Statistics

Statistics

:::info Board Coverage AQA Paper 2 | Edexcel Paper 2 | OCR Paper 3 (Higher) / Paper 6 (Foundation) | WJEC Unit 2 :::

1. Data Collection

1.1 Types of Data

Qualitative (categorical) data describes qualities or characteristics (e.g. Colour, type of Car).

Quantitative data is numerical:

• Discrete: can only take specific values (e.g. Number of siblings: 0, 1, 2, …).
• Continuous: can take any value within a range (e.g. Height, temperature).

1.2 Sampling Methods

Method	Description	Advantage	Disadvantage
Simple random	Every member has equal chance	Unbiased	May not represent subgroups
Systematic	Every $k$-th member selected	Easy to implement	Can coincide with a pattern
Stratified	Sample proportionally from subgroups	Represents all groups	Requires knowledge of groups
Quota	Interviewer fills quotas	Quick	Not truly random
Opportunity	Whoever is available	Convenient	Highly biased

Stratified sampling formula. If a population of size $N$ contains a subgroup of size $n$Then The sample from that subgroup should be:

\mathrm{sample size = \frac{n}{N} \times \mathrm{total sample size

Worked Example. A school has 600 boys and 400 girls. A sample of 50 students is needed. How many Boys and girls should be in the sample?

\mathrm{Boys = \frac{600}{1000} \times 50 = 30, \qquad \mathrm{Girls = \frac{400}{1000} \times 50 = 20

Worked Example (Higher Tier). A population of 1200 is divided into three age groups: Under 18 (300), 18—40 (500), Over 40 (400). A stratified sample of 60 is required.

Under 18: $\frac{300}{1200} \times 60 = 15$.

18—40: $\frac{500}{1200} \times 60 = 25$.

Over 40: $\frac{400}{1200} \times 60 = 20$.

1.3 Bias in Data Collection

A sample is biased if it does not fairly represent the population. Sources of bias include:

• Sampling from a non-representative subset
• Leading questions in surveys
• Non-response (voluntary response bias)
• Using convenience sampling

:::caution A large sample size does not fix a biased sampling method. A sample of 10000 taken from Only one school is still biased if you want to draw conclusions about all schools in the country. :::

2. Averages and Measures of Central Tendency

2.1 Mean, Median, and Mode

Measure	Definition	When to use
Mean	$\bar{x} = \frac{\sum x}{n}$	Uses all data; affected by outliers
Median	Middle value when data is ordered	Not affected by outliers
Mode	Most frequent value	Useful for categorical data

Worked Example. Find the mean, median, and mode of: $3, 5, 2, 7, 5, 8, 5, 1, 9, 4$.

Ordered: $1, 2, 3, 4, 5, 5, 5, 7, 8, 9$.

\mathrm{Mean = \frac{1 + 2 + 3 + 4 + 5 + 5 + 5 + 7 + 8 + 9}{10} = \frac{49}{10} = 4.9

\mathrm{Median = \frac{5 + 5}{2} = 5

\mathrm{Mode = 5

2.2 Mean from a Frequency Table

$\bar{x} = \frac{\sum (f \times x)}{\sum f}$

Worked Example.

Score $x$	Frequency $f$	$f \times x$
1	4	4
2	7	14
3	12	36
4	9	36
5	3	15
Total	35	105

$\bar{x} = \frac{105}{35} = 3$

2.3 Estimated Mean from Grouped Data

For grouped data, use the midpoint of each class as an estimate for $x$.

Worked Example.

Height (cm)	Frequency $f$	Midpoint $x$	$f \times x$
$140 \leq h \lt 150$	5	145	725
$150 \leq h \lt 160$	12	155	1860
$160 \leq h \lt 170$	18	165	2970
$170 \leq h \lt 180$	10	175	1750
$180 \leq h \lt 190$	5	185	925
Total	50		8230

\mathrm{Estimated mean = \frac{8230}{50} = 164.6 \mathrm{ cm

2.4 Estimated Median and Interquartile Range from Grouped Data

To find the median class: find the position $\frac{n}{2}$ and locate it in the cumulative frequency.

