Probability Distributions

Example

A discrete random variable $X$ has PMF:

Example: Finding an unknown parameter

$P(X = x) = kx$ for $x = 1, 2, 3, 4$. Find $k$ and $E(X)$.

$k(1 + 2 + 3 + 4) = 1 \implies 10k = 1 \implies k = 0.1$

$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$

Binomial Distribution

Conditions

A random variable $X$ follows a binomial distribution, $X \sim B(n, p)$When all four conditions Hold:

1. Fixed number of trials: exactly $n$ identical trials.
2. Independent trials: each trial”s outcome does not affect any other.
3. Two outcomes: each trial yields success (probability $p$) or failure (probability $q = 1-p$).
4. Constant probability: $p$ is the same for every trial.

$X$ counts the number of successes in $n$ trials.

Probability Mass Function

Where $\dbinom{n}{x} = \dfrac{n!}{x!(n-x)!}$ counts the arrangements of $x$ successes among $n$ Trials.

Mean and Variance

Shape

• $p = 0.5$: symmetric about $np$.
• $p \lt 0.5$: positively skewed (right tail longer).
• $p \gt 0.5$: negatively skewed (left tail longer).

As $n$ increases the distribution approaches a bell shape (by the Central Limit Theorem). The mode Of $B(n, p)$ is at $\lfloor (n+1)p \rfloor$.

Cumulative Probabilities

On a GDC, $P(X \le k)$ is computed directly. For “at least” problems, use the complement:

Normal Approximation to the Binomial

When $n$ is large and $p$ is not too close to 0 or 1 (rule of thumb: $np \ge 5$ and $n(1-p) \ge 5$), The binomial can be approximated by the normal with matching mean and variance:

A continuity correction is required. For example:

Example

A factory produces bulbs with 3% defect rate. $X \sim B(20, 0.03)$ is the number of defects in a Sample of 20.

$P(X = 2) = \binom{20}{2}(0.03)^2(0.97)^{18} = 190 \times 0.0009 \times 0.5781 \approx 0.0988$

$P(X \le 1) = (0.97)^{20} + 20(0.03)(0.97)^{19} \approx 0.5438 + 0.3364 \approx 0.8802$

$P(X \ge 3) = 1 - P(X \le 2) \approx 1 - 0.8802 - 0.0988 = 0.0210$

$E(X) = 20(0.03) = 0.6$, $\sigma = \sqrt{20(0.03)(0.97)} = \sqrt{0.582} \approx 0.763$

Example: IB Paper 2 style

A multiple choice test has 15 questions with 5 options each. A student guesses all answers.

$X \sim B(15, 0.2)$.

$P(X = 4) = \binom{15}{4}(0.2)^4(0.8)^{11} \approx 0.1876$

$P(X \ge 8) = 1 - P(X \le 7) \approx 0.0042$

To set a pass mark so that guessing gives at most 1% chance of passing:

$P(X \ge 7) \approx 0.0181$ and $P(X \ge 8) \approx 0.0042$So the minimum pass mark is 8 correct.

Poisson Distribution

Conditions

$X \sim \mathrm{Po}(\lambda)$ models the number of events in a fixed interval of time or space when:

1. Events occur singly: no simultaneous events.
2. Independence: events in non-overlapping intervals are independent.
3. Constant rate: events occur at average rate $\lambda$ per unit interval.
4. Randomness: the count is proportional to the interval size.

Probability Mass Function

Where $\lambda \gt 0$ is the mean number of events, and $e \approx 2.71828$.

Mean and Variance

That $E(X) = \mathrm{Var}(X)$ is a distinguishing feature. If observed data has mean approximately Equal to variance, a Poisson model may be appropriate.

Poisson as a Limit of the Binomial

If $n \to \infty$, $p \to 0$While $np = \lambda$ stays constant, then $B(n, p) \to \mathrm{Po}(\lambda)$. The Poisson approximates the binomial when $n$ is large, $p$ is Small, and $np$ is moderate ( $n \ge 50$, $p \le 0.1$).

Additivity

If $X \sim \mathrm{Po}(\lambda_1)$ and $Y \sim \mathrm{Po}(\lambda_2)$ are independent, then:

If the rate is $\lambda$ per unit interval, then over $t$ intervals the count is $\mathrm{Po}(t\lambda)$.

Example

A helpdesk receives $\lambda = 3.5$ calls per hour. $X \sim \mathrm{Po}(3.5)$.

$P(X = 5) = \dfrac{e^{-3.5} \cdot 3.5^5}{5!} \approx 0.1318$

$P(X \le 2) = e^{-3.5}\!\left(1 + 3.5 + \dfrac{12.25}{2}\right) = 10.625 \, e^{-3.5} \approx 0.3208$

Over 2 hours: $Y \sim \mathrm{Po}(7)$, $P(Y \gt 7) = 1 - P(Y \le 7) \approx 0.4013$.

