International Baccalaureate · DP 2: Analysis and Approaches HL

Revision: Probability and Statistics

⌚ ~51 min ▢ 6 blocks ✓ 17 exercises ➣ Prerequisites : Continuous Random Variables and the Normal Distribution, Conditional Probability

I Toolbox: Joint Probability with Trees

Recappreviously studied — click to reread

Proposition — Joint Probability Formula

For $P(E) > 0$ and $P(F) > 0$ respectively: $$P(E \cap F) = P(E) \times \PCond{F}{E}, \quad P(E \cap F) = P(F) \times \PCond{E}{F}.$$

Example

For this probability tree,

find $P(S \cap G)$.

Answer

Path: $S$ to $G$ (highlighted):
Calculate: $$\begin{aligned} P(S \cap G) &= P(S) \times \PCond{G}{S}\\ &= \frac{6}{10} \times \frac{1}{2}\\ &= \frac{3}{10}. \end{aligned}$$

Skills to practice

Calculating Joint Probabilities with Trees

II Toolbox: Finding an Unknown Mean or Standard Deviation

Recappreviously studied — click to reread

Proposition — Standardising a Normal Variable

If $X \sim \mathcal{N}(\mu, \sigma^2)$, then the standardised variable $$Z = \frac{X - \mu}{\sigma}$$ follows the standard normal distribution: $Z \sim \mathcal{N}(0, 1)$. Consequently, for every real number $x$, $$P(X \leq x) = P(Z \leq z) \qquad \text{where} \qquad z = \frac{x - \mu}{\sigma},$$ and, for $x_1 < x_2$ with standardised values $z_1$ and $z_2$, $P(x_1 \leq X \leq x_2) = P(z_1 \leq Z \leq z_2)$.
The number $z$ is the standardised value (or $z$-value) of $x$: the number of standard deviations between $x$ and the mean, positive when $x > \mu$ and negative when $x < \mu$.

Method — Finding an Unknown Mean or Standard Deviation

When $\mu$ or $\sigma$ is unknown, the inverse normal function cannot be used on $X$, since it needs both parameters. Standardising moves the condition to $Z \sim \mathcal{N}(0, 1)$, whose parameters are known.

Write each condition as a left area: $P(X < x) = p$, where a condition $P(X > x) = q$ becomes $P(X < x) = 1 - q$.
Sketch: place $x$ on the side of $\mu$ that $p$ indicates; a left area less than $0.5$ puts $x$ below the mean, so its standardised value is negative.
Standardise and read $z$: $P(X < x) = P\left(Z < \frac{x - \mu}{\sigma}\right) = p$, so $\frac{x - \mu}{\sigma} = z$, where $z$ is given by the inverse normal function with area $p$, $\sigma = 1$ and $\mu = 0$.
Solve for the unknown, from $x - \mu = z\sigma$:
- $\sigma$ known: $\mu = x - z\sigma$;
- $\mu$ known: $\sigma = \frac{x - \mu}{z}$, which requires $z \neq 0$. A condition at the median ($p = 0.5$, so $z = 0$) only says that $x = \mu$ and gives no information on $\sigma$;
- both unknown: two conditions with different values of $z$ give the system $\mu + z_1\sigma = x_1$ and $\mu + z_2\sigma = x_2$; subtracting the equations gives $\sigma = \frac{x_2 - x_1}{z_2 - z_1}$, then $\mu = x_1 - z_1\sigma$.
Keep $z$ unrounded in the calculator's memory, round $\mu$ and $\sigma$ only at the end, and check: $\sigma$ must be positive, and the normal cumulative distribution function with the values found must give back the probabilities of the question.

The mock-up gives the standardised value with left area $0.03$: $z \approx -1.881$, negative because the area is less than $0.5$.

Example

A machine fills jars of honey. The mass $X$ of honey in a jar, in grams, is normally distributed with standard deviation $4$ g.

$3\pourcent$ of the jars contain less than $450$ g. Find the mean mass $\mu$, to $1$ decimal place.
The machine is set to the mean $\mu = 456$ g, and a maintenance can reduce the standard deviation. Find the standard deviation $\sigma$ for which only $1\pourcent$ of the jars contain less than $450$ g, to $3$ significant figures.

Answer

The model is $X \sim \mathcal{N}(\mu, 4^2)$ with $\mu$ unknown, and $P(X < 450) = 0.03$: a small left area, so $450$ lies below the mean. Standardising, $P\left(Z < \frac{450 - \mu}{4}\right) = 0.03$, so $\frac{450 - \mu}{4} = z$, where the inverse normal function with area $0.03$, $\sigma = 1$ and $\mu = 0$ gives $z = -1.8807936\ldots$ Then $$\begin{aligned}[t] \frac{450 - \mu}{4} &= z && \text{(standardised value of 450)}\\ 450 - \mu &= 4z\\ \mu &= 450 - 4z\\ &= 450 - 4 \times (-1.8807936\ldots)\\ &= 457.5231\ldots \end{aligned}$$ The mean mass is $\mu \approx 457.5$ g. Check: with $\mu = 457.5231\ldots$ and $\sigma = 4$, the normal cumulative distribution function gives $P(X < 450) = 0.0300$.
Now $\mu = 456$ is known and $\sigma$ is unknown: $P(X < 450) = 0.01$, so $\frac{450 - 456}{\sigma} = z$, where the inverse normal function with area $0.01$, $\sigma = 1$ and $\mu = 0$ gives $z = -2.3263479\ldots$, which is not zero. Then $$\begin{aligned}[t] \frac{450 - 456}{\sigma} &= z && \text{(standardised value of 450)}\\ \sigma &= \frac{450 - 456}{z}\\ &= \frac{-6}{-2.3263479\ldots}\\ &= 2.5791\ldots \end{aligned}$$ The standard deviation must be reduced to $\sigma \approx 2.58$ g. It is positive, as it must be: $450 - 456$ and $z$ are both negative, since $450$ lies below the mean.

Skills to practice

Finding an Unknown Mean or Standard Deviation

III Exam-Style Questions

Skills to practice

Working without Technology: Probability and Statistics
Working with Technology: Probability and Statistics
Working without Technology: Further Probability
Working with Technology: Further Probability