Interquartile range (IQR): \mathrm{IQR = Q_3 - Q_1.

This is a measure of spread that is not affected by outliers.

Worked Example (Higher Tier). Using the data above, estimate the median height.

Cumulative frequencies: 5, 17, 35, 45, 50.

Median position: $\frac{50}{2} = 25$. This falls in the $160 \leq h \lt 170$ class (between 17 and 35).

Using linear interpolation within the class:

\mathrm{Median = 160 + \frac{25 - 17}{35 - 17} \times 10 = 160 + \frac{8}{18} \times 10 = 160 + 4.44 = 164.4 \mathrm{ cm

2.5 Choosing the Right Average

• Use the mean when data is roughly symmetric and there are no extreme outliers.
• Use the median when data is skewed or has outliers (e.g., income data).
• Use the mode for categorical data (e.g., favourite colour).

Worked Example (Higher Tier). The salaries of five employees are: 18000, 22000, 25000, 28000, 95000 (pounds). Compare the mean and median.

Mean $= \frac{188000}{5} = 37600$. Median $= 25000$.

The mean is heavily distorted by the outlier (95000). The median of 25000 is a much better Representation of the typical salary.

3. Measures of Spread

3.1 Range

\mathrm{Range = \mathrm{maximum value - \mathrm{minimum value

Simple but affected by outliers.

3.2 Interquartile Range

\mathrm{IQR = Q_3 - Q_1

Where $Q_1$ is the lower quartile (25th percentile) and $Q_3$ is the upper quartile (75th Percentile).

Finding quartiles from raw data:

For $n$ data values in order:

• $Q_1$ position: $\frac{n + 1}{4}$
• $Q_3$ position: $\frac{3(n + 1)}{4}$

3.3 Box Plots

A box plot displays the minimum, $Q_1$Median, $Q_3$And maximum.

Worked Example. Data set: $2, 5, 7, 8, 10, 12, 15, 18, 22, 25$.

• Minimum: 2
• $Q_1$ position: $\frac{11}{4} = 2.75$So $Q_1 = 7 + 0.75(8 - 7) = 7.75$
• Median position: 5.5, so median $= \frac{10 + 12}{2} = 11$
• $Q_3$ position: $\frac{33}{4} = 8.25$So $Q_3 = 18 + 0.25(22 - 18) = 19$
• Maximum: 25
• \mathrm{IQR = 19 - 7.75 = 11.25

3.4 Outliers

A common definition: a value is an outlier if it is more than 1.5 \times \mathrm{IQR below $Q_1$ or above $Q_3$.

\mathrm{Lower fence = Q_1 - 1.5 \times \mathrm{IQR \mathrm{Upper fence = Q_3 + 1.5 \times \mathrm{IQR

Worked Example. Using the data above, identify any outliers.

Lower fence $= 7.75 - 1.5 \times 11.25 = 7.75 - 16.875 = -9.125$.

Upper fence $= 19 + 16.875 = 35.875$.

Since all values are between $-9.125$ and $35.875$There are no outliers.

4. Representing Data

4.1 Charts and Diagrams

Chart	Best for
Bar chart	Comparing categories
Pie chart	Showing proportions
Pictogram	Simple visual comparison
Scatter graph	Relationship between two variables
Histogram	Continuous data with unequal class widths
Cumulative frequency	Finding medians and quartiles

4.2 Histograms

In a histogram, the area of each bar represents the frequency, not the height.

\mathrm{Frequency density = \frac{\mathrm{frequency}{\mathrm{class width}

Worked Example.

