Statistics | IB - Wyatt's Notes

Measures of Central Tendency

Mean

The arithmetic mean of a data set $x_1, x_2, \ldots, x_n$ :

\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i = \frac{\sum x_i}{n}

For grouped data with frequencies $f_i$ :

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Median

The median is the middle value when data is arranged in order.

If $n$ is odd: median $= x_{\frac{n+1}{2}}$
If $n$ is even: median $= \dfrac{x_{\frac{n}{2}} + x_{\frac{n}{2}+1}}{2}$

For grouped data, use linear interpolation within the median class.

Mode

The mode is the most frequently occurring value. A data set can be unimodal, bimodal, or have no Mode.

Comparing Measures

Measure	Advantages	Disadvantages
Mean	Uses all data points, algebraic properties	Affected by outliers
Median	Robust to outliers	Does not use all data
Mode	Simple, useful for categorical data	May not exist or be unique

Example Find the mean, median, and mode of: $3, 5, 5, 7, 8, 9, 12, 15, 45$ .

Mean: $\bar{x} = \dfrac{3+5+5+7+8+9+12+15+45}{9} = \dfrac{109}{9} \approx 12.1$

Median: 5th value $= 8$

Mode: $5$ (appears twice)

The mean (12.1) is significantly higher than the median (8) due to the outlier $45$ .

Measures of Spread

Range

\mathrm{Range} = x_{\max} - x_{\min}

Interquartile Range (IQR)

\mathrm{IQR} = Q_3 - Q_1

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

Variance

The variance measures the average squared deviation from the mean.

Population variance:

\sigma^2 = \frac{1}{N}\sum_{i=1}^{N}(x_i - \mu)^2

Sample variance (unbiased estimator):

S^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2

Standard Deviation

\sigma = \sqrt{\sigma^2} \quad \mathrm{or} \quad s = \sqrt{s^2}

Computational Formula

S^2 = \frac{\sum x_i^2 - n\bar{x}^2}{n - 1} = \frac{n\sum x_i^2 - (\sum x_i)^2}{n(n-1)}

Exam Tip Know whether to use the population formula ( $\div N$ ) or the sample formula ( $\div (n-1)$ ). In IB Exams, when data is from a sample, use $s^2$ (dividing by $n-1$ ). Your GDC uses the sample Formula by default.

Grouped Data

Estimating the Mean from Grouped Data

Use the midpoint of each class interval:

\bar{x} \approx \frac{\sum f_i m_i}{\sum f_i}

Where $m_i$ is the midpoint of class $i$ .

Estimating the Median from Grouped Data

Use linear interpolation within the median class:

\mathrm{Median} \approx L + \left(\frac{\frac{n}{2} - F}{f}\right) \times w

Where:

$L$ = lower boundary of median class
$n$ = total frequency
$F$ = cumulative frequency before median class
$f$ = frequency of median class
$w$ = class width

Example

Mass (g)	Frequency
$0 \le m \lt 20$	5
$20 \le m \lt 40$	12
$40 \le m \lt 60$	18
$60 \le m \lt 80$	10
$80 \le m \lt 100$	5

Total $n = 50$ . Median position $= \dfrac{50}{2} = 25$ .

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

Median is in the $40 \le m \lt 60$ class ( $F = 17$ , $f = 18$ ).

\mathrm{Median} \approx 40 + \left(\frac{25 - 17}{18}\right) \times 20 = 40 + \frac{8}{18} \times 20 = 40 + 8.89 = 48.89 \mathrm{ g}

Box-and-Whisker Plots

Components

A box-and-whisker plot displays the five-number summary:

Minimum (or lower whisker)
$Q_1$ (first quartile)
Median ( $Q_2$ )
$Q_3$ (third quartile)
Maximum (or upper whisker)

Outliers

A value is a potential outlier if it falls outside:

Q_1 - 1.5 \times \mathrm{IQR} \quad \mathrm{or} \quad Q_3 + 1.5 \times \mathrm{IQR}

Interpreting Box Plots

The box represents the middle 50% of data (IQR).
The line inside the box is the median.
Whiskers extend to the minimum and maximum (or to the most extreme non-outlier values).
Skewness: if the median is closer to $Q_1$ The data is right-skewed (positively skewed). If closer to $Q_3$ Left-skewed (negatively skewed).

