📐 Statistical Data Analysis — Revision Notes

Walkthrough 1 — building a cumulative frequency table Basic

1. The cumulative frequencies are the running totals 4,   4 + 11 = 15,   15 + 21 = 36,   36 + 16 = 52,   52 + 8 = 60.
   Each entry adds one more class to the ones already counted. The final total is 60, which is the number of leaves — that check is worth doing before anything is written into the table.

2. Length (x cm)	x ≤ 7	x ≤ 9	x ≤ 11	x ≤ 13	x ≤ 15
   Cumulative frequency	4	15	36	52	60

   The class labels change: each running total belongs to the upper boundary of its class, so the headings become "x ≤ 7", "x ≤ 9" and so on. The mark is for those new headings above the running totals.
3. Number of leaves 11 cm or shorter = 36
   "11 cm or shorter" is exactly the entry against x ≤ 11, so it is read straight out of the table with no arithmetic at all.
4. Number longer than 9 cm = 60 − 15 = 45
   "Longer than 9" is everything that is not "9 or shorter", so subtract the cumulative frequency at x ≤ 9 from the total. A cumulative frequency table always counts upwards; every "more than" question is a subtraction from n.
5. Number longer than 9 cm but not longer than 13 cm = 52 − 15 = 37
   A "between" count is the difference of two cumulative frequencies: everything up to 13 minus everything up to 9. x ≤ 13 keeps the 13s in and x ≤ 9 takes the 9s out, which leaves exactly 9 < x ≤ 13.

Walkthrough 2 — drawing the curve Basic

1. Cumulative frequencies: 2,   16,   32,   56,   73,   80
   Running totals first, always: 2, then 2 + 14, then 16 + 16, then 32 + 24, then 56 + 17, then 73 + 7. The last one is 80, the number of students, so the addition is right.
2. Points to plot: (20, 2), (30, 16), (40, 32), (50, 56), (60, 73), (70, 80) — and (10, 0) to start.
   Each running total is plotted against the upper boundary of its class, because that total counts everything up to that boundary. The extra point (10, 0) says that no student took 10 minutes or less, which is what the table's first class means.
3. Choose scales that use most of the grid: t from 10 to 70 on the horizontal axis, cumulative frequency from 0 to 80 on the vertical axis.
   A curve squeezed into a corner cannot be read accurately, and every later answer is a reading off this curve. Pick a scale where 2 cm (or 10 small squares) is a round number of units, so that the readings are easy and honest.

4. Fig. 13.1

   Join the crosses with a single smooth curve drawn in one movement, not with a ruler between consecutive points. The smooth curve is the model of the distribution; it is what makes the readings between the plotted points possible.

Walkthrough 3 — three readings off one curve Intermediate

1. Fig. 13.2

   From the curve, 36 minutes → 25 students and 53 minutes → 62 students.
   Both parts (a) and (b) start from a time, so both are read up first: vertical line to the curve with a ruler, then horizontal line to the axis. Read to the nearest whole student — a count of people cannot have a decimal.
2. Number who finished in 36 minutes or less = 25
   This one is the reading itself. Say "estimate": the curve is a model built from grouped data, so 25 is an estimate, not a count.
3. Number who took longer than 53 minutes = 80 − 62 = 18
   The curve counts upwards, so it gives the 62 who took 53 minutes or less. "Longer than" is the rest of the 80, so the subtraction is the line that answers the question.
4. Percentage = $\frac{18}{80}\times 100\%$ = 22.5%
   The question asked for a percentage, not a number of students, so convert. Dividing by 80 — the total — and not by 62 is the whole of the difficulty here.
5. 90% of 80 = $\frac{90}{100}\times 80$ = 72 students
   Part (c) gives a proportion, not a time, so the reading has to start on the cumulative frequency axis — and that axis is measured in students. Turn the 90% into 72 before touching the graph.