Distance (km)	Frequency	Class width	Frequency density
$0 \leq d \lt 10$	15	10	1.5
$10 \leq d \lt 25$	30	15	2.0
$25 \leq d \lt 40$	45	15	3.0
$40 \leq d \lt 60$	20	20	1.0

The bar heights (frequency densities) are 1.5, 2.0, 3.0, and 1.0 respectively.

Worked Example (Higher Tier). A histogram has a bar of width 5 and height 4. What frequency does This bar represent?

\mathrm{Frequency = \mathrm{frequency density \times \mathrm{class width = 4 \times 5 = 20.

Worked Example (Higher Tier). A histogram has bars with frequency densities 3, 5, 2, and 4 for Classes of width 4, 6, 8, and 3 respectively. Find the total frequency represented.

$3 \times 4 + 5 \times 6 + 2 \times 8 + 4 \times 3 = 12 + 30 + 16 + 12 = 70$.

4.3 Cumulative Frequency

A cumulative frequency curve (ogive) is plotted by adding frequencies as you go. The $x$-axis Shows the upper boundary of each class.

From the curve you can read:

• Median (at $\frac{n}{2}$)
• Lower quartile $Q_1$ (at $\frac{n}{4}$)
• Upper quartile $Q_3$ (at $\frac{3n}{4}$)

Worked Example. Using the distance data above, draw the cumulative frequency curve and estimate The median distance.

Cumulative frequencies: 15, 45, 90, 110.

Total $n = 110$. Median at position 55.

$Q_1$ at position 27.5, $Q_3$ at position 82.5.

From the ogive, the median falls in the $25 \leq d \lt 40$ class.

4.4 Scatter Graphs and Correlation

Correlation describes the relationship between two variables:

Type	Description
Positive	As $x$ increases, $y$ tends to increase
Negative	As $x$ increases, $y$ tends to decrease
No correlation	No apparent relationship

Line of best fit: A straight line drawn through the data that approximately follows the trend. It should pass through the mean point $(\bar{x}, \bar{y})$ and have roughly equal numbers of points On each side.

:::caution Correlation does not imply causation. Two variables may be correlated due to a third Factor, or by coincidence. :::

4.5 Interpolation and Extrapolation

• Interpolation: Estimating values within the range of the data. Generally reliable.
• Extrapolation: Estimating values outside the range of the data. Unreliable.

Worked Example (Higher Tier). A scatter graph of exam score against hours of revision shows a Strong positive correlation. The line of best fit passes through $(2, 35)$ and $(10, 85)$. Estimate The exam score for a student who revised for 7 hours.

Gradient: $m = \frac{85 - 35}{10 - 2} = \frac{50}{8} = 6.25$.

Equation: $y - 35 = 6.25(x - 2)$So $y = 6.25x + 22.5$.

At $x = 7$: $y = 6.25 \times 7 + 22.5 = 43.75 + 22.5 = 66.25$.

The estimated exam score is approximately 66.

5. Probability

5.1 Basic Probability

P(\mathrm{event) = \frac{\mathrm{number of favourable outcomes}{\mathrm{total number of outcomes}

The probability of any event satisfies $0 \leq P(E) \leq 1$.

Complement rule: P(\mathrm{not E) = 1 - P(E)

Mutually exclusive events: P(A \mathrm{ or B) = P(A) + P(B)

Independent events: P(A \mathrm{ and B) = P(A) \times P(B)

Theorem. For any events $A$ and $B$: $0 \leq P(A \cap B) \leq \min(P(A), P(B))$.

Proof. $A \cap B \subseteq A$So $P(A \cap B) \leq P(A)$. Similarly $P(A \cap B) \leq P(B)$. Since probabilities are non-negative, $P(A \cap B) \geq 0$. $\blacksquare$

5.2 Combined Events

For two events $A$ and $B$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

This is the addition rule and holds for all events (not just mutually exclusive ones).

Worked Example. The probability that it rains is 0.3. The probability that a student arrives Late is 0.1. The probability that it rains AND the student arrives late is 0.05. Find the Probability that it rains OR the student arrives late.