Cumulative Frequency

Cumulative Frequency Graph (Ogive)

Plot cumulative frequency against the upper class boundary. From this graph, you can read:

Median: at $\dfrac{n}{2}$
Quartiles: at $\dfrac{n}{4}$ and $\dfrac{3n}{4}$
Percentiles: at the appropriate fraction of $n$

Example Using the grouped data from the previous example:

Upper boundary	Cumulative frequency
20	5
40	17
60	35
80	45
100	50

To find $Q_1$ (at $12.5$ ): interpolate between $(20, 5)$ and $(40, 17)$ .

Q_1 \approx 20 + \frac{12.5 - 5}{17 - 5} \times 20 = 20 + 12.5 = 32.5 \mathrm{ g}

Correlation

Scatter Diagrams

A scatter diagram plots two variables to visually assess the relationship between them.

Pearson”s Correlation Coefficient ( $r$ )

Measures the strength and direction of the linear relationship between two variables.

R = \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum(x_i - \bar{x})^2 \sum(y_i - \bar{y})^2}}

Properties of $r$

Value	Interpretation
$r = 1$	Perfect positive linear correlation
$r = -1$	Perfect negative linear correlation
$r = 0$	No linear correlation
$0 \lt r \lt 1$	Positive linear correlation
$-1 \lt r \lt 0$	Negative linear correlation

Strength Guidelines

$	r	$	Strength
$0.0$ — $0.3$	Weak
$0.3$ — $0.7$	Moderate
$0.7$ — $1.0$	Strong

Computational Formula

R = \frac{n\sum x_iy_i - \sum x_i \sum y_i}{\sqrt{[n\sum x_i^2 - (\sum x_i)^2][n\sum y_i^2 - (\sum y_i)^2]}}

Exam Tip Correlation does NOT imply causation. Two variables may be strongly correlated without one causing The other (they may both be influenced by a third variable).

Linear Regression

Least Squares Regression Line

The line of best fit minimises the sum of squared residuals:

Y = a + bx

Where:

B = \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sum(x_i - \bar{x})^2} = \frac{n\sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2}

A = \bar{y} - b\bar{x}

Regression Line of $x$ on $y$

If predicting $x$ from $y$ :

X = a' + b'y

B' = \frac{n\sum x_iy_i - \sum x_i \sum y_i}{n\sum y_i^2 - (\sum y_i)^2}

Coefficient of Determination ( $r^2$ )

R^2 = \frac{\mathrm{explained variation}}{\mathrm{total variation}}

$r^2$ represents the proportion of variance in $y$ explained by the linear relationship with $x$ .

$r^2 = 1$ : the line explains all the variation.
$r^2 = 0$ : the line explains none of the variation.

Example Given the data:

$x$	1	2	3	4	5
$y$	2.1	3.9	6.2	7.8	10.1

Find the regression line of $y$ on $x$ .

N = 5, \quad \sum x = 15, \quad \sum y = 30.1

\sum x^2 = 55, \quad \sum xy = 110.2

B = \frac{5(110.2) - 15(30.1)}{5(55) - 225} = \frac{551 - 451.5}{275 - 225} = \frac{99.5}{50} = 1.99

\bar{y} = 6.02, \quad \bar{x} = 3

A = 6.02 - 1.99(3) = 6.02 - 5.97 = 0.05

Regression line: $y = 0.05 + 1.99x$ .

Extrapolation and Interpolation

Interpolation: predicting within the range of data (generally reliable).
Extrapolation: predicting outside the range of data (unreliable and potentially misleading).

Exam Tip Never extrapolate beyond the data range without acknowledging the uncertainty. IB exam questions Often ask you to comment on the reliability of a prediction.

