Statistics and Data Analysis for ESAT Mathematics 1

Mean, mode, median and range for ungrouped data

The mean, median, and mode are all types of averages used to describe the centre of a data set.

1. The mean of a list of $n$ numbers is the sum of those numbers divided by $n$. The mean does not necessarily have to be one of the numbers in the list.

$mean = \frac{\text{sum of all values in the list}}{\text{number of values in the list}}$

If the data is in a frequency table, the sum is found by multiplying each value by its frequency and adding the results.

1. The median is the middle value when $n$ numbers are arranged in order of size. It is the value at position $\frac{n+1}{2}$.

If $n$ is odd, the median is a specific number in the list. If $n$ is even, the median falls between two numbers and is calculated as their average. In a frequency table, the median is found by counting through the frequencies to find the position $\frac{n+1}{2}$.

1. The mode is the most frequent value. A data set can have one mode, multiple modes, or no mode at all. It must always be one of the values present in the data.

2. The range measures how spread out the data is. It is the positive difference between the largest and smallest values. When using a frequency table, ensure the range is calculated from the possible values, not the frequencies.

Worked Example: Mean of Ungrouped Data

Find the mean of: 1193, 1194, 1196, 1198, 1199, 1201, 1203, 1204, 1205, 1207.

Method 1 (Direct Calculation): $(1193 + 1194 + 1196 + 1198 + 1199 + 1201 + 1203 + 1204 + 1205 + 1207) \div 10 = 12,000 \div 10 = 1200$.

Method 2 (Subtraction from Smallest): Subtract 1193 from every value: 0, 1, 3, 5, 6, 8, 10, 11, 12, 14. Mean of differences: $70 \div 10 = 7$. Add back: $1193 + 7 = 1200$.

Method 3 (Reference Point): Using 1200 as a central reference, the differences are: -7, -6, -4, -2, -1, 1, 3, 4, 5, 7. As shown in the diagram, these balance each other out to a sum of zero.

The average of these differences is $0 \div 10 = 0$, so the mean is $1200 + 0 = 1200$.

Worked Example: Median, Mode, and Range

Find these values for: 2, 5, 4, 1, 0, 6, 4, 8, 6, 9, 3, 8, 7, 1, 8, 6.

First, order the 16 values: 0, 1, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 8, 8, 9. Median position: $\frac{16+1}{2} = 8.5$. The 8th value is 5 and the 9th is 6, so median = $\frac{5+6}{2} = 5.5$. Mode: There are three 6s and three 8s, so there are two modes: 6 and 8. Range: $9 - 0 = 9$.

Grouped data: Modal class and estimates

Grouped data provides class intervals rather than specific numbers. Because exact values are unknown, statistics must be estimated.

1. The modal class is the interval with the highest frequency.
2. The range is estimated as the difference between the lower limit of the lowest interval and the higher limit of the highest interval.
3. The mean estimate is calculated using the mid interval value ($x$) for each class. It is an estimate because it assumes a symmetric distribution within the classes.

$Mean \approx \frac{\sum (f \times x)}{\sum f}$

1. The median estimate is found by identifying the interval containing the $\frac{n+1}{2}$th item and using linear interpolation.

Worked Example: Shopper Ages

A shopping centre records ages ($x$):

Age Interval	Frequency ($f$)	Mid-interval ($x$)	$f \times x$
$10 \leq x \leq 19$	5	14.5	72.5
$20 \leq x \leq 29$	10	24.5	245
$30 \leq x \leq 39$	12	34.5	414
$40 \leq x \leq 49$	18	44.5	801
$50 \leq x \leq 59$	25	54.5	1362.5
$60 \leq x \leq 69$	35	64.5	2257.5
$70 \leq x \leq 79$	20	74.5	1490

Modal class: $60 \leq x \leq 69$ (frequency 35). Estimated Range: $79 - 10 = 69$. Estimated Mean: Total $\sum (f \times x) = 6642.5$. Total shoppers $\sum f = 125$. Mean = $6642.5 \div 125 = 53.14$, or 53 to the nearest year. Estimated Median: Position $\frac{125+1}{2} = 63$. Cumulative frequency reaches 45 before the $50 \leq x \leq 59$ class. The 63rd age is the $63 - 45 = 18$th age in this group of 25. The estimate uses the rank of 17.5:

This gives 57.2, or 57 to the nearest year.