$P(R \cup L) = 0.3 + 0.1 - 0.05 = 0.35$

Worked Example (Higher Tier). Are the events “it rains” and “the student arrives late” Independent?

$P(R) \times P(L) = 0.3 \times 0.1 = 0.03$.

But $P(R \cap L) = 0.05 \neq 0.03$.

The events are not independent.

5.3 Tree Diagrams

Tree diagrams show all possible outcomes of two or more events.

Rules:

• Multiply along branches (for “and”)
• Add between branches (for “or”)

Worked Example. A bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

First draw: $P(R) = \frac{3}{8}$

Second draw (after removing one red): $P(R) = \frac{2}{7}$

P(\mathrm{both red) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}

Worked Example (Higher Tier). Using the same bag, find the probability that the two balls are Different colours.

P(\mathrm{different) = P(\mathrm{RB) + P(\mathrm{BR) = \frac{3}{8} \times \frac{5}{7} + \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}

Worked Example (Higher Tier). A fair coin is tossed three times. Find the probability of getting Exactly two heads.

There are $2^3 = 8$ equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Favourable: HHT, HTH, THH (3 outcomes).

P(\mathrm{exactly 2 heads) = \frac{3}{8}

5.4 Conditional Probability

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

This reads as “the probability of $A$ given $B$”.

Worked Example. In a class of 30 students, 18 play football, 12 play rugby, and 6 play both. Find the probability that a student plays rugby given that they play football.

$P(R \mid F) = \frac{P(R \cap F)}{P(F)} = \frac{6/30}{18/30} = \frac{6}{18} = \frac{1}{3}$

Worked Example (Higher Tier). A factory produces items in three shifts. Shift A produces 40% of Items with 2% defective, shift B produces 35% with 3% defective, and shift C produces 25% with 1% Defective. An item is found to be defective. What is the probability it came from shift B?

This is Bayes’ theorem. Let $D$ be the event “defective” and $B$ be “from shift B”.

$P(D) = 0.40 \times 0.02 + 0.35 \times 0.03 + 0.25 \times 0.01 = 0.008 + 0.0105 + 0.0025 = 0.021$

$P(B \mid D) = \frac{P(D \mid B) \cdot P(B)}{P(D)} = \frac{0.03 \times 0.35}{0.021} = \frac{0.0105}{0.021} = 0.5$

5.5 Venn Diagrams and Probability

A Venn diagram with two sets $A$ and $B$ has four regions:

Region	Notation	Description
Left only	$A \cap B'$	In $A$ but not in $B$
Right only	$A' \cap B$	In $B$ but not in $A$
Overlap	$A \cap B$	In both
Outside	$(A \cup B)'$	In neither

Worked Example. In a survey of 100 people, 45 like tea, 35 like coffee, and 15 like both. A Person is chosen at random. Find the probability they like neither.

$|T \cup C| = 45 + 35 - 15 = 65$.

P(\mathrm{neither) = \frac{100 - 65}{100} = 0.35.

Worked Example (Higher Tier). Using the same data, given that a person likes at least one drink, Find the probability they like tea but not coffee.

$|T \cap C'| = 45 - 15 = 30$.

$P(T \cap C' \mid T \cup C) = \frac{30}{65} = \frac{6}{13}$.

6. Statistical Distributions (Higher Tier)

6.1 The Normal Distribution

The normal distribution is a bell-shaped curve defined by its mean $\mu$ and standard deviation $\sigma$.

Key properties:

• Symmetric about $\mu$
• Approximately 68% of data lies within $\mu \pm \sigma$
• Approximately 95% of data lies within $\mu \pm 2\sigma$
• Approximately 99.7% of data lies within $\mu \pm 3\sigma$

6.2 Standard Deviation

The standard deviation measures the average distance of data points from the mean.

