Module 27 of 45Lesson 1 of 20 · Unit S06 · The Three Averages
Statistics & Probability · Unit S06 · Foundations
The Three Averages
An average is one number standing in for a whole list. There are three of them, they usually disagree, and choosing between them is most of the skill. This unit works all three out from a raw list, from a frequency table and from grouped classes — and then watches what one extreme value does to each.
The first calculating unit of the strand. S01 to S05 collected data and drew it. From here on the data gets worked on, and the very first thing anyone does to a set of numbers is average it — usually without saying which of the three averages they mean.
By the end of this unit you can
- Calculate the mean, the median and the mode of a raw list of values
- Use the median position (n + 1) ÷ 2 for an odd and for an even number of values
- Calculate a mean straight from a frequency table as Σfx ÷ Σf, without re-listing the data
- State the modal class of grouped data and estimate its mean from the class midpoints
- Describe why a mean from grouped data is an estimate and never the exact answer
- Recompute the mean and the median after one extreme value joins, and describe which one moved
- Choose the right average for a situation, and describe the reason
- Round an average sensibly and state its unit
The one idea behind all of it
Three averages, three different questions. The mean asks “if everyone had the same, how much would each get?” The median asks “who is standing in the middle?” The mode asks “which answer came up most often?” None of them is the average; each is an average. So the useful question is never “what is the average?” but “which average, and why that one?”
mean = Σfx ÷ Σf
and every average must land between the smallest and the largest value — that is the check
Three questions decide everything on this page. Which average? — mean, median or mode. What shape is the data in? — a raw list, a frequency table or grouped classes. Is there an extreme value? — because if there is, the mean and the median will disagree, and choosing between them becomes the whole question.
Where this unit sits
What this unit assumes. Adding a list and dividing by how many there are (Number N02 and N03 — fraction and decimal arithmetic), and rounding an answer sensibly to 3 significant figures (Number N04 Round It Right). Neither is retaught here; both are used on every page.
What it builds on in this strand. S04 Picture the Data taught you to read values out of a chart. Here you average the values you read. The frequency tables are exactly the tables S01 built.
What comes next. S07 How Spread Out? adds the second half of every comparison: an average on its own says where the data sits, but not how tightly it is packed. S17 returns to grouped data when the values are measured rather than counted.
Where the numbers in this unit come from. Eleven marks, printed in full; a frequency table of 30 learners, printed in full; a grouped table of 30 players, printed in full; nine sponsored-walk amounts, printed in full. Every average on this page can be checked by hand against the data it came from, and that is deliberate.
1
Three averages, one list
mean, median and mode
The mean is the total of all the values divided by how many values there are — the “share it out equally” average. The median is the middle value once the list is in order. The mode is the value that occurs most often. All three are called averages, all three are single numbers standing in for the whole list, and on almost any real dataset all three are different. Write down which one you are finding before you start calculating, because the three take three different routes.
EXAMPLEThe raw list, in full. Eleven learners sat a mental-maths quiz marked out of 20. Their marks, in the order the teacher read them out, were:
16, 8, 10, 11, 9, 16, 10, 9, 13, 16, 14.
16, 8, 10, 11, 9, 16, 10, 9, 13, 16, 14.
Every dot is one learner. The tallest stack is the mode; the two dashed lines are the mean and the median, and they are not in the same place.
Worked example 1 · Three answers, one list
1
Write the eleven marks in order, smallest first.
8, 9, 9, 10, 10, 11, 13, 14, 16, 16, 16
2
Add them all up.
total = 132
3
Divide the total by how many values there are.
mean = 132 ÷ 11 = 12 marks
4
Find the median position: (n + 1) ÷ 2 with n = 11.
position 6
5
Count in to the 6th value of the ordered list.
median = 11 marks
6
Look for the value that occurs most often.
mode = 16 marks
7
Check all three lie between the smallest and the largest mark.
8 to 16
Three averages, one list, three different numbers — and not one of them is wrong. Set the working out in this order every time: order the list, then mean, then median, then mode, then the check that all three sit inside the data.
