to specify the probability of each possible outcome. It assigns a probability to each value of a discrete random variable. The PMF is a crucial concept in IB DP Mathematics as it helps in calculating probabilities and understanding the distribution of discrete random variables, such as in the binomial distribution.

What role does the cumulative distribution function (CDF) play in probability theory?

The cumulative distribution function (CDF) plays a vital role in probability theory by providing the probability that a random variable takes on a value less than or equal to a specific value. It is applicable to both discrete and continuous probability distributions. In the IB DP curriculum, understanding the CDF is essential for analyzing the behavior of random variables and for solving problems related to probability and statistics.

Why is "Computing probabilities without writing $X \sim B(n, p)$ or $X \sim N(\mu, \sigma^2)$." a common mistake in Probability Distribution?

Why it happens: Jumping to the GDC. How to avoid it: STATE the distribution and parameters — this earns AO1 method marks even if the arithmetic slips.

Why is "Using variance where the GDC requires standard deviation." a common mistake in Probability Distribution?

Why it happens: N(, ^2) has ^2 but GDC takes . How to avoid it: N(170, 64) = 8, NOT 64.

Why is "$P(X \ge 7) = 1 - $ binomCdf$(n, p, 7)$ for a discrete variable." a common mistake in Probability Distribution?

Why it happens: Off-by-one — Cdf(n, p, 7) = P(X 7). How to avoid it: P(X 7) = 1 - P(X 6) = 1 - binomCdf(n, p, 6).

ProbabilityIB Maths AA HL

Probability Distribution

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Probability Distributions — IB Maths AA HL: discrete and continuous, $E(X)$, $\text{Var}(X)$, binomial, normal

AA HL extends SL with continuous random variables (probability density functions) and richer normal-distribution applications. This note covers discrete and continuous distributions, expected value, variance, and the two named distributions $B(n, p)$ and $N(\mu, \sigma^2)$ .

What you’ll learn

Mapped to the IB DP Maths AA HL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — Define discrete and continuous random variables; state expected value formulas.
AO1 — Apply binomial and normal formulas.
AO2 — Compute probabilities and inverse normals via GDC.
AO2 — Find $E(X)$ and $\text{Var}(X)$ for continuous distributions via integration.
AO3 — Justify the choice of distribution with the four binomial conditions or normal context.

Discrete and continuous random variables

Sum or integrate.

Discrete. $X$ takes values $x_1, x_2, \ldots$ with probabilities $p_i$ summing to 1.

$E(X) = \sum x_i p_i, \qquad \text{Var}(X) = \sum (x_i - E(X))^2 p_i = E(X^2) - [E(X)]^2.$

Continuous. $X$ has probability density function (pdf) $f(x)$ with:

$f(x) \ge 0$ for all $x$ .
$\int_{-\infty}^{\infty} f(x)\,dx = 1$ .
$P(a \le X \le b) = \int_a^b f(x)\,dx$ .

$E(X) = \int_{-\infty}^{\infty} x f(x)\,dx, \qquad \text{Var}(X) = E(X^2) - [E(X)]^2 = \int x^2 f(x)\,dx - [E(X)]^2.$

Worked example (discrete). $X$ takes values $1, 2, 3$ with probabilities $0.2, 0.5, 0.3$ .

$E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1$ .

$E(X^2) = 1(0.2) + 4(0.5) + 9(0.3) = 4.9$ .

$\text{Var}(X) = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49$ . SD $= 0.7$ .

Worked example (continuous). $f(x) = 3x^2$ for $0 \le x \le 1$ , else 0. Verify pdf and find $E(X)$ .

$\int_0^1 3x^2\,dx = [x^3]_0^1 = 1$ ✓.

$E(X) = \int_0^1 x \cdot 3x^2\,dx = \int_0^1 3x^3\,dx = [3x^4/4]_0^1 = 3/4$ .

Discrete: probabilities SUM to 1.
Continuous: pdf INTEGRATES to 1.
$E(X^2) \ne [E(X)]^2$ — gives variance.

Binomial and Normal distributions

$B(n, p)$ for counts, $N(\mu, \sigma^2)$ for measurements.

