✦ Unit 3 — Statistics ✦
Question 1
Mean / Median / Mode
📖 Concept: Measures of Central Tendency
Mean: \(\bar{x} = \dfrac{\sum x}{n}\) — Median: middle value when data sorted — Mode: most frequent value.
For a grouped frequency table, the mean is estimated as \(\bar{x} = \dfrac{\sum fx}{\sum f}\) where \(x\) is the midpoint of each class.
For a grouped frequency table, the mean is estimated as \(\bar{x} = \dfrac{\sum fx}{\sum f}\) where \(x\) is the midpoint of each class.
📝 Worked Example
Data: 3, 7, 7, 8, 10, 12.
Mean \(= \dfrac{3+7+7+8+10+12}{6} = \dfrac{47}{6} \approx 7.83\) | Median \(= \dfrac{7+8}{2} = 7.5\) | Mode \(= 7\).
Mean \(= \dfrac{3+7+7+8+10+12}{6} = \dfrac{47}{6} \approx 7.83\) | Median \(= \dfrac{7+8}{2} = 7.5\) | Mode \(= 7\).
Question
The data set \(4,\, 6,\, 6,\, 9,\, 11,\, 14\) has mean \(\bar{x}\) and median \(M\). Which statement is correct?
The data set \(4,\, 6,\, 6,\, 9,\, 11,\, 14\) has mean \(\bar{x}\) and median \(M\). Which statement is correct?
Question 2
IQR & Outliers
📖 Concept: Interquartile Range & Outlier Rule
\(\text{IQR} = Q_3 - Q_1\).
A value \(x\) is an outlier if \(x < Q_1 - 1.5\times\text{IQR}\) or \(x > Q_3 + 1.5\times\text{IQR}\).
A value \(x\) is an outlier if \(x < Q_1 - 1.5\times\text{IQR}\) or \(x > Q_3 + 1.5\times\text{IQR}\).
📝 Worked Example
\(Q_1=10,\; Q_3=20\Rightarrow \text{IQR}=10\).
Lower fence: \(10-15=-5\). Upper fence: \(20+15=35\). Value 40 is an outlier; value 2 is not.
Lower fence: \(10-15=-5\). Upper fence: \(20+15=35\). Value 40 is an outlier; value 2 is not.
Question
For a data set with \(Q_1 = 15\) and \(Q_3 = 31\), which value would be classified as an outlier?
For a data set with \(Q_1 = 15\) and \(Q_3 = 31\), which value would be classified as an outlier?
Question 3
Standard Deviation
📖 Concept: Standard Deviation
Population SD: \(\sigma = \sqrt{\dfrac{\sum(x-\bar{x})^2}{n}}\). Sample SD: \(s = \sqrt{\dfrac{\sum(x-\bar{x})^2}{n-1}}\).
A larger SD means data is more spread out from the mean.
A larger SD means data is more spread out from the mean.
📝 Worked Example
Dataset A: 5,5,5,5 → \(\sigma=0\). Dataset B: 1,3,7,9 → much larger \(\sigma\). If each value increases by a constant \(k\), the mean shifts but \(\sigma\) stays the same.
Question
A dataset has mean \(\bar{x}=20\) and standard deviation \(\sigma=4\). Every value in the dataset is multiplied by 3. What are the new mean and standard deviation?
A dataset has mean \(\bar{x}=20\) and standard deviation \(\sigma=4\). Every value in the dataset is multiplied by 3. What are the new mean and standard deviation?
Question 4
Cumulative Frequency & Percentiles
📖 Concept: Cumulative Frequency
Cumulative frequency is a running total of frequencies. To estimate the median from a cumulative frequency graph, read off the value at \(\dfrac{n}{2}\). For \(Q_1\): at \(\dfrac{n}{4}\), for \(Q_3\): at \(\dfrac{3n}{4}\).
📝 Worked Example
\(n=80\) students. Median \(\Rightarrow\) read at cum. freq. = 40. \(Q_1 \Rightarrow\) read at 20. \(Q_3 \Rightarrow\) read at 60.
Question
A cumulative frequency curve is drawn for 120 students' test scores. At what cumulative frequency value should you read to estimate the 75th percentile (\(Q_3\))?
A cumulative frequency curve is drawn for 120 students' test scores. At what cumulative frequency value should you read to estimate the 75th percentile (\(Q_3\))?
Question 5
Histogram — Frequency Density
📖 Concept: Frequency Density
When class widths differ, use: \(\text{Frequency Density} = \dfrac{\text{Frequency}}{\text{Class Width}}\).
In a histogram, the area of each bar \(=\) frequency (not the height).