EXAMPLEThe eleven marks in the dot plot above give three different answers: the mean is 12, the median is 11 and the mode is 16. Same eleven learners, same eleven marks, three honest answers to three different questions.
Practice · Three averages, one list · 10 questions
P1Calculate the mean of 5, 8, 11, 12, 14.Fluency
Answer10
P2State the mode of 3, 7, 7, 9, 12.Fluency
Answer7
P3The list 4, 6, 9, 11, 20 is already in order. State the median.Fluency
Answer9
P4The eleven marks are printed above. State the mode.Fluency
Answer16 marks
P5Calculate the mean of the eleven marks, and state the total you divided.Applied
Answer12 marksAdd all eleven marks first: the total is 132. There are 11 marks, so the mean is 132 ÷ 11 = 12 marks. The total is worth writing down on its own line, because a mean that is not checked against a total is a mean nobody can mark.
P6State the median of the eleven marks and describe how you found it.Applied
Answer11 marksOrder the eleven marks: 8, 9, 9, 10, 10, 11, 13, 14, 16, 16, 16. With n = 11 the median position is (11 + 1) ÷ 2 = 6, so count in to the 6th value, which is 11 marks. Counting in from either end lands on the same value, which is a quick check.
P7Calculate the mean of 12, 15, 18, 21 and describe why an average need not be one of the values.Applied
Answer16.5The total is 66 and there are 4 values, so the mean is 66 ÷ 4 = 16.5. No one scored 16.5. The mean is what each would have if the total were shared out equally, and an equal share very often lands between two of the actual values.
P8A list of 6 values has a mean of 7. Calculate the total of the six values.Applied
Answer42The mean is the total divided by how many values there are, so the total is the mean multiplied by how many: 6 × 7 = 42. Running the mean backwards like this is how nearly every “find the missing value” question is unlocked.
P9Five values have a mean of 9. Four of them are 6, 8, 10 and 12. Calculate the fifth value.Challenge
Answer9Five values with mean 9 must have a total of 5 × 9 = 45. The four known values add to 6 + 8 + 10 + 12 = 36. The fifth value is therefore 45 − 36 = 9. Check by adding all five: 6 + 8 + 9 + 10 + 12 = 45, and 45 ÷ 5 = 9.
P10Describe the difference between what the mean uses and what the median uses.Challenge
Answerthe mean uses every value; the median uses only the position of the middle valueThe mean is built from the total, so changing any single value changes it. The median is built from the order: it only asks which value is in the middle, so changing the largest value to something much larger leaves the median exactly where it was. That single difference is why the two averages disagree whenever a dataset has an extreme value.
2
Finding the middle
order first, then count in
The median needs the list in order — smallest to largest — before anything else happens. Then the middle value sits at position (n + 1) ÷ 2. With n odd that lands on a value: for 11 values it is the 6th. With n even it lands halfway between two positions: for 10 values it is position 5.5, so the median is the mean of the 5th and 6th values, and it need not be a value in the list at all. The mode needs no ordering, but it can be shared by two values, and if every value occurs once we say there is no mode (that is the convention this course uses).
EXAMPLEFor the eleven marks, (11 + 1) ÷ 2 = 6, so the median is the 6th value of the ordered list, which is 11. For the ten sponsored-walk amounts later in this unit, (10 + 1) ÷ 2 = 5.5, so the median is halfway between the 5th and the 6th values.
Practice · Finding the middle · 10 questions
P11State the median of 12, 9, 15.Fluency
Answer12
P12Calculate the median of 4, 7, 9, 10.Fluency
Answer8
P13A list has 15 values in order. State the position of the median.Fluency
Answer8th
P14State the mode of 2, 3, 3, 5, 7, 7.Fluency
Answer3 and 7
P15The nine sponsored-walk amounts are printed below. Calculate their median.Applied
Answer45 pulaIn order the nine amounts are 10, 20, 35, 40, 45, 50, 50, 50, 60. With n = 9 the median position is (9 + 1) ÷ 2 = 5, so the median is the 5th amount, 45 pula. Four amounts lie below it and four lie above it, which is the check.