Binomial $X \sim B(n, p)$ . $n$ independent trials, each success with probability $p$ .

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \qquad E(X) = np, \qquad \text{Var}(X) = np(1-p).$

Four conditions: fixed $n$ , two outcomes, constant $p$ , independent trials.

Normal $X \sim N(\mu, \sigma^2)$ . Symmetric bell-curve with mean $\mu$ and variance $\sigma^2$ .

68-95-99.7 rule:

$P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.68$ .
Within $\pm 2\sigma \approx 0.95$ .
Within $\pm 3\sigma \approx 0.997$ .

Standardisation. $Z = (X - \mu)/\sigma$ has $N(0, 1)$ .

GDC commands.

$P(a \le X \le b)$ : normCdf $(a, b, \mu, \sigma)$ .
Find $x$ with $P(X \le x) = p$ : invNorm $(p, \mu, \sigma)$ .
$P(X = k)$ for binomial: binomPdf $(n, p, k)$ .
$P(X \le k)$ : binomCdf $(n, p, k)$ .

Worked example. Heights $X \sim N(170, 8^2)$ cm. Find $P(165 \le X \le 180)$ and the cutoff height for the top 10%.

Probability: normCdf(165, 180, 170, 8) $\approx 0.628$ .

Top 10%: invNorm $(0.90, 170, 8) \approx 180.25$ cm.

Worked example (binomial). Coin biased $P(H) = 0.6$ . 10 flips. $P(X \ge 7)$ ?

$1 -$ binomCdf(10, 0.6, 6) $\approx 0.382$ .

Binomial: 4 conditions; mean $np$ , var $np(1-p)$ .
Normal: 68-95-99.7; standardise with $Z = (X - \mu)/\sigma$ .
GDC routines — STATE the distribution and parameters.

How it’s examined

Paper 1: discrete distribution tables, small- $n$ binomial, exact normal values via standardisation. Paper 2: GDC arithmetic. Paper 3: extended-response inference and modelling.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

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Step-by-step worked examples — Probability Distribution

Step-by-step solutions to past-paper-style questions on probability distribution, written exactly the way a tutor would explain them at the board.

1Expected value from a table
Getting started• discrete
Question
$X$ takes values $0, 1, 2, 3$ with probabilities $0.1, 0.3, 0.4, 0.2$ . Find $E(X)$ and $\text{Var}(X)$ .
Step-by-step solution
1. Step 1
  $E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 1.7$ .
2. Step 2
  $E(X^2) = 0 + 0.3 + 1.6 + 1.8 = 3.7$ .
3. Step 3
  $\text{Var}(X) = 3.7 - 1.7^2 = 3.7 - 2.89 = 0.81$ .
Answer
$E(X) = 1.7$ , $\text{Var}(X) = 0.81$ .
2Binomial — at least one
Building confidence• binomial
Question
A quality control test passes parts with probability 0.95. From 20 parts, find $P(X \ge 19)$ where $X$ is the number passing.
Step-by-step solution
1. Step 1
  $X \sim B(20, 0.95)$ — state the distribution.
2. Step 2
  $P(X \ge 19) = P(X = 19) + P(X = 20)$ .
3. Step 3
  $P(X = 19) = \binom{20}{19}(0.95)^{19}(0.05) \approx 0.3774$ .
4. Step 4
  $P(X = 20) = (0.95)^{20} \approx 0.3585$ .
5. Step 5
  Sum $\approx 0.7358$ . GDC: $1 -$ binomCdf $(20, 0.95, 18)$ confirms.
Answer
$\approx 0.736$ .
3Normal — central probability and cutoff
Building confidence• normal, GDC
Question
Exam scores $X \sim N(62, 12^2)$ . (a) Find $P(X \ge 70)$ . (b) Find the score above which the top 5% lie.
Step-by-step solution
1. Step 1
  (a) normCdf $(70, 10^{99}, 62, 12) \approx 0.2525$ .
2. Step 2
  (b) invNorm $(0.95, 62, 12) \approx 81.7$ .
Answer
(a) $\approx 0.253$ . (b) $\approx 81.7$ .
4Continuous: verify pdf and find variance
Stretch• continuous, integration
Question
$f(x) = kx$ for $0 \le x \le 2$ , else 0. Find $k$ , $E(X)$ and $\text{Var}(X)$ .
Step-by-step solution
1. Step 1
  $\int_0^2 kx\,dx = k[x^2/2]_0^2 = 2k = 1 \Rightarrow k = 1/2$ .
2. Step 2
  $E(X) = \int_0^2 x \cdot (x/2)\,dx = [x^3/6]_0^2 = 8/6 = 4/3$ .
3. Step 3
  $E(X^2) = \int_0^2 x^2 \cdot (x/2)\,dx = [x^4/8]_0^2 = 16/8 = 2$ .
4. Step 4
  $\text{Var}(X) = 2 - (4/3)^2 = 2 - 16/9 = 2/9$ .
Answer
$k = 1/2$ , $E(X) = 4/3$ , $\text{Var}(X) = 2/9$ .