In a histogram, the area of each bar \(=\) frequency (not the height).
📝 Worked Example
Class \([10, 20)\): freq \(=30\), width \(=10\) → freq. density \(=3\).
Class \([20, 50)\): freq \(=45\), width \(=30\) → freq. density \(=1.5\).
Class \([20, 50)\): freq \(=45\), width \(=30\) → freq. density \(=1.5\).
Question
A histogram bar for the class \([30, 50)\) has a frequency density of 2.5. How many data values fall in this class?
A histogram bar for the class \([30, 50)\) has a frequency density of 2.5. How many data values fall in this class?
Question 6
Correlation Coefficient
📖 Concept: Pearson's r
The correlation coefficient \(r\) satisfies \(-1 \le r \le 1\).
\(r \approx +1\): strong positive; \(r \approx -1\): strong negative; \(r \approx 0\): no linear correlation.
Rule of thumb: \(|r| > 0.7\) strong, \(0.3 < |r| \le 0.7\) moderate, \(|r| \le 0.3\) weak.
\(r \approx +1\): strong positive; \(r \approx -1\): strong negative; \(r \approx 0\): no linear correlation.
Rule of thumb: \(|r| > 0.7\) strong, \(0.3 < |r| \le 0.7\) moderate, \(|r| \le 0.3\) weak.
📝 Worked Example
\(r = -0.91\): strong negative correlation — as \(x\) increases, \(y\) tends to decrease significantly.
Question
A scatter diagram shows points that roughly follow a line sloping downward to the right, but fairly scattered. Which value of \(r\) is most likely?
A scatter diagram shows points that roughly follow a line sloping downward to the right, but fairly scattered. Which value of \(r\) is most likely?
Question 7
Linearisation — Log Transformation
📖 Concept: Linearising \(y = ab^x\)
Take \(\log\) of both sides: \(\log y = \log a + x\log b\).
Plot \(\log y\) vs \(x\) → straight line with gradient \(\log b\) and \(y\)-intercept \(\log a\).
Plot \(\log y\) vs \(x\) → straight line with gradient \(\log b\) and \(y\)-intercept \(\log a\).
📝 Worked Example
\(y = 5 \cdot 3^x \Rightarrow \log y = \log 5 + x\log 3\). Plot \(X = x\), \(Y = \log y\) → gradient \(= \log 3 \approx 0.477\), intercept \(= \log 5 \approx 0.699\).
Question
The model \(y = a \cdot b^x\) is linearised by plotting \(\log_{10} y\) against \(x\). The resulting line has gradient 0.301 and \(y\)-intercept 1. What are \(a\) and \(b\)?
The model \(y = a \cdot b^x\) is linearised by plotting \(\log_{10} y\) against \(x\). The resulting line has gradient 0.301 and \(y\)-intercept 1. What are \(a\) and \(b\)?
Question 8
Spearman Rank Correlation
📖 Concept: Spearman's \(r_s\)
\(r_s = 1 - \dfrac{6\sum d^2}{n(n^2-1)}\) where \(d\) = difference in ranks and \(n\) = number of pairs.
Used when data is ordinal or non-normal. \(r_s = +1\) means perfect agreement in rankings.
Used when data is ordinal or non-normal. \(r_s = +1\) means perfect agreement in rankings.
📝 Worked Example
\(n=5\), \(\sum d^2 = 10\): \(r_s = 1 - \dfrac{6\times10}{5\times24} = 1 - \dfrac{60}{120} = 0.5\).
Question
For \(n=6\) pairs with \(\sum d^2 = 14\), calculate Spearman's rank correlation coefficient.
For \(n=6\) pairs with \(\sum d^2 = 14\), calculate Spearman's rank correlation coefficient.
Question 9
Hypothesis Testing — t-test
📖 Concept: t-test & p-value
The one-sample t-test checks whether a sample mean differs significantly from a hypothesised value \(\mu_0\):
\(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\).
If \(p\text{-value} < \alpha\) (significance level), reject \(H_0\); otherwise, fail to reject.
\(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\).
If \(p\text{-value} < \alpha\) (significance level), reject \(H_0\); otherwise, fail to reject.
📝 Worked Example
\(H_0: \mu=50\). Sample gives \(p=0.03\). At \(\alpha=0.05\): \(0.03<0.05\) → reject \(H_0\). At \(\alpha=0.01\): \(0.03>0.01\) → fail to reject \(H_0\).
Question
A t-test gives a p-value of 0.08. At a significance level of \(\alpha = 0.05\), what is the correct conclusion?
A t-test gives a p-value of 0.08. At a significance level of \(\alpha = 0.05\), what is the correct conclusion?