Advantages and disadvantages of summary data

1. Mean: Includes every data value, but is heavily skewed by outliers.
2. Mode: Can be used for non numerical data, but may fall at an extreme edge rather than the centre.
3. Median: Not affected by outliers, but only usable for ordered data.
4. Range: Shows the full spread, but is sensitive to extreme values.
5. Interquartile Range (IQR): Measures the spread of the middle 50% (difference between upper quartile $Q3$ and lower quartile $Q1$). It is resistant to outliers but ignores 50% of the data.

Worked Example: Wage Distribution

A factory has 6 cleaners (£x), 36 machinists (£2x), 3 managers (£5x), and 1 owner (£50x). The mean is approximately £3.1x, which is unrepresentative of the majority who earn £2x. The median and mode (£2x) better represent the central trend. The IQR effectively shows the wage of the majority, while the range highlights the massive disparity created by the owner salary.

Graphical estimation: Cumulative frequency

From a cumulative frequency graph:

1. The median is found at frequency $\frac{n}{2}$.
2. Quartiles are found at $\frac{n}{4}$ (Lower Quartile) and $\frac{3n}{4}$ (Upper Quartile).
3. The modal class is the interval where the graph is steepest.
4. The range is estimated from the horizontal spread of the curve.

Worked Example: History Test

For 70 students and a test out of 40:

Median: At 35 students, the mark is approximately 17.5. Quartiles: $Q1$ (17.5 students) is 12.25 marks; $Q3$ (52.5 students) is 21.5 marks. $IQR = 21.5 - 12.25 = 9.25$.

Mean Estimation: Divide the graph into intervals, find the mid intervals, and multiply by the frequency increase in each segment.

Interval	Freq	Midpoint	F x Mid
$5 < m \leq 10$	12	7.5	90
$10 < m \leq 15$	11	12.5	137.5
$15 < m \leq 20$	25	17.5	437.5
$20 < m \leq 25$	14	22.5	315
$25 < m \leq 30$	6	27.5	165
$30 < m \leq 35$	1	32.5	32.5
$35 < m \leq 40$	1	37.5	37.5
Total	70		1215

Mean estimate = $1215 \div 70 = 17.4$.

Graphical estimation: Histograms

In a histogram, frequency density is defined as $\frac{\text{frequency}}{\text{class width}}$. The area of the bar represents the frequency.

Worked Example: Plant Heights

Total number of plants: Sum of (frequency density $\times$ width). $1 \times 5 + 1.6 \times 5 + 2.4 \times 5 + 3 \times 5 + 2 \times 10 = 5 + 8 + 12 + 15 + 20 = 60$.

Mean estimate: $(5 \times 7 + 8 \times 12 + 12 \times 17 + 15 \times 22 + 20 \times 29.5) \div 60 = 20.9$. Modal class: The highest bar, $19.5 \leq x < 24.5$. Median class: Position $\frac{60+1}{2} = 30.5$. Cumulative frequencies are 5, 13, 25, 40. The 30.5th item is in the $19.5 \leq x < 24.5$ class.

Using median and interquartile range to compare distributions

Comparing datasets involves looking at central tendency and spread like for like.

For example, when comparing two factories, Factory A (Range 4, IQR 2) and Factory B (Range 11, IQR 6), Factory A is more consistent because its measures of spread are smaller. If a buyer needs a consistent supply of at least 11 units, they would choose Factory A if its lower bound (Upper Quartile minus Range) is at least 11. Specifically, if Factory A has an Upper Quartile of 15 and a Range of 4, its minimum value is at least $15 - 4 = 11$.