Key Definitions and Keywords — Probability Distribution

Definitions to memorise and the exact keywords mark schemes credit for probability distribution answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA HL sitting.

Binomial distribution $B(n, p)$
Examiner keyword
Count of successes in $n$ independent trials, constant $p$ per trial. $E(X) = np$ , $\text{Var}(X) = np(1-p)$ .
Normal distribution $N(\mu, \sigma^2)$
Examiner keyword
Symmetric continuous distribution with mean $\mu$ and variance $\sigma^2$ . Standardise to $Z = (X - \mu)/\sigma$ .
Probability density function (pdf)
Examiner keyword
Non-negative function $f$ with $\int f = 1$ ; $P(a \le X \le b) = \int_a^b f(x)\,dx$ .

Common Mistakes and Misconceptions — Probability Distribution

The traps other students keep falling into on probability distribution questions — taken from recent IB DP Maths AA HL examiner reports and mark schemes — and how to avoid them.

✕Computing probabilities without writing $X \sim B(n, p)$ or $X \sim N(\mu, \sigma^2)$ .
Why it happens
Jumping to the GDC.
How to avoid it
STATE the distribution and parameters — this earns AO1 method marks even if the arithmetic slips.
✕Using variance where the GDC requires standard deviation.
Why it happens
$N(\mu, \sigma^2)$ has $\sigma^2$ but GDC takes $\sigma$ .
How to avoid it
$N(170, 64) \Rightarrow \sigma = 8$ , NOT 64.
✕ $P(X \ge 7) = 1 -$ binomCdf $(n, p, 7)$ for a discrete variable.
Why it happens
Off-by-one — Cdf $(n, p, 7) = P(X \le 7)$ .
How to avoid it
$P(X \ge 7) = 1 - P(X \le 6) = 1 -$ binomCdf $(n, p, 6)$ .

Past paper style quiz

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Probability Distribution — frequently asked questions

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Probability Distribution

Short Study Notes in the form of Slides

Short Study Notes — Probability Distribution

Detailed Notes

What you’ll learn

How it’s examined

1Expected value from a table

2Binomial — at least one

3Normal — central probability and cutoff

4Continuous: verify pdf and find variance

Binomial distribution B(n,p)B(n, p)B(n,p)

Normal distribution N(μ,σ2)N(\mu, \sigma^2)N(μ,σ2)

Probability density function (pdf)

✕Computing probabilities without writing X∼B(n,p)X \sim B(n, p)X∼B(n,p) or X∼N(μ,σ2)X \sim N(\mu, \sigma^2)X∼N(μ,σ2).

✕Using variance where the GDC requires standard deviation.

✕P(X≥7)=1−P(X \ge 7) = 1 - P(X≥7)=1− binomCdf(n,p,7)(n, p, 7)(n,p,7) for a discrete variable.

Past paper style quiz

Probability Distribution — frequently asked questions

Binomial distribution $B(n, p)$

Normal distribution $N(\mu, \sigma^2)$

✕Computing probabilities without writing $X \sim B(n, p)$ or $X \sim N(\mu, \sigma^2)$ .

✕ $P(X \ge 7) = 1 -$ binomCdf $(n, p, 7)$ for a discrete variable.