Question 10
Chi-Square Test — Independence
📖 Concept: \(\chi^2\) Test of Independence
\(\chi^2 = \sum \dfrac{(O-E)^2}{E}\) where \(O\) = observed, \(E\) = expected.
Degrees of freedom: \(\nu = (\text{rows}-1)(\text{cols}-1)\).
If \(\chi^2 > \chi^2_{\text{critical}}\), reject \(H_0\) (variables are NOT independent).
Degrees of freedom: \(\nu = (\text{rows}-1)(\text{cols}-1)\).
If \(\chi^2 > \chi^2_{\text{critical}}\), reject \(H_0\) (variables are NOT independent).
📝 Worked Example
2×3 table: \(\nu=(2-1)(3-1)=2\). At \(\alpha=0.05\), \(\chi^2_{\text{crit}}=5.991\).
If calculated \(\chi^2=7.5>5.991\), reject \(H_0\) — the variables are dependent.
If calculated \(\chi^2=7.5>5.991\), reject \(H_0\) — the variables are dependent.
Question
In a \(\chi^2\) test of independence on a 3×4 contingency table, what are the degrees of freedom?
In a \(\chi^2\) test of independence on a 3×4 contingency table, what are the degrees of freedom?
Question 11
Box Plot & Five-Number Summary
📖 Concept: Box Plot
A box plot displays: Min, \(Q_1\), Median (\(Q_2\)), \(Q_3\), Max.
The box spans \(Q_1\) to \(Q_3\) (the IQR). Outliers are plotted separately beyond the fences. Skewness: if median is closer to \(Q_1\) → positive skew; closer to \(Q_3\) → negative skew.
The box spans \(Q_1\) to \(Q_3\) (the IQR). Outliers are plotted separately beyond the fences. Skewness: if median is closer to \(Q_1\) → positive skew; closer to \(Q_3\) → negative skew.
📝 Worked Example
Five-number summary: 5, 12, 20, 28, 40.
Median (20) is equidistant from \(Q_1\) and \(Q_3\) → roughly symmetric distribution.
Median (20) is equidistant from \(Q_1\) and \(Q_3\) → roughly symmetric distribution.
Question
A box plot has \(Q_1 = 10\), median \(= 14\), \(Q_3 = 30\). What does this suggest about the skewness of the distribution?
A box plot has \(Q_1 = 10\), median \(= 14\), \(Q_3 = 30\). What does this suggest about the skewness of the distribution?
Question 12
Mean from Frequency Table
📖 Concept: Estimated Mean from Grouped Data
\(\bar{x} = \dfrac{\sum f \cdot m}{\sum f}\) where \(m\) is the midpoint of each class and \(f\) is the frequency.
📝 Worked Example
Class [0,10): \(f=5,\,m=5\); Class [10,20): \(f=10,\,m=15\).
\(\bar{x} = \dfrac{5\times5 + 10\times15}{15} = \dfrac{25+150}{15} = \dfrac{175}{15} \approx 11.67\).
\(\bar{x} = \dfrac{5\times5 + 10\times15}{15} = \dfrac{25+150}{15} = \dfrac{175}{15} \approx 11.67\).
Question
A grouped frequency table shows: \([0,10)\!:\!f\!=\!4;\;[10,20)\!:\!f\!=\!6;\;[20,30)\!:\!f\!=\!10\). Estimate the mean.
A grouped frequency table shows: \([0,10)\!:\!f\!=\!4;\;[10,20)\!:\!f\!=\!6;\;[20,30)\!:\!f\!=\!10\). Estimate the mean.
✦ Unit 4 — Geometry, Trigonometry & Vectors ✦
Question 13
2D Coordinate Geometry
📖 Concept: Distance & Midpoint
Distance: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
Midpoint: \(M = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
Gradient: \(m = \dfrac{y_2-y_1}{x_2-x_1}\). Perpendicular gradient: \(m_\perp = -\dfrac{1}{m}\).
Midpoint: \(M = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
Gradient: \(m = \dfrac{y_2-y_1}{x_2-x_1}\). Perpendicular gradient: \(m_\perp = -\dfrac{1}{m}\).
📝 Worked Example
\(A(1,3)\), \(B(5,7)\): \(d=\sqrt{16+16}=4\sqrt{2}\), midpoint \(=(3,5)\), gradient \(=1\), perp. gradient \(=-1\).
Question
Line \(AB\) has gradient \(\dfrac{3}{4}\). A line perpendicular to \(AB\) passes through \((2, 5)\). What is its equation?