P16A list has 12 values. Describe where its median lies.Applied
Answerhalfway between the 6th and the 7th values(12 + 1) ÷ 2 = 6.5, so the median position falls between the 6th and the 7th values of the ordered list. The median is the mean of those two values, and with an even number of values it need not be one of the values at all.
P17Order the list 14, 8, 22, 11, 9, 30, 17 and state the median.Applied
Answer14In order: 8, 9, 11, 14, 17, 22, 30. There are 7 values, so the median position is (7 + 1) ÷ 2 = 4, and the 4th value is 14. Note that the middle of the list as written was 11, which is not the median at all.
P18One more amount of 130 pula joins the nine. Calculate the median of the ten amounts.Applied
Answer47.5 pulaIn order the ten amounts are 10, 20, 35, 40, 45, 50, 50, 50, 60, 130. With n = 10 the median position is (10 + 1) ÷ 2 = 5.5, so the median is halfway between the 5th and 6th values: (45 + 50) ÷ 2 = 47.5 pula.
P19A learner reads the median of 14, 8, 22, 11, 9, 30, 17 straight off as 11. Describe the error and state the correct median.Challenge
Answer14They took the middle of the list as it was written down. The median is defined on the ordered list: 8, 9, 11, 14, 17, 22, 30, whose 4th value is 14. The raw list almost never arrives in order, so ordering it is the first move, not an optional tidy-up.
P20A set of 8 values has a median of 12, and the two middle values are equal. State those two values and describe why.Challenge
Answerboth are 12With 8 values the median is the mean of the 4th and 5th values. If those two values are equal, their mean is that value itself, so both must be 12. If they were different — say 11 and 13 — their mean would still be 12, so “equal” is the extra piece of information that pins them both down.
3
Averages from a table
Σfx ÷ Σf
A frequency table is a list with the repeats collected up, so you never have to re-write it out. Add a third column, f × x: each value multiplied by how many times it occurred. Then mean = Σfx ÷ Σf — the total of the f × x column divided by the total of the frequency column. The symbol Σ just means “add up”. The mode is the value with the largest frequency, not the frequency itself, and the median is found by counting down the frequency column until you reach the middle position.
The f × x column does the repeating for you, so the 30 scores are never written out again.
Work each row out from the table itself. The total is the check, not the source.
Worked example 2 · From the table
1
Add an f × x column: each score multiplied by its frequency.
six rows
2
Row x = 0: 0 × 2.
f × x = 0
3
Row x = 2: 2 × 10.
f × x = 20
4
Row x = 5: 5 × 1.
f × x = 5
5
Add the f × x column.
Σfx = 69
6
Add the frequency column.
Σf = 30
7
Divide: mean = Σfx ÷ Σf.
69 ÷ 30 = 2.3 correct answers
The whole point is that the 30 scores are never written out again. The f × x column does the repeating for you, and the divisor is Σf = 30 — the number of learners — never the number of rows.
EXAMPLEFrom the spelling table: Σfx = 69 and Σf = 30, so the mean is 69 ÷ 30 = 2.3 correct answers. Notice that the mean is not a possible score — no learner got 2.3 answers right — and that is perfectly normal.
Practice · Averages from a table · 10 questions
P21The spelling table is printed above. State the frequency of the score 2.Fluency
Answer10
P22The spelling table is printed above. Calculate f × x for the row x = 4.Fluency
Answer20
P23The spelling table is printed above. State the mode.Fluency
Answer2 correct answers
P24The spelling table is printed above. State Σf.Fluency
Answer30
P25Calculate Σfx for the spelling table.Applied
Answer69Multiply each score by its frequency and add the column: 0 + 6 + 20 + 18 + 20 + 5 = 69. That total is the sum of all 30 learners’ scores, without any of them being written out again.
P26Calculate the mean score of the 30 learners.Applied
Answer2.3 correct answersmean = Σfx ÷ Σf = 69 ÷ 30 = 2.3 correct answers. The divisor is the number of learners, 30, and never the number of rows in the table.
P27Calculate the median score of the 30 learners from the table.Applied
Answer2With 30 values the median position is (30 + 1) ÷ 2 = 15.5, so it lies between the 15th and 16th scores. Counting down the frequency column: 2 learners scored 0, then 6 more reach 8, then 10 more reach 18. Both the 15th and the 16th learner are in that row, so the median is 2.
