International Baccalaureate · DP 2: Analysis and Approaches HL

Revision: Mixed Practice

⌚ ~35 min ▢ 4 blocks ✓ 13 exercises ➣ Prerequisites : Revision: Probability and Statistics, Revision: Integration and Kinematics

I Toolbox: The Sum of an Infinite Geometric Series

Recappreviously studied — click to reread

Proposition — Convergence of a Geometric Series

An infinite geometric series (with non-zero first term) converges to a finite sum if and only if its common ratio $r$ is between -1 and 1, i.e., $\boldsymbol{|r| < 1}$.
If the series converges, its sum to infinity, denoted $S_\infty$, is given by the formula:$$\boldsymbol{S_\infty = \frac{u_1}{1-r}}$$ where $u_1$ is the first term and $|r|<1$.If the sequence starts at $u_0$, the formula is:$$\boldsymbol{S_\infty = \frac{u_0}{1-r}}$$

Example

Find the sum of the infinite geometric series$$16 + 8 + 4 + 2 + 1 + \cdots$$

Answer

The first term is $u_1 = 16$ and the common ratio is $r = \dfrac{8}{16} = \dfrac{1}{2}$.
Since $|r| = \dfrac{1}{2} < 1$, the series converges and the sum-to-infinity formula applies:$$\begin{aligned}[t] S_\infty &= \frac{u_1}{1 - r} && {\scriptstyle(|r| < 1)} \\ &= \frac{16}{1 - \tfrac{1}{2}} = \frac{16}{\tfrac{1}{2}} \\ &= 32. \end{aligned}$$

Skills to practice

Calculating the Sum of an Infinite Geometric Series

II Toolbox: The Inverse Normal Function

Recappreviously studied — click to reread

Definition — Quantile

The $p$-quantile of a random variable $X$ is the value $x_p$ such that $P(X \leq x_p) = p$. For example, the $0.95$-quantile (or 95th percentile) is the value $x_{0.95}$ below which 95$\pourcent$ of the distribution's values lie.

Method — Finding a Value with the Inverse Normal Function

The reverse question gives a probability and asks for a value of $X$. The inverse normal function of the calculator (Inverse Normal, invNorm) returns the value $k$ such that $P(X < k) = p$, for $0 < p < 1$, from the area $p$ to the left of $k$, $\sigma$ and $\mu$.

State the model $X \sim \mathcal{N}(\mu, \sigma^2)$ and name the unknown value $k$.
Sketch and shade: place $k$ on the side of $\mu$ that the probability indicates — a left area less than $0.5$ puts $k$ below the mean — and shade the given area.
Turn the condition into a left area. The condition $P(X < k) = p$ is entered as it is; $P(X > k) = q$ becomes $P(X < k) = 1 - q$. An interval centred on $\mu$ that contains the proportion $p$ of the values leaves $\frac{1 - p}{2}$ in each tail: its ends are the values with left areas $\frac{1 - p}{2}$ and $\frac{1 + p}{2}$.
Enter the area, $\mu$ and $\sigma$, each in its field (the command invNorm takes the area, $\mu$, then $\sigma$; the tail is set to the left, on the models that ask for one), then check the position of $k$ against the sketch and report it to $3$ significant figures, or to the precision of the data.

The mock-up finds $k$ such that $P(X < k) = 0.9$ for $X \sim \mathcal{N}(64, 5^2)$: a left area of $0.9$ puts $k$ above the mean, and $k \approx 70.4$.

Example

The scores $X$ on an entrance test are normally distributed with mean $500$ and standard deviation $80$.

A scholarship is offered to the top $12\pourcent$ of the candidates. Find the minimum score for a scholarship.
The lowest $5\pourcent$ of the candidates are offered extra support. Find the score below which a candidate is offered support.

Answer

The model is $X \sim \mathcal{N}(500, 80^2)$.

Let $k$ be the minimum score: $P(X > k) = 0.12$. The right tail is small, so $k$ lies above the mean. As a left area, $$\begin{aligned}[t] P(X < k) &= 1 - P(X > k) && \text{(complementary event)}\\ &= 1 - 0.12\\ &= 0.88. \end{aligned}$$ The inverse normal function with area $0.88$, $\sigma = 80$ and $\mu = 500$ gives $593.9989\ldots$: the minimum score for a scholarship is about $594$.
Let $m$ be the score: $P(X < m) = 0.05$, a left tail, so $m$ lies below the mean. The inverse normal function with area $0.05$, $\sigma = 80$ and $\mu = 500$ gives $368.4117\ldots$: a candidate who scores below about $368$ is offered support.

Skills to practice

Finding a Value with the Inverse Normal Function

III Exam-Style Questions

Skills to practice

Working without Technology: Mixed Practice
Working with Technology: Mixed Practice