Hypothesis Testing

Key Concepts

Null hypothesis ( $H_0$ ): The statement being tested ( “no effect” or “no difference”).
Alternative hypothesis ( $H_1$ ): The statement we suspect might be true.
Significance level ( $\alpha$ ): The threshold for rejecting $H_0$ (commonly 0.05 or 0.01).
Test statistic: A value computed from the sample data.
$p$ -value: The probability of observing the test statistic (or more extreme) assuming $H_0$ is true.
Critical value: The boundary value(s) that define the rejection region.

Decision Rule

If $p$ -value $\lt \alpha$ : reject $H_0$ .
If $p$ -value $\ge \alpha$ : do not reject $H_0$ .

Types of Errors

Error Type	Description
Type I	Rejecting $H_0$ when it is true (false positive). Probability $= \alpha$ .
Type II	Failing to reject $H_0$ when it is false (false negative). Probability $= \beta$ .

One-Tailed vs Two-Tailed Tests

Test	$H_1$	Rejection Region
Two-tailed	Parameter $\neq$ value	Both tails
Right-tailed	Parameter $\gt$ value	Upper tail
Left-tailed	Parameter $\lt$ value	Lower tail

Hypothesis Test for Correlation

To test whether the population correlation coefficient $\rho$ is zero:

$H_0: \rho = 0$ and $H_1: \rho \neq 0$ (or $\rho \gt 0$ or $\rho \lt 0$ ).
Compute $r$ from the sample.
Compare $|r|$ with the critical value (or compute the $p$ -value).
Make a conclusion.

The test statistic for large samples ( $n \gt 30$ ):

T = r\sqrt{\frac{n-2}{1-r^2}}

Which follows a $t$ -distribution with $n - 2$ degrees of freedom.

Example A sample of 12 students gives a correlation coefficient of $r = 0.85$ between hours studied and exam Score. Test at the 5% significance level whether there is a positive correlation.

$H_0: \rho = 0$ vs $H_1: \rho \gt 0$ .

This is a one-tailed test. The critical value for $n = 12$ at 5% level is approximately $0.497$ .

Since $r = 0.85 \gt 0.497$ We reject $H_0$ .

There is sufficient evidence at the 5% level to conclude a positive correlation between hours Studied and exam score.

Chi-Squared Test for Independence

Used to determine whether two categorical variables are independent.

Test statistic:

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

Where $O_i$ are observed frequencies and $E_i$ are expected frequencies.

Expected frequency:

E_`\{ij}` = \frac{(\mathrm{row } i \mathrm{ total}) \times (\mathrm{column } j \mathrm{ total})}{\mathrm{grand total}}

Degrees of freedom: $\nu = (r-1)(c-1)$ where $r$ is the number of rows and $c$ is the number of Columns.

Example Test whether gender and favourite subject are independent:

	Maths	Science	English	Total
Male	30	25	15	70
Female	20	20	40	80
Total	50	45	55	150

Expected frequencies:

$E(\mathrm{Male, Maths}) = \dfrac{70 \times 50}{150} = 23.33$

$E(\mathrm{Male, Science}) = \dfrac{70 \times 45}{150} = 21.00$

$E(\mathrm{Male, English}) = \dfrac{70 \times 55}{150} = 25.67$

$E(\mathrm{Female, Maths}) = \dfrac{80 \times 50}{150} = 26.67$

$E(\mathrm{Female, Science}) = \dfrac{80 \times 45}{150} = 24.00$

$E(\mathrm{Female, English}) = \dfrac{80 \times 55}{150} = 29.33$

\chi^2 = \frac{(30-23.33)^2}{23.33} + \frac{(25-21)^2}{21} + \frac{(15-25.67)^2}{25.67} + \frac{(20-26.67)^2}{26.67} + \frac{(20-24)^2}{24} + \frac{(40-29.33)^2}{29.33}

= 1.91 + 0.76 + 4.44 + 1.67 + 0.67 + 3.88 = 13.33

Degrees of freedom: $(2-1)(3-1) = 2$ .