6. Fig. 13.3

   T = 59.1 (1 d.p.), so 90% of the students finished within about 59 minutes.
   Across from 72 to the curve, then straight down; 1 decimal place is about what the grid supports. Then say what the number means in minutes and students, since "T = 59.1" on its own is not yet an answer.

Walkthrough 4 — an even number of values Basic

1. In ascending order: 6,  7,  8,  9,  10,  11,  12,  13,  14,  17,  21,  25
   Everything in this walkthrough is a position in an ordered list, so ordering is not a preliminary — it is the first line of the solution. Count the values afterwards to be sure none was dropped: there are 12.
2. Position of median = $\frac{12+1}{2}$ = 6.5,  so median = $\frac{11+12}{2}$ = 11.5 books
   Position 6.5 means "halfway between the 6th and the 7th value", so the median is the mean of 11 and 12. With an even n the median need not be — and here is not — one of the data values.
3. Lower half: 6, 7, 8, 9, 10, 11  →  $Q_1$ = $\frac{8+9}{2}$ = 8.5 books
   With n even the 12 values split cleanly into two sixes. Q₁ is the median of the lower six, found the same way as before: the mean of that half's own middle pair.
4. Upper half: 12, 13, 14, 17, 21, 25  →  $Q_3$ = $\frac{14+17}{2}$ = 15.5 books
   Same procedure on the upper six. Notice Q₃ sits further from the median (4 above) than Q₁ does (3 below) — the big values 21 and 25 are pulling the top half apart.
5. IQR = $Q_3-Q_1$ = 15.5 − 8.5 = 7 books
   One subtraction, one number, with units. This says the middle half of the group's borrowing is spread over a band 7 books wide.
6. Range = 25 − 6 = 19 books
   Largest minus smallest. It is nearly three times the IQR, and the reason is the single member who borrowed 25 — one unusual reader stretches the range but leaves the interquartile range almost untouched.

Walkthrough 5 — an odd number of values Basic

1. In ascending order: 19,  22,  24,  26,  28,  29,  30,  33,  35,  38,  41
   Eleven values, so there is a genuine middle one. Ordering first, again — and counting to check that all eleven survived the reordering.
2. Position of median = $\frac{11+1}{2}$ = 6,  so median = 29 push-ups
   With n odd the position is a whole number, so the median is an actual data value — the 6th one — and no averaging is needed.
3. Lower half: 19, 22, 24, 26, 28  →  $Q_1$ = 24   (the median itself is not in this half)
   When n is odd the middle value is left out of both halves, leaving five values below and five above. Five is odd, so Q₁ is the middle one of those five, 24; including the 29 would give six values and a different answer.
4. Upper half: 30, 33, 35, 38, 41  →  $Q_3$ = 35
   The upper five, treated identically: the middle one is 35. The two halves must always contain the same number of values — if they do not, the median has been counted twice or dropped.
5. IQR = 35 − 24 = 11 push-ups
   The middle half of the class is spread over 11 push-ups. Give the units: a bare "11" does not say what has been measured.

Walkthrough 6 — median and quartiles off the curve Intermediate

1. n = 80, so read across at $\frac{80}{4}$ = 20, $\frac{80}{2}$ = 40 and $\frac{3\times 80}{4}$ = 60.
   Work out the three cumulative frequencies before touching the graph. They are positions on the vertical axis, so all three readings are "read across, then down".

2. Fig. 13.5

   $Q_1$ = 32.7 min,   median = 43.5 min,   $Q_3$ = 52.0 min
   Three separate readings, each with its own pair of ruled construction lines. Read to 1 decimal place — that is about what the grid supports, and a second decimal place would claim a precision the graph does not have.
3. Median = 43.5 minutes
   Half the students finished within about 43.5 minutes and half took longer. Say it in minutes: the number alone is not an answer about students.
4. IQR = 52.0 − 32.7 = 19.3 minutes
   Subtract, and give one number with its units. The middle 50% of the students are spread across a band 19.3 minutes wide — that is what an interquartile range is telling you.
5. Range = 70 − 10 = 60 minutes
   The curve runs from t = 10, where it leaves the axis, to t = 70, where it reaches 80, so those class boundaries are the best estimates of the fastest and slowest times. Both ends are estimates.