$\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$

For frequency data:

$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$

Computational formula (avoids calculating deviations individually):

$\sigma = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}$

Worked Example. Find the standard deviation of: $4, 8, 6, 5, 3, 9, 7$.

$\bar{x} = \frac{42}{7} = 6$

$\sigma = \sqrt{\frac{(4-6)^2 + (8-6)^2 + (6-6)^2 + (5-6)^2 + (3-6)^2 + (9-6)^2 + (7-6)^2}{7}}$

$\sigma = \sqrt{\frac{4 + 4 + 0 + 1 + 9 + 9 + 1}{7}} = \sqrt{\frac{28}{7}} = \sqrt{4} = 2$

Worked Example (Higher Tier). If every value in a data set is increased by 5, what happens to The mean and standard deviation?

The new mean is $\bar{x} + 5$. The standard deviation is unchanged, because adding a constant shifts All values by the same amount but does not change their spread.

If every value is multiplied by $k$The new mean is $k\bar{x}$ and the new standard deviation is $|k|\sigma$.

7. Comparing Distributions (Higher Tier)

7.1 Using Box Plots

When comparing two box plots:

• Compare the medians (measure of central tendency)
• Compare the IQRs (measure of spread)
• Compare the ranges
• Comment on skewness (if the median is closer to $Q_1$ or $Q_3$)

Worked Example. Class A has median 65, $Q_1 = 50$, $Q_3 = 75$. Class B has median 60, $Q_1 = 45$, $Q_3 = 70$.

Class A has a higher median (65 vs 60), so on average they scored better. Class A also has a Narrower IQR ($75 - 50 = 25$ vs $70 - 45 = 25$), so both classes have the same spread by this Measure.

7.2 Using Histograms

When comparing histograms, compare the shapes (symmetric, skewed, bimodal), the locations of peaks, And the overall spread.

Common Pitfalls

• Using frequency instead of frequency density in histograms. The bar height is frequency density, not frequency.
• Forgetting that cumulative frequency uses upper class boundaries, not midpoints.
• Assuming correlation means causation. This is one of the most common errors in statistics.
• Not subtracting $P(A \cap B)$ in the addition rule for non-mutually exclusive events.
• Confusing “with replacement” and “without replacement” in probability questions. This changes the probabilities on the second draw.
• Counting the overlap twice when adding probabilities from a Venn diagram.
• Using the wrong quartile formula. For $n$ data points, $Q_1$ is at position $\frac{n+1}{4}$ not $\frac{n}{4}$.
• Forgetting that the area under a histogram bar equals the frequency, not the height.
• Misinterpreting the median from grouped data. The estimated median uses linear interpolation within the median class, not just the midpoint.

9.1 Independence vs. Mutual Exclusivity

A very common confusion is between independent events and mutually exclusive events. These are Different concepts:

• Independent events: The occurrence of one does not affect the probability of the other. $P(A \cap B) = P(A) \times P(B)$.
• Mutually exclusive events: They cannot occur at the same time. $P(A \cap B) = 0$.

In fact, if two events have positive probability and are mutually exclusive, they cannot be Independent (because $P(A \cap B) = 0 \neq P(A) \times P(B)$ when both are positive).

Example. A card is drawn from a standard 52-card deck. Let $A$ be “the card is a heart” and $B$ Be “the card is a king”. These events are NOT mutually exclusive (the king of hearts is both), but They ARE independent: $P(A) = 13/52 = 1/4$, $P(B) = 4/52 = 1/13$And $P(A \cap B) = 1/52 = (1/4)(1/13)$.

9.2 Why the Standard Deviation Uses Squared Differences

The standard deviation uses squared differences from the mean rather than absolute differences for Two key reasons:

1. Mathematical convenience. Squared differences are differentiable everywhere, enabling the use of calculus. Absolute values have a “corner” at zero.
2. Additivity. The variance of a sum of independent random variables equals the sum of their variances. This does not hold for mean absolute deviation.