P28Describe why the mode of the spelling table is 2 and not 10.Applied
Answer2 is the value; 10 is only how many learners had it10 is the largest number in the frequency column, so it is the frequency of the mode, not the mode itself. The mode is the score that occurred most often, which is 2 correct answers. An answer to “what is the mode?” is always in the units of the variable.
P29A learner computes the mean of the spelling table as 69 ÷ 6 = 11.5. Describe the error and state the correct mean.Challenge
Answer2.3 correct answersThey divided by 6, the number of rows, instead of by Σf = 30, the number of learners. The correct mean is 69 ÷ 30 = 2.3. The check is that an average must lie between the smallest and the largest value: 11.5 is not even a possible score out of 5.
P30Calculate how many of the 30 learners scored 3 or more, and state which rows you added.Challenge
Answer12“3 or more” means the rows x = 3, x = 4 and x = 5, so add those three frequencies: 6 + 5 + 1 = 12 learners. The row x = 3 is included, because “or more” keeps the value itself.
4
When the data is grouped
midpoints, and the word estimate
Grouped data has been sorted into classes and the raw values thrown away. You can still say which class is fullest — the modal class — and you can still get a mean, but only by assuming every value in a class sits at that class’s midpoint. The estimated mean is Σfm ÷ Σf, where m is the midpoint of each class. It is an estimate, and the word is part of the answer: the real values are gone, so the real mean cannot be recovered from this table.
Five classes, five midpoints. The estimated mean is the f × m total divided by 30.
Worked example 3 · Grouped goals
1
Find the midpoint of each class: halfway between its two boundaries.
2.5, 7.5, 12.5, 17.5, 22.5
2
Class 0 ≤ x < 5: midpoint 2.5, frequency 2, so f × m.
f × m = 5
3
Class 10 ≤ x < 15: midpoint 12.5, frequency 11, so f × m.
f × m = 137.5
4
Add the f × m column.
Σfm = 395
5
Divide by the total frequency.
395 ÷ 30 = 13.2 goals (3 s.f.)
6
Name the fullest class.
modal class 10 ≤ x < 15
7
Say why the mean is only an estimate.
the raw goal counts are gone
Every number in that table is a class, not a value, so the mean it gives is an estimate. Say the word in the answer. The class 0 ≤ x < 5 is written with boundaries, so its midpoint is 2.5 — and since goals are whole numbers, that midpoint is itself an assumption, which is one more reason the answer is an estimate.
EXAMPLEFrom the grouped table: Σfm = 395 and Σf = 30, so the estimated mean is 395 ÷ 30 = 13.2 goals to 3 significant figures. The modal class is 10 ≤ x < 15. Both answers carry the word “estimated” or the word “class” for a reason.
Practice · When the data is grouped · 10 questions
P31The grouped goals table is printed above. State the modal class.Fluency
Answer10 ≤ x < 15
P32State the midpoint of the class 5 ≤ x < 10.Fluency
Answer7.5
P33State how many players are in the class 15 ≤ x < 20.Fluency
Answer8
P34The grouped goals table is printed above. State Σf.Fluency
Answer30
P35Calculate f × m for the class 10 ≤ x < 15.Applied
Answer137.5The midpoint of 10 ≤ x < 15 is (10 + 15) ÷ 2 = 12.5, and its frequency is 11, so f × m = 11 × 12.5 = 137.5. Every class contributes its midpoint that many times.
P36Calculate Σfm for the grouped goals table.Applied
Answer395Multiply each midpoint by its frequency and add: 5 + 45 + 137.5 + 140 + 67.5 = 395. That total is the estimated total number of goals scored by all 30 players.
P37Calculate the estimated mean number of goals, to 3 significant figures.Applied
Answer13.2 goalsestimated mean = Σfm ÷ Σf = 395 ÷ 30 = 13.2 goals to 3 significant figures. The word estimated belongs in the answer: the table no longer holds the players’ real goal counts.