Walkthrough 7 — percentiles off the same curve Intermediate

1. 30% of 80 = $\frac{30}{100}\times 80$ = 24   and   90% of 80 = $\frac{90}{100}\times 80$ = 72
   Both percentages become numbers of students, because that is what the cumulative frequency axis counts. Doing this arithmetic first stops the classic error of looking for "30" on the time axis.

2. Fig. 13.6

   Two "read across" moves on the same graph, each with its own ruled lines. Keep them on the diagram — they are the working for this question.
3. $P_{30}$ = 35.3 minutes
   30% of the students — 24 of them — finished within about 35.3 minutes. To 1 decimal place, which is what this grid supports.
4. $P_{90}$ = 59.1 minutes
   The second reading, from 72. Notice it is far to the right of the median: the last 40% of the students are strung out across the second half of the time axis, which is what makes the top of the curve flatten off.
5. The slowest 10% are the 8 students above $P_{90}$, so the least time taken by a student in that group is 59.1 minutes.
   "The slowest 10%" and "the 90th percentile" are the same cut seen from the two sides: 90% at or below, 10% above. Recognising that saves you a second reading — and the answer is a time, so it is quoted in minutes.

Walkthrough 8 — working backwards from the median Advanced

1. For the mode to be 1, we need x < 5, so the greatest possible value is x = 4.
   The mode is the value with the highest frequency, and 1 occurs 5 times. If x reached 5 there would be two modes; if it passed 5 the mode would be 3. So x must stop at 4.
2. Total number of students n = 3 + 5 + 4 + x + 2 + 1 = x + 15
   Everything about a median is about position, and position depends on n, so n has to be written in terms of x before anything else can be said.
3. Cumulative frequencies:  0 → 3,  1 → 8,  2 → 12,  3 → 12 + x,  4 → 14 + x,  5 → 15 + x
   The running totals say which values sit in which positions: positions 9 to 12 hold the value 2, positions 13 to 12 + x hold the value 3. That turns a statement about the median into a statement about a position.
4. Position of median = $\frac{n+1}{2}$ = $\frac{x+16}{2}$
   This is listed, discrete data — the students are counted one by one — so the position rule is $\frac{n+1}{2}$, not $\frac{n}{2}$.
5. For the median to be 2, this position must lie in positions 9 to 12: $\frac{x+16}{2}\le 12$,  so x + 16 ≤ 24 and x ≤ 8.
   Positions 9 to 12 are the ones holding the value 2. Pushing x larger drags the median position rightwards into the 3s, so the largest x is the one that leaves the median position at 12, the last position still holding a 2.
6. Greatest possible value: x = 8   (check: n = 23, position 12, and the 12th value is 2 ✓)
   Test the value just found: with x = 8 the median position is 12 and the 12th student joined 2 clubs. With x = 9 the position is 12.5 and the median becomes $\frac{2+3}{2}$ = 2.5, so 8 is the largest.

Walkthrough 9 — drawing a box-and-whisker plot Basic

1. In ascending order: 18,  21,  24,  26,  28,  30,  31,  33,  36,  39,  44,  47
   A box plot is built entirely out of positions in an ordered list, so ordering is the first line. Twelve values, so the smallest is 18 and the largest is 47 — two of the five numbers are already settled.
2. Median = $\frac{30+31}{2}$ = 30.5   (position $\frac{12+1}{2}$ = 6.5)
   Twelve values means the median falls between the 6th and 7th, so it is their mean. It need not be a possible mark, and here it is not — 30.5 out of 50 was nobody's score.
3. Lower half 18, 21, 24, 26, 28, 30 → $Q_1$ = $\frac{24+26}{2}$ = 25;  upper half 31, 33, 36, 39, 44, 47 → $Q_3$ = $\frac{36+39}{2}$ = 37.5
   Each quartile is the median of its own half of six values. Writing both halves out is worth the two seconds: it is where a miscount would show up.
4. Five-number summary: 18,  25,  30.5,  37.5,  47
   List the five in order before drawing anything. Anything out of order here means an error above, and the picture would be drawn from wrong numbers.