Example: Poisson approximation to Binomial

A typesetter makes errors at a rate of 1 per 500 characters. In a passage of 2000 characters, find The probability of at most 2 errors.

Exact: $X \sim B(2000, 1/500)$With $\lambda = 2000/500 = 4$.

Approximate: $X \approx \mathrm{Po}(4)$.

$P(X \le 2) = e^{-4}\!\left(1 + 4 + \dfrac{16}{2}\right) = 13e^{-4} \approx 0.2381$

Using exact binomial: $P(X \le 2) = (499/500)^{2000} + 2000(1/500)(499/500)^{1999} + \binom{2000}{2}(1/500)^2(499/500)^{1998}$

This is computationally intensive but gives a result extremely close to 0.2381.

Normal Distribution

Definition and Properties

$X \sim N(\mu, \sigma^2)$ has probability density function:

Key properties: bell-shaped, symmetric about $x = \mu$Asymptotic to the $x$-axis, total area = 1, Inflection points at $x = \mu \pm \sigma$. The mean, median, and mode all equal $\mu$.

$E(X) = \mu$ and $\mathrm{Var}(X) = \sigma^2$.

For any normal variable, $P(X = a) = 0$ for any specific value $a$ (continuous distribution).

The Empirical Rule (68-95-99.7)

Standard Normal Distribution

The standard normal is $Z \sim N(0, 1)$. Any normal variable standardises via:

The CDF is $\Phi(z) = P(Z \le z)$. Key properties:

Probability Calculations

For $X \sim N(\mu, \sigma^2)$To find $P(a \lt X \lt b)$Convert to $z$-scores:

On the GDC these are computed directly without manual standardisation.

Inverse Normal

Given probability $p$The inverse normal finds $x$ such that $P(X \le x) = p$. For the standard Normal, $z = \Phi^{-1}(p)$. For a general normal: $x = \mu + z\sigma$.

Finding Unknown Parameters

When $\mu$ or $\sigma$ is unknown, use standardisation with a known probability to set up Simultaneous equations. Each known probability gives one equation in two unknowns; two probabilities Are needed.

Example

Bags of flour: $X \sim N(1000, 225)$ (mean 1000 g, $\sigma = 15$ g).

$P(985 \lt X \lt 1020) = P(-1 \lt Z \lt 1.333) = \Phi(1.333) - \Phi(-1) \approx 0.9088 - 0.1587 = 0.7501$

$P(X \lt 970) = P(Z \lt -2) = 0.0228$So about 2.28% are rejected.

For the mass exceeded by only 5%: $P(X \le x) = 0.95$, $x = 1000 + 1.645(15) = 1024.67$ g.

Example: Unknown parameters

Test scores are normal. 15% score above 80, 10% score below 45. Find $\mu$ and $\sigma$.

$\dfrac{80 - \mu}{\sigma} = 1.036$ and $\dfrac{45 - \mu}{\sigma} = -1.282$.

Subtracting: $35 = 2.318\sigma$So $\sigma \approx 15.1$ and $\mu = 80 - 1.036(15.1) \approx 64.4$.

Example: Normal approximation to Binomial

$X \sim B(80, 0.4)$. Find $P(X \le 30)$ using a normal approximation.

\mu = 80(0.4) = 32$$\sigma^2 = 80(0.4)(0.6) = 19.2$$\sigma = 4.382.

With continuity correction: $P(X \le 30) \approx P\!\left(Z \le \dfrac{30.5 - 32}{4.382}\right) = P(Z \le -0.342)$.

$\approx 0.3665$

Exact binomial: $P(X \le 30) \approx 0.3642$. The approximation is very close.

Continuous Uniform Distribution (AHL)

Definition

$X \sim U(a, b)$ has PDF:

And $f(x) = 0$ otherwise. The PDF is constant over $[a, b]$Meaning all values in the interval are Equally likely.

Mean and Variance

CDF

For any $[c, d] \subseteq [a, b]$: $P(c \le X \le d) = \dfrac{d - c}{b - a}$.

Example

A bus arrives every 15 minutes. $X \sim U(0, 15)$ is the waiting time.

$P(X \gt 10) = 5/15 = 1/3$

$E(X) = 7.5$ minutes, $\sigma = \dfrac{15}{2\sqrt{3}} = \dfrac{5\sqrt{3}}{2} \approx 4.33$ minutes.

Given 5 minutes already waited, the remaining wait is $U(0, 10)$: $P(\mathrm{wait} \ge 8) = 2/10 = 1/5$.

