Methods of Analysing Data

Mean. Add up all the values, divide by how many there are. $\bar{x} = \dfrac{\sum x}{n}.$

Median. Sort the values from low to high. The median is the middle.

• $n$ ODD: median is the $\dfrac{n+1}{2}$-th value.
• $n$ EVEN: median is the average of the $\dfrac{n}{2}$-th and $(\dfrac{n}{2} + 1)$-th values.

Mode. The value(s) that occur most often. A data set can be:

• Unimodal: one mode.
• Bimodal: two values tied for highest frequency.
• No mode: every value occurs the same number of times.

Range. Maximum minus minimum.

Worked. Data: $4, 7, 2, 9, 4, 5, 8$.

• Sort: $2, 4, 4, 5, 7, 8, 9$.
• $n = 7$, median = $\dfrac{8}{2} = 4$th value = $5$.
• Mean: $\dfrac{2 + 4 + 4 + 5 + 7 + 8 + 9}{7} = \dfrac{39}{7} \approx 5.57$.
• Mode: $4$ (appears twice).
• Range: $9 - 2 = 7$.

When data is summarised in a frequency table: $\bar{x} = \dfrac{\sum f x}{\sum f}.$

Worked. Number of pets per household:

Pets ($x$)	Frequency ($f$)
0	5
1	12
2	8
3	3
4	2

• $\sum f = 5 + 12 + 8 + 3 + 2 = 30$.
• $\sum f x = (0)(5) + (1)(12) + (2)(8) + (3)(3) + (4)(2) = 0 + 12 + 16 + 9 + 8 = 45$.
• Mean: $\dfrac{45}{30} = 1.5$ pets per household.

Median from a frequency table. Find the cumulative frequency. Locate the value at position $\dfrac{n+1}{2}$ (or the average of $\dfrac{n}{2}$ and $\dfrac{n}{2}+1$ for $n$ even). Use the cumulative column to find which $x$ value sits there.

Mode from a frequency table. The value with the highest frequency.

When data is grouped into class intervals (e.g. $0 < t \le 10$, $10 < t \le 20$, …) we don't know the exact values within each class. We approximate by assuming each value sits at the midpoint.

Steps.

1. Find the midpoint of each class: $x = \dfrac{\text{lower} + \text{upper}}{2}$.
2. Compute $f x$ for each row.
3. Estimate of mean: $\dfrac{\sum f x}{\sum f}$.

Worked. Time spent on homework (minutes):

Class	Midpoint $x$	Frequency $f$	$f x$
$0 < t \le 30$	15	4	60
$30 < t \le 60$	45	12	540
$60 < t \le 90$	75	8	600
$90 < t \le 120$	105	6	630

• $\sum f = 30$, $\sum f x = 1830$.
• Estimate of mean: $\dfrac{1830}{30} = 61$ minutes.

Why "estimate"? We assumed every value in $0 < t \le 30$ was exactly $15$ — usually false. The result is an APPROXIMATION.

Modal class. The class with the HIGHEST frequency. Cambridge expects you to write the full class interval (e.g. "$30 < t \le 60$"), not just the midpoint.

When asked to compare two data sets, Cambridge wants comments on TWO aspects:

1. Average / centre. Use mean or median.

• Higher mean: data set is on average higher.
• Outliers can pull the mean — median is more robust.

2. Spread / consistency. Use range or interquartile range (IQR — see Cumulative Frequency).

• Smaller range: more consistent / less variable.
• Smaller IQR: middle 50% is more tightly grouped.

Worked. Two classes' test scores:

• Class A: mean $= 65$, range $= 30$.
• Class B: mean $= 65$, range $= 12$.

Comment: "Both classes have the same mean ($65$), so on average they performed equally. However, Class B has a much smaller range ($12$ vs $30$), meaning Class B's scores are more consistent."

Cambridge mark scheme expectation. TWO comparison statements: one about average, one about spread. Each must reference the actual numerical statistics — vague statements like "Class B is more consistent" without citing the smaller range lose marks.