Critical value at $\alpha = 0.05$ with $\nu = 2$ : $5.99$ .

Since $13.33 \gt 5.99$ We reject $H_0$ . Gender and favourite subject are not independent.

Exam Tip For the chi-squared test, always check that all expected frequencies are at least 5. If any $E_i \lt 5$ Combine categories or note the limitation.

IB Exam-Style Questions

Question 1 (Paper 1 style)

The marks of 8 students in Maths and Physics are:

Student	Maths ( $x$ )	Physics ( $y$ )
`A`	72	68
`B`	85	82
`C`	60	58
`D`	90	88
`E`	78	74
`F`	65	70
`G`	88	85
`H`	76	72

(a) Calculate Pearson’s correlation coefficient.

Using a GDC: $r \approx 0.960$ .

(b) Interpret this value.

There is a very strong positive linear correlation between Maths and Physics marks.

(c) Find the equation of the regression line of $y$ on $x$ .

Using a GDC: $y \approx 5.10 + 0.917x$ .

Question 2 (Paper 2 style)

The reaction times (in seconds) of 20 drivers were measured:

0.42, 0.55, 0.61, 0.48, 0.72, 0.38, 0.65, 0.51, 0.44, 0.59, 0.67, 0.53, 0.46, 0.71, 0.39, 0.58, 0.63, 0.50, 0.57, 0.66

(a) Find the mean and standard deviation.

$\bar{x} \approx 0.555$ , $s \approx 0.099$ .

(b) Construct a box-and-whisker plot.

Ordered data: 0.38, 0.39, 0.42, 0.44, 0.46, 0.48, 0.50, 0.51, 0.53, 0.55, 0.57, 0.58, 0.59, 0.61, 0.63, 0.65, 0.66, 0.67, 0.71, 0.72.

Median: $\dfrac{0.55 + 0.57}{2} = 0.56$ .

$Q_1$ : median of lower half $= \dfrac{0.46 + 0.48}{2} = 0.47$ .

$Q_3$ : median of upper half $= \dfrac{0.63 + 0.65}{2} = 0.64$ .

$\mathrm{IQR} = 0.64 - 0.47 = 0.17$ .

Lower fence: $0.47 - 1.5(0.17) = 0.215$ . Minimum $= 0.38$ .

Upper fence: $0.64 + 1.5(0.17) = 0.895$ . Maximum $= 0.72$ .

No outliers.

Question 3 (Paper 1 style)

A researcher claims that the mean height of a population is $170\mathrm{ cm}$ . A sample of 25 people Gives $\bar{x} = 173\mathrm{ cm}$ with $s = 8\mathrm{ cm}$ . Test this claim at the 5% significance Level.

$H_0: \mu = 170$ vs $H_1: \mu \neq 170$ .

T = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{173 - 170}{8/\sqrt{25}} = \frac{3}{1.6} = 1.875

Degrees of freedom $= 24$ .

Critical value (two-tailed, 5%) $\approx 2.064$ .

Since $|1.875| \lt 2.064$ We do not reject $H_0$ .

There is insufficient evidence at the 5% level to reject the claim that the mean height is $170\mathrm{ cm}$ .

flowchart TD
    A[2_Statistics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Concept	Formula
Mean	$\bar{x} = \dfrac{\sum x_i}{n}$
Sample variance	$s^2 = \dfrac{\sum(x_i - \bar{x})^2}{n-1}$
IQR	$Q_3 - Q_1$
Correlation	$r = \dfrac{n\sum x_iy_i - \sum x_i \sum y_i}{\sqrt{[n\sum x_i^2 - (\sum x_i)^2][n\sum y_i^2 - (\sum y_i)^2]}}$
Regression slope	$b = \dfrac{n\sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2}$
Chi-squared	$\chi^2 = \displaystyle\sum \dfrac{(O_i - E_i)^2}{E_i}$

Exam Strategy For …/4-statistics-and-probability/2_statistics questions in Paper 2, always show your working. State hypotheses for Hypothesis tests. When using your GDC, note what function you used and the inputs. Interpret results In context — never leave a numerical answer without explaining what it means.