Geometric Distribution (AHL)

Definition

$X \sim \mathrm{Geo}(p)$ models the number of trials needed for the first success in independent Bernoulli trials with success probability $p$.

Probability Mass Function

The first $x-1$ trials must be failures, and trial $x$ must succeed. This is the probability of Exactly $x-1$ consecutive failures followed by one success.

Mean and Variance

Useful shortcut

The first $n$ trials must all be failures. Similarly $P(X \ge n) = (1-p)^{n-1}$.

Example

A basketball player has free-throw success rate 72%. $X \sim \mathrm{Geo}(0.72)$.

$P(X = 3) = (0.28)^2(0.72) = 0.0784 \times 0.72 \approx 0.05645$

$P(X \gt 5) = (0.28)^5 \approx 0.00172$

$E(X) = 1/0.72 \approx 1.389$ attempts.

Negative Binomial Distribution (AHL)

Definition

$X \sim \mathrm{NB}(r, p)$ models the number of trials needed to obtain exactly $r$ successes. The Geometric distribution is the special case $\mathrm{NB}(1, p)$.

Probability Mass Function

In the first $x-1$ trials there are $r-1$ successes (in $\dbinom{x-1}{r-1}$ ways), and trial $x$ is The $r$-th success.

Mean and Variance

Note the parallel with geometric: multiplying $r$ by a factor scales both $E(X)$ and $\mathrm{Var}(X)$ by the same factor.

Example

A coin has $P(\mathrm{heads}) = 0.4$. $X \sim \mathrm{NB}(3, 0.4)$ counts flips for 3 heads.

$P(X = 7) = \dbinom{6}{2}(0.4)^3(0.6)^4 = 15 \times 0.064 \times 0.1296 \approx 0.1244$

$E(X) = 3/0.4 = 7.5$, $\mathrm{Var}(X) = 3(0.6)/0.16 = 11.25$ $\sigma = \sqrt{11.25} \approx 3.354$.

Central Limit Theorem (AHL)

Statement

If $X_1, X_2, \ldots, X_n$ are independent and identically distributed with mean $\mu$ and variance $\sigma^2$Then for large $n$:

This holds regardless of the shape of the original distribution. The rule of thumb is $n \ge 30$.

Distribution of the Sum

The sum $S_n = X_1 + \cdots + X_n$ is approximately $S_n \sim N(n\mu, n\sigma^2)$ for large $n$.

Standard Error

As $n$ increases, the standard error decreases: larger samples give more precise estimates of the Population mean.

Example

Apple masses: mean 150 g, $\sigma = 20$ g. Sample of 36. Find $P(\bar{X} \gt 155)$.

$\bar{X} \sim N(150, 400/36)$. $P\!\left(Z \gt \dfrac{5}{20/6}\right) = P(Z \gt 1.5) = 0.0668$.

Example: Sum of uniform variables

$X \sim U(2, 10)$. Sample of 50 observations. Find $P(\mathrm{sum} \gt 310)$.

$\mu = 6$, $\sigma^2 = 64/12 = 16/3$. Sum has mean $300$ and variance $50(16/3) = 800/3$.

$P\!\left(Z \gt \dfrac{10}{\sqrt{800/3}}\right) = P(Z \gt 0.612) \approx 0.2704$.

Confidence Intervals (AHL)

Concept

A $C\%$ confidence interval gives a range of plausible values for an unknown population parameter. If the sampling process were repeated many times, approximately $C\%$ of constructed intervals would Contain the true parameter. The confidence level does not mean there is a $C\%$ probability that the Parameter lies in any particular interval.

Confidence Interval for the Mean ($\sigma$ known)

Where $z_{\alpha/2}$ satisfies $P(Z \gt z_{\alpha/2}) = \alpha/2$ and $\alpha = 1 - C/100$.

When $\sigma$ is unknown and $n$ is large ($n \ge 30$), replace $\sigma$ with the sample standard Deviation $s$.

Margin of Error and Sample Size

Margin of error: $E = z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}$. To halve $E$Quadruple $n$.

Required sample size for margin $E$: $n = \left(\dfrac{z_{\alpha/2} \cdot \sigma}{E}\right)^2$ (round up to the next integer).

Example

Bottle volumes: $N(\mu, 25)$, $\sigma = 5$ ml. Sample of 25 gives $\bar{x} = 498$ ml.

95% CI: $498 \pm 1.960 \times 5/\sqrt{25} = 498 \pm 1.96$So $(496.04, 499.96)$ ml.

For margin 1 ml at 95%: $n = (1.960 \times 5/1)^2 = 96.04$Round up to 97.