P38Describe why the mean from the grouped table is only an estimate.Applied
Answerthe raw values are gone, so every value in a class is assumed to sit at the midpoint, which makes the answer an estimateGrouping threw the individual goal counts away. To get a total at all you have to assume every player in a class scored the class midpoint, and almost none of them did. The arithmetic is exact; the assumption is not, so the answer is an estimate and is written as one.
P39Determine which class holds the median of the 30 players, and describe how you found it.Challenge
Answer10 ≤ x < 15The median position is (30 + 1) ÷ 2 = 15.5, so it lies between the 15th and 16th players. Counting down the frequency column: 2, then 6 reaches 8, then 11 reaches 19 — so both the 15th and the 16th player are in the class 10 ≤ x < 15. The table cannot say where in that class the median sits, only which class holds it.
P40A learner writes the estimated mean as 13.1666667 goals. Describe what is wrong and state the answer to 3 significant figures.Challenge
Answer13.2 goalsNine figures of accuracy from a table that no longer holds the real goal counts is a false claim of precision. The division 395 ÷ 30 does not stop, so the answer is rounded: 13.2 goals to 3 significant figures. Rounding an estimate is not sloppiness; quoting it unrounded is dishonesty about what the table knows.
5
One extreme value
what an outlier does to each average
One value far away from the rest is an extreme value, or outlier. The mean is built from the total, so every value pulls on it and a far-away value pulls hard. The median is built from a position, so a far-away value only shifts it along the order by one place. That is the whole difference: an outlier moves the mean a long way and the median hardly at all. Neither answer is wrong — but only one of them still describes a typical value.
EXAMPLEThe raw list, in full. Nine learners walked for the school fund. The money each of them raised, in pula, was:
50, 20, 40, 60, 45, 50, 50, 35, 10. A tenth learner then hands in a card for 130 pula.
50, 20, 40, 60, 45, 50, 50, 35, 10. A tenth learner then hands in a card for 130 pula.
The two markers are already apart, because the amounts are not evenly spread.
Only one of the two markers really moves — and it moves past most of the data.
Worked example 4 · The visitor
1
Write the nine amounts in order.
10, 20, 35, 40, 45, 50, 50, 50, 60
2
Mean before: total ÷ 9.
360 ÷ 9 = 40 pula
3
Median before: (9 + 1) ÷ 2 = position 5.
45 pula
4
Add the tenth amount of 130 pula and re-order.
10, 20, 35, 40, 45, 50, 50, 50, 60, 130
5
Mean after: total ÷ 10.
490 ÷ 10 = 49 pula
6
Median after: (10 + 1) ÷ 2 = position 5.5, so halfway between the 5th and 6th values.
47.5 pula
7
Describe which average moved, and by how much.
mean by 9, median by 2.5
One value moved the mean 9 pula and the median 2.5 pula. The mean is built from the total, so it feels every pula of the newcomer; the median is built from a position, so it only slides along the order.
EXAMPLENine amounts have mean 40 pula and median 45 pula. One more amount of 130 pula arrives, and the mean jumps to 49 while the median moves only to 47.5. The mean moved 9 pula; the median moved 2.5.
Practice · One extreme value · 10 questions
P41The nine sponsored-walk amounts are printed above. State their mean.Fluency
Answer40 pula
P42The nine sponsored-walk amounts are printed above. State their median.Fluency
Answer45 pula
P43One more amount of 130 pula joins the nine. Calculate the new mean.Fluency
Answer49 pula
P44One more amount of 130 pula joins the nine. State the new median.Fluency
Answer47.5 pula
P45Calculate how far the mean moved when the 130 pula amount joined.Applied
Answer9 pulaThe mean was 40 pula and is now 49 pula, so it moved 49 − 40 = 9 pula. The new total is 490 and there are now 10 amounts, which is where the new mean comes from.
P46Calculate how far the median moved when the 130 pula amount joined.Applied
Answer2.5 pulaThe median was 45 pula and is now 47.5 pula, so it moved 2.5 pula. The new value only pushed the middle position along by half a place: it is now halfway between the 5th and 6th amounts instead of on the 5th.
