Data collection — primary vs secondary, tally charts, frequency tables: Exercises

Before we start, one reminder about the tools you will use:

A tally mark is a single stroke $∣$ meaning "one thing happened". Four strokes then a diagonal across them ( $∣∣∣∣$ ) means a bundle of five. We bundle in fives because our eyes cannot reliably count a long row of loose strokes.
Frequency $f$ = how many times a category appeared (just a whole number count).
Total $\sum f$ = add up all the frequencies. The symbol $\sum$ (Greek capital "sigma") is shorthand for "add these all together" — nothing more.

For each source, say whether it gives primary or secondary data: (a) You time how long each of your friends holds their breath. (b) You copy last year's rainfall figures from the weather department's website. (c) A newspaper table of yesterday's cricket scores that you use in a project. (d) You measure the mass of 20 apples with your own kitchen scale.

Recall Solution L1.1

The test: did I gather it firsthand for my purpose? If yes → primary. If someone else gathered it → secondary.

(a) Primary — you did the timing yourself.
(b) Secondary — the weather department collected it, you reused it.
(c) Secondary — the newspaper collected the scores.
(d) Primary — you did the measuring.

Exercise L1.2

A tally shows: $∣∣∣∣ ∣∣∣∣ ∣∣∣$ . What frequency does it represent?

Recall Solution L1.2

Two full bundles of five, then three loose strokes. $f = 2 \times 5 + 3 = 13.$ Answer: 13.

Exercise L1.3

Draw the tally marks for a frequency of 9.

Recall Solution L1.3

Nine = one bundle of five + four loose: $9 = 1 \times 5 + 4 \Rightarrow ∣∣∣∣ ∣∣∣∣.$ Note: the four extra strokes stay loose because a bundle only forms on the fifth mark.

Level 2 — Application

Exercise L2.1

A dice was rolled 20 times. Results in order: $3, 5, 1, 6, 3, 2, 5, 5, 4, 1, 6, 3, 2, 5, 1, 3, 6, 4, 5, 3.$ Build a tally chart and frequency table for faces 1–6, then check the total.

Recall Solution L2.1

Go through the list once, adding one mark per roll to that face's row. Counting each face:

1: rolls at positions 3, 10, 15 → 3
2: positions 6, 13 → 2
3: positions 1, 5, 12, 16, 20 → 5
4: positions 9, 18 → 2
5: positions 2, 7, 8, 14, 19 → 5
6: positions 4, 11, 17 → 3

Face	Tally	Frequency
1	$∥∥∥$	3
2	$∥∥$	2
3	$∥∥∥∥$	5
4	$∥∥$	2
5	$∥∥∥∥$	5
6	$∥∥∥$	3
Total		20

Check: $3 + 2 + 5 + 2 + 5 + 3 = 20$ ✓ matches the 20 rolls.

Exercise L2.2

From the fruit survey (Apple 11, Banana 8, Orange 4, total 23), find the relative frequency of Banana as a percentage, rounded to one decimal place.

Recall Solution L2.2

Relative frequency = this category's count ÷ grand total, then ×100 to make it a percentage. $\frac{8}{23} \times 100% = 34.782...% \approx 34.8%.$ Answer: $34.8%$ .

Exercise L2.3

20 students' scores are grouped. Fill the cumulative frequency column:

Range	Frequency	Cumulative
0–20	2	?
21–40	3	?
41–60	7	?
61–80	6	?
81–100	2	?

Recall Solution L2.3

Cumulative frequency is a running total: each row = its own frequency plus everything above it. This answers "how many scored up to and including this range?"

0–20: $2$
21–40: $2 + 3 = 5$
41–60: $5 + 7 = 12$
61–80: $12 + 6 = 18$
81–100: $18 + 2 = 20$

The last cumulative value $= 20 = \sum f$ , which is always true — the running total must end at the grand total.

Level 3 — Analysis

Exercise L3.1

A frequency table for pet ownership is partly damaged:

Pet	Frequency
Dog	14
Cat	?
Fish	5
None	9
Total	40

Find the missing frequency for Cat, and then the relative frequency of Cat as a percentage.

Recall Solution L3.1

The total constrains everything: all frequencies must add to 40. $14 + Cat + 5 + 9 = 40 \Rightarrow Cat = 40 - 28 = 12.$ Relative frequency of Cat: $\frac{12}{40} \times 100% = 30%.$ Answers: Cat $= 12$ , relative frequency $= 30%$ .

Exercise L3.2

Using the cumulative frequencies from Exercise L2.3, answer: (a) How many students scored 60 or below? (b) How many scored more than 60?

Recall Solution L3.2

(a) "60 or below" is exactly the cumulative frequency at the 41–60 row: 12 students. (b) "More than 60" is everyone else: $\sum f - 12 = 20 - 12 = 8 students .$ Answers: (a) 12, (b) 8. (Notice 12 + 8 = 20 ✓.)

Exercise L3.3

See the bar chart below. It shows frequencies for four ice-cream flavours. Read off each frequency, and find what fraction of all responses chose Chocolate.

Figure — Data collection — primary vs secondary, tally charts, frequency tables

Recall Solution L3.3

Reading each bar's height:

Vanilla = 6, Chocolate = 10, Strawberry = 4, Mango = 5.