Transformations of Data

Linear Transformations

If every data value is transformed by $y_i = ax_i + b$ :

| Statistic | Original | Transformed | | ----------------------- | ----------- | ------------------------------------------------ | --- | ------------ | | Mean | $\bar{x}$ | $a\bar{x} + b$ | | | | Standard deviation | $s_x$ | $| a | s_x$ | | Variance | $s_x^2$ | $a^2 s_x^2$ | | | | Median | $Q_2$ | $aQ_2 + b$ | | | | IQR | $Q_3 - Q_1$ | $| a | (Q_3 - Q_1)$ | | Correlation coefficient | $r$ | $r$ (unchanged if $a \gt 0$ Negated if $a \lt 0$ ) | | |

Standardised Scores (z-scores)

The z-score measures how many standard deviations a value is from the mean:

Z = \frac{x - \bar{x}}{s}

Example In a test with mean 65 and standard deviation 8, a student scores 81. Find the z-score.

Z = \frac{81 - 65}{8} = 2.0

The student scored 2 standard deviations above the mean.

Non-Linear Regression

Transformations for Non-Linear Data

When data shows a non-linear pattern, transform the variables to linearise:

Relationship	Transformation	Linear Form
$y = a x^b$	$\log y = \log a + b \log x$	Plot $\log y$ vs $\log x$
$y = a e^{bx}$	$\ln y = \ln a + bx$	Plot $\ln y$ vs $x$
$y = a + b\ln x$	$y = a + b\ln x$	Plot $y$ vs $\ln x$
$y = \frac{a}{x} + b$	$y = a\!\left(\frac{1}{x}\right) + b$	Plot $y$ vs $1/x$

Power Law

If $\log y$ vs $\log x$ gives a straight line, then $y = ax^b$ where:

$b$ is the gradient
$\log a$ is the $y$ -intercept

Exponential Law

If $\ln y$ vs $x$ gives a straight line, then $y = ae^{bx}$ where:

$b$ is the gradient
$\ln a$ is the $y$ -intercept

Example Data suggests $y$ is related to $x$ by $y = ax^b$ . A plot of $\log y$ vs $\log x$ has gradient $1.5$ And $y$ -intercept $0.7$ . Find the relationship.

B = 1.5, \quad \log a = 0.7 \implies a = 10^{0.7} \approx 5.01

Y \approx 5.01x^{1.5}

Additional Exam-Style Questions

Question 4 (Paper 2 style)

Two groups of students took a test. The results are summarised below:

Group A: $n = 30$$\bar{x} = 72$$s = 8$

Group B: $n = 25$$\bar{x} = 68$$s = 10$

(a) Find the overall mean.

\bar{x}_{\mathrm{overall}} = \frac{30 \times 72 + 25 \times 68}{55} = \frac{2160 + 1700}{55} = \frac{3860}{55} = 70.2

(b) Comment on the spread of the two groups.

Group B has a larger standard deviation (10 vs 8), meaning the scores are more spread out in Group B. Group A’s scores are more tightly clustered around the mean.

Question 5 (Paper 2 style)

A scientist investigates the relationship between temperature ( $x$ In $\degree$ C) and reaction rate ( $y$ In mol/L/s). The following data was collected:

$x$	10	20	30	40	50	60
$y$	0.4	1.1	2.5	5.2	10.8	22.0

(a) Explain why a linear regression model may not be appropriate.

The data appears to show exponential growth — as temperature increases, the rate increases by an Increasing amount. A plot of $y$ vs $x$ would show a curve, not a straight line.

(b) By plotting $\ln y$ against $x$ Determine whether the relationship is of the form $y = ae^{bx}$ .

$x$	10	20	30	40	50	60
$\ln y$	$-0.916$	$0.095$	$0.916$	$1.649$	$2.380$	$3.091$

A plot of $\ln y$ vs $x$ would show an approximately linear relationship with a positive gradient, Confirming the exponential model is appropriate.

