The median of 3, 7, 9, 15, 21 is:

9. The data is already in order and there are five values, so the middle one is the third: 9. Option B is the mean, which is a different statistic.

The class boundaries of the class 30–39 are:

29.5 and 39.5. Boundaries lie half a unit outside the limits so that consecutive classes meet without a gap. The class 30–39 really covers everything from 29.5 up to 39.5.

Which average is most affected by an extreme value?

Mean. The mean is calculated from every value, so a single very large or very small one shifts it substantially. The median depends only on position, and the mode only on frequency.

For grouped data, the mean is calculated using:

class midpoints. The original values are lost after grouping, so every value in a class is assumed to sit at its midpoint. This is why the answer is an estimate rather than the exact mean.

A shopkeeper wants to know which size of shirt to order most of. He should use the:

mode. The mode is the most frequently occurring size, which is exactly what "order most of" asks for. A mean size of 39.6 would not correspond to any shirt he can buy.

The range of 14, 22, 9, 30, 18 is:

21. Largest 30 minus smallest 9 gives 21. The range is a single number, not an interval, so writing "9 to 30" would not be accepted.

Two data sets have the same mean but different standard deviations. This tells you:

one is more spread out than the other. The standard deviation measures spread. A larger one means values sit typically further from the mean, even though the centres of the two sets coincide.

Why are deviations squared when calculating variance?

Because the deviations would otherwise total zero. By the definition of the mean, the positive and negative deviations cancel exactly, so their sum is always zero. Squaring makes them all positive so the spread survives the addition.

Information Handling — Mathematics Notes, Worked Examples and Quiz

From a list of numbers to a table you can read

Raw data — forty test marks written down in the order the papers were collected — tells you almost nothing. Organising it does. A frequency table records how often each value, or each class of values, occurs, and every calculation in this chapter starts there.

When the values are few, tally each one separately. When they are spread out, group them into classes of equal width. The chapter uses three terms that examiners test directly: the class limits are the values written in the table (say 10–19), the class boundaries are the values where one class actually ends and the next begins (9.5 and 19.5), and the class mark or midpoint is their average (14.5).

Why boundaries are half a unit outside the limits

Marks are recorded as whole numbers, so a class written 10–19 really covers everything from 9.5 up to 19.5 — a mark of 19.4 would round to 19. The boundaries close the gap between one class and the next, which is why a histogram has no spaces between its bars while a bar chart does.

The three averages, and when each one lies to you

An average is a single number standing in for the whole set. Three are in the syllabus, and the exam question is usually not "calculate it" but "which one is appropriate and why".

Average	Strength	Weakness
Mean	uses every value; best for further calculation	a single extreme value drags it badly
Median	unaffected by extremes; good for skewed data such as income	ignores the actual sizes of the other values
Mode	the only average usable for categories like shoe size or colour	may not exist, or there may be several

mean x̄ = Σx / n(ungrouped)mean x̄ = Σfx / Σf(grouped, x = class midpoint)median = the middle value when the data is put in ordermode = the value occurring most oftenfor an even number of values the median is the mean of the two middle ones

The question examiners actually ask

Nine salaries of 30 000 rupees and one of 3 000 000. The mean is about 327 000 — a figure nobody in the room earns. The median is 30 000, which describes the situation honestly. Whenever a data set has one wild value, the median is the appropriate average, and saying why is worth the mark.

Calculating the mean of grouped data

Once data is grouped, the individual values are gone; all you know is which class each one fell into. So you assume every value sits at the midpoint of its class. That makes the mean an estimate, and questions almost always use the word "estimate" for exactly this reason.

Worked example

Estimate the mean of: 0–10 (f = 2), 10–20 (f = 5), 20–30 (f = 9), 30–40 (f = 12), 40–50 (f = 8), 50–60 (f = 4).

Find each class midpoint: 5, 15, 25, 35, 45, 55.Midpoint = (lower boundary + upper boundary) ÷ 2. Every value in the class is treated as sitting there.
Multiply each midpoint by its frequency: 10, 75, 225, 420, 360, 220.This is the fx column, and setting it out as a column is what earns the method mark.
Σf = 2 + 5 + 9 + 12 + 8 + 4 = 40 and Σfx = 1310.Always total the frequency column too — it is the denominator and a useful check on your table.
x̄ = 1310 / 40 = 32.75.Sanity check: the modal class is 30–40, so a mean of about 33 is exactly where it should be.

Estimated mean = 32.75

The dashed line is the mean. Compare the bar chart with the histogram: the bars are identical, but the histogram has no gaps because its classes are continuous and meet at their boundaries.

Measuring the spread

Two classes can have the same mean and be nothing alike — one clustered tightly around it, the other scattered from top to bottom. A measure of spread is what tells them apart.

The range is the simplest: largest value minus smallest. It is quick, but it depends entirely on the two most extreme values and ignores everything in between. For grouped data it is taken as the difference between the highest and lowest class boundaries.

The variance and standard deviation use every value. The standard deviation is the more useful of the two because it comes back to the original units: a standard deviation of 8 marks means "typically about 8 marks away from the mean".

range = largest − smallestvariance σ² = Σ(x − x̄)² / nstandard deviation σ = √(σ²)square the distances so that values above and below the mean cannot cancel out

Why square the deviations?

If you simply averaged (x − x̄), the answer would always be exactly zero — the positives and negatives cancel by the very definition of the mean. Squaring makes every deviation positive, so they cannot cancel; taking the square root at the end brings the answer back into the units you started with.

Before you leave this chapter

Class limits are what is written; class boundaries are where the classes really meet, half a unit outside.
Grouped mean = Σfx / Σf, using midpoints — and it is an estimate, not the true mean.
The median is the appropriate average whenever an extreme value would distort the mean.
The mode is the only average available for non-numerical categories.
Range is quick but ignores everything between the extremes; standard deviation uses every value and keeps the original units.

Pie charts

A pie chart shows how a total splits into parts. Each category gets a sector whose angle is its share of the full 360°, so the chart shows proportions well and absolute sizes not at all — two pie charts drawn the same size can represent totals of 40 and 4 000.

The calculation is one line per category, and the marks come from showing it.

sector angle = (frequency / total frequency) × 360°the angles must add to 360° — check this before you draw anything

Worked example

In a survey of 60 students, 24 chose cricket, 18 hockey, 12 football and 6 badminton. Find the angle of each sector.

Cricket: (24/60) × 360 = 144°.Cricket is 40% of the students, and 40% of 360 is 144.
Hockey: (18/60) × 360 = 108°.
Football: (12/60) × 360 = 72°; badminton: (6/60) × 360 = 36°.
Check: 144 + 108 + 72 + 36 = 360° ✓If the angles do not total 360°, one of them is wrong, and you have caught it before drawing.

144°, 108°, 72° and 36°

Information Handling