P47Describe which average the extra amount moved more, and why.Applied
Answerthe mean, because it is built from the totalThe mean moved 9 pula and the median only 2.5 pula. Every value feeds into the mean through the total, so a value far from the rest pulls it hard. The median only cares about position, so one extra value at the far end slides it along the order by half a place and no further.
P48Calculate the range of the nine amounts and of the ten amounts.Applied
Answer50 pula and 120 pulaRange = largest − smallest. Before: 60 − 10 = 50 pula. After: 130 − 10 = 120 pula. One value more than doubled the range while the median barely moved — a first look at why S07 measures spread separately from centre.
P49The 130 pula amount is found to belong to another school and is taken out again. Calculate the mean of the remaining nine amounts and describe what that shows.Challenge
Answer40 pulaRemoving it takes the total back to 360 over 9 amounts, so the mean returns to exactly 40 pula. That is the point: the whole 9 pula shift was the work of one value out of ten. An average that one entry can move that far should always be quoted with the size of the dataset beside it.
P50Determine what a tenth amount would have to be for the mean of the ten to stay at 40 pula.Challenge
Answer40 pulaTen amounts with a mean of 40 need a total of 10 × 40 = 400. The nine amounts already total 360, so the tenth must be 400 − 360 = 40 pula — the mean itself. Adding a value equal to the mean never moves the mean, which is the cleanest way to see what the mean actually balances.
6
Which average, and why
choosing, and saying why
Mode for words and categories, and for anything where “most popular” is the question — a shop orders the modal shoe size, not the mean one. Median when one or two extreme values would drag the mean somewhere no one actually is — pay, prices, house sizes. Mean when the values are reasonably even and you want every one of them to count. Say which one you chose and why: an average quoted without its name is an invitation to be misled, which is where S03 came in.
Worked example 5 · Which average?
1
A shop must decide which shoe sizes to order for next term.
mode — the size that sells most
2
Give the reason.
the mean size might not be a size the shop can even buy
3
A small workshop is asked what a worker there typically earns, and one person earns many times what the others do.
median — it ignores the extreme value
4
Give the reason.
the mean is dragged up by the one large pay
5
A teacher wants one number for a class test where the marks are spread evenly with no extreme values.
mean — every mark counts
6
Give the reason.
no value is far enough out to distort the total
Choosing is the assessed skill, and the reason earns as much as the choice. Categories take the mode; extreme values take the median; even, complete data takes the mean.
EXAMPLEWhich average would you quote for the ten walk amounts? The mean is 49, the median is 47.5 and the mode is 50. If you want to describe what a typical learner raised, the mean is the one number of the three that no learner is anywhere near — one big sponsorship has moved it.
Practice · Which average, and why · 10 questions
P51State which average can be used for a variable whose answers are words, such as favourite colour.Fluency
Answerthe mode
P52State which average is least affected by one extreme value.Fluency
Answerthe median
P53State which average uses every value in the dataset.Fluency
Answerthe mean
P54Every value in a list occurs exactly once. State which average does not exist for it.Fluency
Answerthe mode
P55A shop must decide which shoe sizes to order. Describe which average it should use and why.Applied
Answerthe modeThe shop wants the size it will sell most of, which is the mode. The mean shoe size could easily come out as 7.4, and no shop can order a size 7.4 shoe. When the question is “which one is most popular?”, the mode is the only average that answers it.
P56A small workshop reports what a worker there typically earns. One person earns many times what everyone else does. Describe which average to use and why.Applied
Answerthe medianThe one very large pay drags the mean up to a figure nobody actually earns. The median ignores how far out that value is and reports the middle worker, so it still describes a typical pay. The mean is not wrong arithmetic — it is simply the wrong question.
P57A class test has marks spread evenly with no extreme values. Describe which average to use and why.Applied
Answerthe meanWith no extreme values there is nothing to distort the total, so the mean is safe and it has the advantage of using every mark. The median would throw away most of the information the class produced.
P58A shop says “the average shirt size we sell is 8.4”. Describe why that number is not useful.Applied
Answershirt sizes are counted in fixed steps, so 8.4 is not a size; the mode is what the shop needsThe shop has quoted a mean of a discrete variable. There is no size 8.4 to order, and the number does not tell the shop which sizes to stock. The mode — the size that sells most — is the average this question wants.