(c) Using a GDC, find the equation of the regression line of $\ln y$ on $x$ .

The regression gives approximately $\ln y = -1.40 + 0.074x$ .

So $y = e^{-1.40}e^{0.074x} \approx 0.247e^{0.074x}$ .

Question 6 (Paper 1 style)

The correlation coefficient between study hours and exam scores for 50 students is $r = 0.72$ .

(a) Calculate the coefficient of determination and interpret it.

R^2 = 0.5184

About $51.8\%$ of the variation in exam scores is explained by the linear relationship with study Hours.

(b) The $p$ -value for testing $H_0: \rho = 0$ is $0.0001$ . What conclusion can be drawn?

Since $p = 0.0001 \lt 0.05$ We reject $H_0$ . There is strong evidence of a positive correlation Between study hours and exam scores.

Measures of Shape: Skewness and Kurtosis

Skewness

Skewness measures the asymmetry of the distribution.

Type	Description	Mean vs Median
Positive (right) skew	Long tail to the right	Mean $\gt$ Median
Negative (left) skew	Long tail to the left	Mean $\lt$ Median
Symmetric	No skew	Mean = Median

The Pearson coefficient of skewness:

\mathrm{Skewness} = \frac{3(\bar{x} - Q_2)}{s}

Kurtosis

Kurtosis measures the “tailedness” of the distribution compared to a normal distribution.

Type	Description
Leptokurtic	Heavy tails, sharp peak (kurtosis $\gt 3$ )
Mesokurtic	Same as normal (kurtosis $= 3$ )
Platykurtic	Light tails, flat peak (kurtosis $\lt 3$ )

Standardised Scores and Normal Distribution Tables

Using z-Scores

Given a population with mean $\mu$ and standard deviation $\sigma$ :

Convert raw score to z-score: $z = \dfrac{x - \mu}{\sigma}$
Use the standard normal table (or GDC) to find probabilities.
Convert back: $x = \mu + z\sigma$

Finding Percentiles

The $p$ -th percentile is the value below which $p\%$ of the data falls.

X_p = \mu + z_p \cdot \sigma

Where $z_p$ is the z-score such that $P(Z \lt z_p) = p/100$ .

Data Collection Methods

Types of Data

Type	Description	Examples
Qualitative (categorical)	Labels or names	Colour, gender, nationality
Quantitative	Numerical	Height, mass, temperature
Discrete	Integer values	Number of siblings, goals scored
Continuous	Any value in a range	Height, time, mass

Sampling Methods

Method	Description	Advantages	Disadvantages
Simple random	Every member has equal chance	Unbiased	Difficult for large populations
Systematic	Every $k$ -th member selected	Easy to implement	May be periodic bias
Stratified	Population divided into strata	Ensures representation	Complex to organise
Quota	Fixed numbers from each group	Quick	Not random
Convenience	accessible members	Easy	Likely biased

Reliability and Validity

Reliability: consistency of measurements (repeatable).
Validity: measures what it claims to measure.
A measurement can be reliable without being valid, but not valid without being reliable.

contains the hardest questions within the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Statistics with other IB mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Intuition

Statistics is the science of making decisions under uncertainty. The mean tells you the center of the data, but the standard deviation tells you how much trust to place in that center — a low standard deviation means the data clusters tightly around the mean, while a high one means the data spreads widely. Correlation measures the strength of a linear relationship, but it does not prove causation — two variables may move together because a third hidden variable drives both. Hypothesis testing is a courtroom argument: you assume innocence (the null hypothesis) and check whether the evidence is strong enough to convict. The p-value is the probability of seeing evidence this extreme if innocence is true.

Common Pitfalls

Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.
Confusing $P(A|B)$ with $P(B|A)$ . These are related by Bayes’ theorem but are not equal in general.

Cross-References

Topic	Site	Link
[Statistics]	IB	View
[Statistics]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.