9.3 Effect of Coding on Statistical Measures

When data is transformed by a linear coding $y = ax + b$:

• The new mean is $\bar{y} = a\bar{x} + b$.
• The new standard deviation is $s_y = |a| \cdot s_x$.
• Adding a constant $b$ shifts the data but does not change the spread. Multiplying by $a$ scales the spread.

Worked Example. Data has mean 50 and standard deviation 8. After coding $y = \frac{x - 50}{8}$: $\bar{y} = \frac{50 - 50}{8} = 0$ and $s_y = \frac{1}{8} \times 8 = 1$.

This process is called standardisation and produces data with mean 0 and standard deviation 1.

9.4 Outliers and Their Impact

An outlier is a value that lies far from the other data points. There are several ways to detect Outliers:

• Values below Q_1 - 1.5 \times \mathrm{IQR or above Q_3 + 1.5 \times \mathrm{IQR.
• Values more than 2 standard deviations from the mean (for approximately normal data).

Impact on measures:

• The mean is heavily affected by outliers because every value contributes equally.
• The median is resistant to outliers because only on the middle value(s).
• The standard deviation is inflated by outliers.
• The IQR is resistant to outliers.

Example. Consider the data set $\{2, 3, 4, 5, 100\}$. The mean is $114/5 = 22.8$Heavily pulled Up by the outlier 100. The median is 4, which better represents the “typical” value. The standard Deviation is very large due to the outlier. The IQR is $5 - 3 = 2$Unaffected by the outlier.

9.5 Tree Diagrams for Multiple Events

Tree diagrams are useful for visualising multi-stage experiments where events are sequential.

Worked Example. A bag contains 4 red and 6 blue counters. Two counters are drawn without Replacement. Find the probability that both are the same colour.

First draw: P(\mathrm{red) = 4/10 = 0.4, P(\mathrm{blue) = 6/10 = 0.6.

Second draw (if first was red): P(\mathrm{red) = 3/9 = 1/3, P(\mathrm{blue) = 6/9 = 2/3.

Second draw (if first was blue): P(\mathrm{red) = 4/9, P(\mathrm{blue) = 5/9.

P(\mathrm{both red) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} P(\mathrm{both blue) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} P(\mathrm{same colour) = \frac{2}{15} + \frac{1}{3} = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}

9.6 Scatter Graphs and Lines of Best Fit

A scatter graph plots pairs of data values to investigate the relationship between two Variables.

Types of correlation:

Correlation	Description
Positive	As one variable increases, so does the other
Negative	As one variable increases, the other decreases
None	No visible linear pattern

The line of best fit is drawn by eye so that it passes through the middle of the data, with Roughly equal numbers of points on each side. It can be used to make predictions:

• Interpolation: Predicting within the range of the data (reliable).
• Extrapolation: Predicting outside the range of the data (unreliable).

:::caution Never extrapolate far beyond the data range. The relationship may not hold outside the Observed values. :::

9.7 Frequency Polygons

A frequency polygon is created by plotting the class midpoint against the frequency density (for Grouped data) or frequency (for ungrouped data), and joining the points with straight lines.

The area under a frequency polygon equals the total frequency (just like a histogram).

Practice Questions

1. The mean of six numbers is 12. When one number is removed, the mean of the remaining five is 10. Find the removed number.

2. A grouped frequency table has classes 0 \leq x \lt 20$$20 \leq x \lt 40$$40 \leq x \lt 60 $60 \leq x \lt 80$. Explain why an estimated mean calculated from this table is only an approximation.

3. Draw a box plot for the data: $3, 5, 6, 8, 9, 11, 14, 15, 22, 35$. Identify any outliers.

4. A bag contains 4 green and 6 yellow counters. Three counters are drawn without replacement. Find the probability that all three are the same colour.