P59The ten walk amounts have mean 49, median 47.5 and mode 50 pula. Describe why “the average was 49 pula” is not enough on its own.Challenge
Answerit does not say which average, and the three disagreeThree averages, three different numbers: 49, 47.5 and 50 pula. Quoting “the average” without naming it lets the writer pick whichever number suits them — here the mean is the largest, and it is the one a fundraising poster would print. Naming the average, and saying how many values it came from, is the honest form.
P60A tuck shop wants to know which drink to order more of next term. Describe the average it needs and state what data it must collect.Challenge
Answerthe mode; a record of which drink each sale was“Which drink” is a category, so no mean or median exists — only the mode. What the shop must collect is one row per sale naming the drink, which is the tally-and-frequency-table job of S01. The most common drink in that table is the answer, and no other average is available for words.
The averaging routine
Four moves, whatever shape the data arrives in.
1
Which average is asked for?Mean, median or mode. Write the word down first: the three take three different routes and mixing them up costs the whole mark.
2
Put the list in order.Only the median strictly needs it, but ordering also makes the mode obvious and makes a silly mean easy to spot.
3
Do the arithmetic that average asks for.Mean: total ÷ how many. From a table: Σfx ÷ Σf. Grouped: Σfm ÷ Σf, and call it an estimate. Median: position (n + 1) ÷ 2. Mode: biggest frequency, then read off the value.
4
Check it sits inside the data, and give it its unit.An average must lie between the smallest and the largest value. If it does not, the arithmetic is wrong. Then say what it is: 12 marks, 2.3 correct answers, 49 pula.
The mistakes that cost the most marks
1Dividing Σfx by the number of rows instead of Σf
Wrong“Σfx = 69 and the table has 6 rows, so the mean is 69 ÷ 6 = 11.5.”
RightThe divisor is Σf = 30, the number of learners: 69 ÷ 30 = 2.3.
A frequency table has one row per value and one column of how many. The number of rows is a fact about the table; the mean is a fact about the learners. The check that catches it every time: an average must lie between the smallest and the largest value, and 11.5 is not even a possible score.
2Taking the middle of the list as it was written down
Wrong“For 12, 5, 20, 9, 7 the middle one is 20, so the median is 20.”
RightOrder the list first: 5, 7, 9, 12, 20. The middle value is now 9, so the median is 9.
The median is defined on the ordered list, and a raw list almost never arrives in order. Ordering costs ten seconds and is the only thing standing between you and a wrong median. Write the ordered list out on the answer line — it is also evidence of method.
3Reporting the frequency instead of the value as the mode
Wrong“The biggest number in the spelling table is 10, so the mode is 10.”
Right10 is the frequency. The mode is the value that goes with it: 10 learners scored 2, so the mode is 2 correct answers.
Say the sentence out loud: “the most common score was …”. The mode always answers in the units of the variable — correct answers, shoe sizes, goals — never in learners. The same slip turns up in bar charts, where the tallest bar is read as the answer instead of the label underneath it.
Tasks to submit
Criterion B · Investigating patterns
The newcomer
A list of n values has a mean of exactly 10. One new value, x, joins the list. The new mean depends on two things: how big x is, and how many values were already there. This investigation is about the second one.
Part 1 · the structured investigation
- Predict what the new mean will be when n = 9 and x = 22. Write the prediction down before you work anything out.
- Take n = 4 values with a mean of 10 — for example 7, 9, 11, 13 — and tabulate the new mean when x = 10, 16, 22.
- Identify the pattern in your table. Say what happens to the new mean each time x goes up by 6.
- Determine the general rule for the new mean when n values of mean 10 are joined by one value x, and write it as a formula.
- Test and verify the rule: repeat the whole table for n = 9, using a list of 9 values you choose yourself with mean 10, and compare every answer with your formula.
- Justify why the same value of x shifts the mean less when n is larger.
| New value, x | New mean when n = 4 | New mean when n = 9 |
|---|---|---|
| 10 | 10 | |
| 16 | ||
| 22 | ||
| x |
The first entry is done for you: joining a value of 10 to four values whose mean is already 10 leaves the mean at 10. The rest of the table is yours, and the last row is the general rule.