Line \(AB\) has gradient \(\dfrac{3}{4}\). A line perpendicular to \(AB\) passes through \((2, 5)\). What is its equation?
Question 14
Sine Rule
📖 Concept: Sine Rule
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
Use when you know: (i) two angles and a side (AAS/ASA), or (ii) two sides and a non-included angle (SSA — beware the ambiguous case!).
Use when you know: (i) two angles and a side (AAS/ASA), or (ii) two sides and a non-included angle (SSA — beware the ambiguous case!).
📝 Worked Example
\(a=8,\; A=35°,\; B=62°\). Find \(b\):
\(\dfrac{b}{\sin 62°}=\dfrac{8}{\sin 35°}\Rightarrow b = \dfrac{8\sin 62°}{\sin 35°} \approx 12.3\).
\(\dfrac{b}{\sin 62°}=\dfrac{8}{\sin 35°}\Rightarrow b = \dfrac{8\sin 62°}{\sin 35°} \approx 12.3\).
Question
In triangle \(ABC\), \(a = 10\), \(A = 45°\), \(B = 75°\). Find side \(b\) to 3 s.f.
In triangle \(ABC\), \(a = 10\), \(A = 45°\), \(B = 75°\). Find side \(b\) to 3 s.f.
Question 15
Cosine Rule
📖 Concept: Cosine Rule
To find a side: \(a^2 = b^2 + c^2 - 2bc\cos A\).
To find an angle: \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\).
Use when: (i) all three sides known (SSS), or (ii) two sides and the included angle (SAS).
To find an angle: \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\).
Use when: (i) all three sides known (SSS), or (ii) two sides and the included angle (SAS).
📝 Worked Example
\(b=7,\;c=9,\;A=50°\):
\(a^2 = 49+81-2(7)(9)\cos50° \approx 130-80.8 = 49.2 \Rightarrow a \approx 7.01\).
\(a^2 = 49+81-2(7)(9)\cos50° \approx 130-80.8 = 49.2 \Rightarrow a \approx 7.01\).
Question
In a triangle, \(b = 5\), \(c = 8\), \(A = 60°\). Find side \(a\).
In a triangle, \(b = 5\), \(c = 8\), \(A = 60°\). Find side \(a\).
Question 16
Vectors — Magnitude & Direction
📖 Concept: Vector Operations
For \(\vec{v} = \begin{pmatrix}a\\b\end{pmatrix}\): magnitude \(|\vec{v}| = \sqrt{a^2+b^2}\).
Unit vector: \(\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}\).
Dot product: \(\vec{u}\cdot\vec{v} = u_1v_1+u_2v_2 = |\vec{u}||\vec{v}|\cos\theta\).
Unit vector: \(\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}\).
Dot product: \(\vec{u}\cdot\vec{v} = u_1v_1+u_2v_2 = |\vec{u}||\vec{v}|\cos\theta\).
📝 Worked Example
\(\vec{u}=\begin{pmatrix}3\\4\end{pmatrix}\): \(|\vec{u}|=5\), unit vector \(=\begin{pmatrix}0.6\\0.8\end{pmatrix}\).
Question
Find the angle between \(\vec{a} = \begin{pmatrix}1\\2\end{pmatrix}\) and \(\vec{b} = \begin{pmatrix}3\\1\end{pmatrix}\).
Find the angle between \(\vec{a} = \begin{pmatrix}1\\2\end{pmatrix}\) and \(\vec{b} = \begin{pmatrix}3\\1\end{pmatrix}\).
Question 17
Unit Circle & Exact Values
📖 Concept: Exact Trigonometric Values
Key exact values:
\(\sin 30°=\tfrac{1}{2},\;\cos 30°=\tfrac{\sqrt3}{2},\;\tan 30°=\tfrac{1}{\sqrt3}\)
\(\sin 45°=\tfrac{\sqrt2}{2},\;\cos 45°=\tfrac{\sqrt2}{2},\;\tan 45°=1\)
\(\sin 60°=\tfrac{\sqrt3}{2},\;\cos 60°=\tfrac{1}{2},\;\tan 60°=\sqrt3\)
\(\sin 30°=\tfrac{1}{2},\;\cos 30°=\tfrac{\sqrt3}{2},\;\tan 30°=\tfrac{1}{\sqrt3}\)
\(\sin 45°=\tfrac{\sqrt2}{2},\;\cos 45°=\tfrac{\sqrt2}{2},\;\tan 45°=1\)
\(\sin 60°=\tfrac{\sqrt3}{2},\;\cos 60°=\tfrac{1}{2},\;\tan 60°=\sqrt3\)
📝 Worked Example
\(\sin^2\theta + \cos^2\theta = 1\) always. So \(\sin^2 60° + \cos^2 60° = \tfrac{3}{4}+\tfrac{1}{4}=1\). ✓
Question
Simplify \(\dfrac{\sin 60° \cdot \cos 30°}{\tan 45°}\).