5. In a survey, 60% of people like tea, 40% like coffee, and 25% like both. A person is chosen at random. Given that they like tea, find the probability that they also like coffee.

6. Calculate the standard deviation of the following data: $5, 7, 3, 9, 6, 4, 8$.

7. Explain why the mode is the only useful average for categorical data.

8. A histogram has a bar of width 4 and height 3.5. What frequency does this bar represent?

9. Two events $A$ and $B$ are such that P(A) = 0.6$$P(B) = 0.5And $P(A \cup B) = 0.8$. Are $A$ and $B$ independent? Justify your answer.

10. The heights of 100 students are summarised in a cumulative frequency table. Explain how you would estimate the interquartile range from this table.

11. A biased dice has P(\mathrm{even) = 0.6 and P(\mathrm{odd) = 0.4. It is rolled twice. Find the probability that the sum is greater than 8.

12. Explain the difference between a histogram and a bar chart, and when each is most appropriate.

13. A set of data has $\bar{x} = 20$ and $\sigma = 4$. If every value is increased by 5, find the new mean and standard deviation.

14. Two classes took the same test. Class A: median 55, IQR 20. Class B: median 52, IQR 30. Compare the performance of the two classes.

15. In a Venn diagram for events $A$ and B$$P(A) = 0.7$$P(B) = 0.5And $P(A \mid B) = 0.6$. Find $P(A \cap B)$ and $P(A \cup B)$.

16. A fair coin is tossed 5 times. Find the probability of getting at least 3 heads.

17. The mean height of 8 students is 162 cm. When a ninth student joins, the mean becomes 163 cm. Find the height of the ninth student.

18. From a histogram, the first three bars have frequency densities 2, 4, and 3 with class widths 5, 5, and 10. Estimate the total frequency and the mean.

19. A bag contains $n$ red and $n$ blue balls. Two balls are drawn at random without replacement. Show that the probability of drawing two balls of the same colour is $\frac{n - 1}{2n - 1}$.

20. Explain why the standard deviation is always non-negative, and state when it equals zero.

Extended Practice (Higher Tier)

21. The probability that it rains on any given day is 0.3. Find the probability that it rains on exactly 2 out of the next 5 days.

22. Two events $A$ and $B$ are mutually exclusive. If $P(A) = 0.35$ and $P(A \cup B) = 0.65$Find $P(B)$ and $P(A \cap B)$.

23. A set of data has a mean of 50 and a standard deviation of 8. After applying the coding $y = \frac{x - 50}{8}$Find the new mean and standard deviation.

24. The table below shows the distribution of exam scores for a class of 40 students:

Score range	Frequency
$0 \le s \lt 20$	3
$20 \le s \lt 40$	8
$40 \le s \lt 60$	14
$60 \le s \lt 80$	10
$80 \le s \lt 100$	5

Draw a cumulative frequency curve and estimate the median and interquartile range.

25. A bag contains 3 red, 5 blue, and 2 green marbles. Three marbles are drawn at random without replacement. Find the probability that exactly two are the same colour.

26. Explain why the mean is affected by outliers but the median is not. Use the data set $2, 3, 4, 5, 100$ to illustrate your answer.

27. A fair spinner has the numbers 1, 2, 3, 4. It is spun twice. Find the probability that the product of the two numbers is even.

28. The standard deviation of a data set is zero. What does this tell you about the data?

29. Two classes sit the same exam. Class A has 20 students with mean 65 and standard deviation 8. Class B has 30 students with mean 70 and standard deviation 10. Find the overall mean.

30. A card is drawn from a standard deck of 52 cards. Find the probability that it is a heart or a face card (or both).

Worked Examples

Example 1:

A typical exam question on Statistics requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Statistics often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the mathematical techniques and concepts related to statistics, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• measures of central tendency and spread
• probability distributions (binomial, normal)
• hypothesis testing
• correlation and regression
• sampling methods

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.