Part 2 · your own investigation
Investigate the same question for the median. A list of n values has a median of 10; one new value x joins it. Predict, tabulate, find the pattern, state a general rule if there is one, test it and justify what you find. You choose the lists, the values of x, the table and the way you present the result — and be careful: the answer may depend on things the mean did not care about.
Criterion B · Investigating patterns. Levels 1–2: works out some new means correctly for n = 4. Levels 3–4: completes the table correctly and describes the pattern in words. Levels 5–6: states a correct general rule for n values and x, and verifies it against the n = 9 table. Levels 7–8: justifies the rule from the structure of the mean — the new total is the old total plus x, and the shift is (x − 10) shared between n + 1 values, so it shrinks as n grows — and carries Part 2 through to a tested general statement about the median.
Remember, you should communicate in an organized and coherent manner.
Criterion D · Applying mathematics in real-life contexts
The average pay at the brickworks
A small brickworks in Serowe employs eleven people. The owner works there too and pays herself out of the same business. An advertisement for a new job is about to say what the average pay at the brickworks is, and the wage sheet below is all the information there is. The sheet carries more columns than the question needs.
Wage sheet · every person paid by the business
| Role | Monthly pay (P) | Years worked | Shift | Clock no. |
|---|---|---|---|---|
| Moulder | 2550 | 3 | day | 101 |
| Kiln loader | 3450 | 7 | day | 102 |
| Machine operator | 3600 | 5 | day | 103 |
| Yard hand | 2150 | 1 | day | 104 |
| Yard hand | 2050 | 2 | night | 105 |
| Driver | 3850 | 9 | day | 106 |
| Supervisor | 4150 | 12 | day | 107 |
| Store clerk | 4050 | 4 | day | 108 |
| Moulder | 2900 | 6 | night | 109 |
| Kiln loader | 3150 | 8 | night | 110 |
| Cleaner | 2300 | 2 | day | 111 |
| Owner | 24000 | 18 | day | 100 |
This wage sheet is an invented illustrative example written for the task. It is not taken from any business, price list or publication, and no organisation, person or source is named or implied.
- Interpreting the situation. Describe which column of the sheet answers “what is the average pay here?”, and which columns are surplus to that question. Say what each surplus column would be useful for instead.
- Selecting the mathematics. State which averages can be worked out from a pay column, and describe why the mode is of little use here.
- Applying the mathematics. Calculate the mean pay and the median pay of all 12 people on the sheet. Then calculate the mean pay of the eleven workers with the owner’s pay left out.
- Accuracy. Check your mean against the total you added, and state both averages to the nearest pula. Say how many values you ordered to find the median, and which two you used.
- Sense in context. Describe which average the advertisement would print, and which one honestly describes what a worker there earns. Then describe one thing this sheet cannot tell a job applicant.
Criterion D · Applying mathematics in real-life contexts. Levels 1–2: identifies the pay column and makes a start on one average. Levels 3–4: selects the right column, calculates the mean correctly and orders the sheet to find the median. Levels 5–6: all three figures correct with the total shown, the surplus columns identified, and a described recommendation naming the median as the honest figure. Levels 7–8: in addition, describes clearly that the mean of P4850 is higher than any of the eleven workers actually earns, so a single value can make an average describe nobody — and describes something the sheet cannot show, such as hours worked for that pay, which no pay column can ever reveal.
When you are ready
The mastery quiz is 10 questions drawn at random from a bank of 25. You need 70% to pass and you have three attempts. Type an average as digits only, with a decimal point and no unit — 2.3, not “2,3 correct answers”. Where an answer does not come out exactly, give it to 3 significant figures. A multi-gap answer separates its gaps with the bar character, in the order the gaps appear. Work through the practice first, especially Concept 4, where the word estimate is part of the answer.
Printable version
Download Unit S06 as a PDF — the same lesson, laid out for printing and for working offline.
Download Unit S06 as a PDF — the same lesson, laid out for printing and for working offline.