Combining Random Variables

Linear Combinations

For any random variables X$$Y and constants a$$b:

This is the linearity of expectation and holds always, even without independence.

Variance of Sums

For independent $X$ and $Y$:

Note the plus sign even for differences: subtracting a variable still adds variability.

The general formula (not necessarily independent):

Where $\mathrm{Cov}(X, Y) = E(XY) - E(X)E(Y) = 0$ when $X$ and $Y$ are independent.

Example

X \sim B(10, 0.3)$$Y \sim B(15, 0.4)Independent.

$E(X + Y) = 3 + 6 = 9$

$\mathrm{Var}(X + Y) = 10(0.3)(0.7) + 15(0.4)(0.6) = 2.1 + 3.6 = 5.7$

$\mathrm{Var}(2X - 3Y) = 4(2.1) + 9(3.6) = 8.4 + 32.4 = 40.8$

Example: Normal combinations

Bus ride $X \sim N(25, 16)$Walk $Y \sim N(10, 9)$Independent.

$X + Y \sim N(35, 25)$. $P(X + Y \gt 40) = P(Z \gt 1) = 0.1587$.

Machine A produces rods: $X \sim N(50.0, 0.04)$. Machine B: $Y \sim N(50.2, 0.09)$.

$X - Y \sim N(-0.2, 0.13)$. $P(X - Y \gt 0) = P\!\left(Z \gt \dfrac{0.2}{\sqrt{0.13}}\right) = P(Z \gt 0.555) \approx 0.2894$.

IB Exam-Style Questions

Question 1 (Paper 1)

$X \sim B(20, 0.35)$. Find $P(5 \le X \le 8)$.

$P(5 \le X \le 8) = P(X \le 8) - P(X \le 4) \approx 0.7625 - 0.1260 = 0.6365$

Question 2 (Paper 1)

$X \sim \mathrm{Po}(4.2)$. Find $P(X \ge 3)$.

$P(X \ge 3) = 1 - P(X \le 2) = 1 - e^{-4.2}(1 + 4.2 + 8.82) \approx 1 - 0.2103 = 0.7897$

Question 3 (Paper 2)

Daily rainfall: $X \sim N(2.8, 1.44)$ (mean 2.8 mm, $\sigma = 1.2$ mm).

$P(X \gt 4) = P(Z \gt 1) = 0.1587$

Expected days per year exceeding 4 mm: $365 \times 0.1587 \approx 58$ days.

Rainfall exceeded on only 5% of days: $x = 2.8 + 1.645(1.2) = 4.774$ mm.

Question 4 (Paper 2, AHL)

$X \sim U(0, a)$ has $P(X \gt 3) = 0.4$. Find $a$ and $\mathrm{Var}(X)$.

$\dfrac{a-3}{a} = 0.4 \implies 0.6a = 3 \implies a = 5$

$\mathrm{Var}(X) = 25/12 \approx 2.083$

Question 5 (Paper 2, AHL)

$X \sim \mathrm{Geo}(0.15)$. Find the smallest $n$ with $P(X \le n) \ge 0.8$.

$P(X \le n) = 1 - 0.85^n \ge 0.8 \implies 0.85^n \le 0.2$

$n \ge \ln(0.2)/\ln(0.85) \approx 9.90$So $n = 10$.

Question 6 (Paper 2, AHL)

Component lengths: N(\mu, 0.25)$$\sigma = 0.5 mm. Sample of 30 gives $\bar{x} = 100.2$ mm.

90% CI: $100.2 \pm 1.645 \times 0.5/\sqrt{30} = 100.2 \pm 0.150$So $(100.05, 100.35)$ mm.

The claim $\mu = 100$ mm is not supported at 90% confidence, since 100 falls below the interval.

Question 7 (Paper 2, AHL)

$X \sim \mathrm{NB}(4, 0.25)$. Find $P(X = 10)$ and $E(X)$.

$P(X = 10) = \dbinom{9}{3}(0.25)^4(0.75)^6 = 84 \times 0.003906 \times 0.1780 \approx 0.0584$

$E(X) = 4/0.25 = 16$

Question 8 (Paper 2, AHL)

The masses of male students are $N(72, 36)$ and female students are $N(58, 25)$Independent. Find The probability that a randomly chosen male is heavier than a randomly chosen female.

Let $M \sim N(72, 36)$ and $F \sim N(58, 25)$. Then $D = M - F \sim N(72-58, 36+25) = N(14, 61)$.

$P(D \gt 0) = P\!\left(Z \gt \dfrac{0 - 14}{\sqrt{61}}\right) = P(Z \gt -1.793) = \Phi(1.793) \approx 0.9636$

Summary of Distributions

Discrete Distributions

Continuous Distributions