Simplify \(\dfrac{\sin 60° \cdot \cos 30°}{\tan 45°}\).
Question 18
Pythagorean Identity
📖 Concept: Trig Identities
\(\sin^2\theta + \cos^2\theta = 1\)
\(\Rightarrow \tan^2\theta + 1 = \sec^2\theta\)
\(\Rightarrow 1 + \cot^2\theta = \csc^2\theta\)
These are used to simplify expressions or solve equations.
\(\Rightarrow \tan^2\theta + 1 = \sec^2\theta\)
\(\Rightarrow 1 + \cot^2\theta = \csc^2\theta\)
These are used to simplify expressions or solve equations.
📝 Worked Example
Solve \(2\sin^2\theta - \sin\theta - 1 = 0\) for \(0°\le\theta\le360°\).
Factor: \((2\sin\theta+1)(\sin\theta-1)=0 \Rightarrow \sin\theta=-\tfrac{1}{2}\) or \(\sin\theta=1\).
\(\theta = 90°, 210°, 330°\).
Factor: \((2\sin\theta+1)(\sin\theta-1)=0 \Rightarrow \sin\theta=-\tfrac{1}{2}\) or \(\sin\theta=1\).
\(\theta = 90°, 210°, 330°\).
Question
If \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in the first quadrant, what is \(\cos\theta\)?
If \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in the first quadrant, what is \(\cos\theta\)?
Question 19
Vectors in Proofs — Collinearity
📖 Concept: Collinearity via Vectors
Points \(P, Q, R\) are collinear if \(\overrightarrow{PQ} = k\,\overrightarrow{PR}\) for some scalar \(k\).
Equivalently, \(\overrightarrow{PQ} \parallel \overrightarrow{QR}\) (one is a scalar multiple of the other) and they share a common point.
Equivalently, \(\overrightarrow{PQ} \parallel \overrightarrow{QR}\) (one is a scalar multiple of the other) and they share a common point.
📝 Worked Example
\(A(1,2), B(3,6), C(5,10)\).
\(\overrightarrow{AB}=\begin{pmatrix}2\\4\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}4\\8\end{pmatrix} = 2\overrightarrow{AB}\). So A, B, C are collinear.
\(\overrightarrow{AB}=\begin{pmatrix}2\\4\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}4\\8\end{pmatrix} = 2\overrightarrow{AB}\). So A, B, C are collinear.
Question
Points \(A(1,3)\), \(B(4,7)\), \(C(7,k)\) are collinear. Find \(k\).
Points \(A(1,3)\), \(B(4,7)\), \(C(7,k)\) are collinear. Find \(k\).
Question 20
Solving Trig Equations
📖 Concept: Solving \(\sin\theta = k\) over an Interval
Step 1: Find principal value \(\alpha = \arcsin(k)\).
Step 2: Use symmetry of the sine curve — if \(\alpha\) is a solution, so is \(180°-\alpha\) (for \(0°\le\theta\le360°\)).
For \(\cos\theta=k\): solutions are \(\pm\alpha\) (i.e. \(\alpha\) and \(360°-\alpha\)).
For \(\tan\theta=k\): solutions repeat every \(180°\).
Step 2: Use symmetry of the sine curve — if \(\alpha\) is a solution, so is \(180°-\alpha\) (for \(0°\le\theta\le360°\)).
For \(\cos\theta=k\): solutions are \(\pm\alpha\) (i.e. \(\alpha\) and \(360°-\alpha\)).
For \(\tan\theta=k\): solutions repeat every \(180°\).
📝 Worked Example
Solve \(\cos\theta = -\tfrac{1}{2}\) for \(0°\le\theta\le360°\).
Principal: \(\arccos(\tfrac{1}{2})=60°\). Since cosine is negative → 2nd and 3rd quadrants.
\(\theta = 180°-60°=120°\) and \(\theta=180°+60°=240°\).
Principal: \(\arccos(\tfrac{1}{2})=60°\). Since cosine is negative → 2nd and 3rd quadrants.
\(\theta = 180°-60°=120°\) and \(\theta=180°+60°=240°\).
Question
Solve \(\sin\theta = \dfrac{\sqrt{2}}{2}\) for \(0° \le \theta \le 360°\).
Solve \(\sin\theta = \dfrac{\sqrt{2}}{2}\) for \(0° \le \theta \le 360°\).