Total $= 6 + 10 + 4 + 5 = 25$ . Fraction choosing Chocolate: $\frac{10}{25} = \frac{2}{5} = 0.4 = 40%.$ Answer: $\frac{2}{5}$ (40%).

Level 4 — Synthesis

Exercise L4.1

You survey 30 people on how many books they read last month. Raw data: $0, 2, 1, 3, 0, 1, 2, 4, 1, 0, 2, 1, 3, 2, 0, 1, 2, 1, 4, 0, 2, 3, 1, 2, 0, 1, 2, 1, 3, 2.$ Build a complete frequency table with columns: value (0–4), frequency, cumulative frequency, and relative frequency (%). Verify the total.

Recall Solution L4.1

Count each value by tallying through the list once:

0: positions 1,5,10,15,20,25 → 6
1: positions 3,6,9,12,16,18,23,26,28 → 9
2: positions 2,7,11,14,17,21,24,27,30 → 9
3: positions 4,13,22,29 → 4
4: positions 8,19 → 2

Check count so far: $6 + 9 + 9 + 4 + 2 = 30$ ✓.

Books	Frequency	Cumulative	Relative (%)
0	6	6	$6/30 = 20.0%$
1	9	15	$9/30 = 30.0%$
2	9	24	$9/30 = 30.0%$
3	4	28	$4/30 \approx 13.3%$
4	2	30	$2/30 \approx 6.7%$
Total	30	—	100%

Cumulative ends at 30 ✓; percentages sum to $\approx 100%$ ✓ (rounding makes it $99.9%$ – $100%$ ). The mode (most common) is a tie: 1 and 2 books, both frequency 9. See Measures of central tendency — mean median mode.

Exercise L4.2

For the book data in L4.1, someone claims "half the people read at most 1 book." Is this true? Justify using cumulative frequency.

Recall Solution L4.2

"At most 1 book" means values 0 or 1 — that is the cumulative frequency at row "1", which is 15. Half of 30 is 15. $15 = \frac{1}{2} \times 30 \Rightarrow exactly half.$ The claim is true — precisely 50% read 0 or 1 book.

Level 5 — Mastery

Exercise L5.1

A researcher wants "average height of 14-year-olds in India." (a) Describe a primary method and a secondary method. (b) State one advantage and one risk of the secondary method. (c) Which would you choose and why?

Recall Solution L5.1

(a) Primary: measure the heights of a chosen sample of 14-year-olds yourself with a measuring tape (this needs Sampling methods to pick a fair sample). Secondary: use published national health-survey data on adolescent heights. (b) Advantage of secondary: huge sample, national coverage, no travel or cost — you could never measure the whole country alone. Risk: it may be outdated, or defined differently (e.g. it might report ages 13–15 grouped, not exactly 14), so it may not match your question precisely. (c) Secondary is the practical choice for a country-wide figure: one person cannot fairly sample all of India (a primary attempt would be a tiny, biased sample). We accept secondary data provided we check its date, source, and definitions.

Exercise L5.2

Two schools report exam pass data. School A: 30 passed out of 40. School B: 45 passed out of 75. A headline says "School A had more passes, so it teaches better." Use relative frequency to critique this.

Recall Solution L5.2

Counts alone mislead because the schools have different sizes — this is exactly why relative frequency exists. $A: \frac{30}{40} = 0.75 = 75%, B: \frac{45}{75} = 0.60 = 60%.$ School A has fewer and more students, but the fair comparison is the rate: 75% vs 60%. On rate, A does pass a higher proportion — but the headline's reason ("more passes") is wrong logic. If instead A were 30/60 = 50%, A would have the same raw passes yet a lower rate. Always compare proportions, not raw counts, across different-sized groups. See Probability from frequency — experimental vs theoretical.

Exercise L5.3

Continuous ages of 8 people are recorded: $12, 19, 25, 33, 41, 47, 58, 20.$ Group them into class intervals $0 - 19, 20 - 39, 40 - 59$ and build a frequency table. What is the relative frequency of the middle class?

Recall Solution L5.3

Sort each age into its interval (each endpoint here is unambiguous — see Grouped data and class intervals for boundary rules):

0–19: 12, 19 → 2
20–39: 25, 33, 20 → 3
40–59: 41, 47, 58 → 3

Class	Frequency
0–19	2
20–39	3
40–59	3
Total	8

Check: $2 + 3 + 3 = 8$ ✓. Middle class (20–39) relative frequency: $\frac{3}{8} = 0.375 = 37.5%.$ Answer: $37.5%$ .

Recall Quick self-test

Primary vs secondary — the deciding question? ::: Did I collect it firsthand for my own purpose (primary) or did someone else collect it (secondary)? Tally for the number 12 ::: $∣∣∣∣ ∣∣∣∣ ∣∣$ Formula for relative frequency ::: $f_{i} \div \sum f$ (times 100 for a percentage) Cumulative frequency of the last row always equals ::: the grand total $\sum f$ Why compare proportions not counts across groups? ::: Because groups may differ in size; only proportions share a common base of 100%.

Exercises — Data collection — primary vs secondary, tally charts, frequency tables

Level 1 — Recognition

Exercise L1.1