5. Fig. 13.8

   Scale first, then the five marks measured against it. The label Mark on the axis is part of the answer, and the median line sitting left of centre shows that the top half of the class is more spread out than the bottom half.

Walkthrough 10 — getting five answers out of one box plot Intermediate

1. Five-number summary read off the scale: 8,  26,  41,  58,  84
   Read all five before answering anything — the two whisker ends, the two box edges and the line inside the box. Every part of the question is then just arithmetic on these five numbers.
2. Range = 84 − 8 = 76 years
   Largest minus smallest: the two ends of the whiskers, not the ends of the box. Units, because an age is a measurement.
3. Median age = 41 years
   The line drawn inside the box, read straight off the scale. Half the members are younger than 41 and half are older.
4. IQR = 58 − 26 = 32 years
   The width of the box. The middle half of the membership is spread over 32 years — a wide band, which is what you would expect of a club open to all ages.
5. 58 is $Q_3$, so 25% of the members are older: $\frac{25}{100}\times 200$ = 50 members
   The upper quartile has three quarters of the data at or below it, so a quarter above. This is the only kind of counting question a box plot can answer, and only because the value asked about is exactly a quartile.
6. The oldest 25% of members (from 58 to 84) are spread over 26 years, while the youngest 25% (from 8 to 26) are spread over only 18 years, so the ages of the oldest quarter are more varied than those of the youngest quarter.
   Both whiskers hold the same 25% of the members, so the longer one means the same number of people spread more thinly. Say it about members and ages rather than about "the whisker" — the comment has to be about the club.

Walkthrough 11 — comparing two box plots Intermediate

1. Route A: 18, 24, 27, 33, 44  →  median = 27 min,  IQR = 33 − 24 = 9 min
   Read the five numbers off the shared scale, then do the one subtraction. Reading both plots against the same axis is the whole reason they were drawn this way.
2. Route B: 21, 27, 30, 33, 38  →  median = 30 min,  IQR = 33 − 27 = 6 min
   The same five readings for the second plot. Both boxes end at 33, so it is the left edges and the median lines that will do the work in the comparison.
3. Route A has a lower median (27 min against 30 min), so on average a journey on Route A is about 3 minutes quicker than one on Route B.
   Comment one: the average. Quote both figures, say which is larger, and then translate it — "quicker journeys", not "a smaller median". A comparison that never leaves the language of statistics has not compared the buses.
4. Route A has a larger interquartile range (9 min against 6 min), so the journey times on Route A are more spread out and less consistent than those on Route B.
   Comment two: the spread. Two comments means two different things — one about the middle and one about the variation. Two remarks about the median would score once.
5. Route B, because its times are far more consistent: its slowest journey was 38 minutes against Route A's 44, and its middle half spans only 6 minutes. A commuter who must not be late needs a reliable time more than a short average one.
   The decision hangs on the spread, not the average, because being late is caused by the bad days, not the typical ones. Name the route and give the reason: an answer that says only "Route B" has not been justified.

Walkthrough 12 — from cumulative frequency curve to box plot Advanced

1. Smallest value ≈ 10 min and largest value ≈ 70 min, from the two ends of the curve.
   Grouped data never reveals the actual extremes, so the class boundaries are the best estimates available: nobody finished at or below 10 minutes and nobody took more than 70. Both are estimates, and the answer should say so.
2. Five-number summary: 10,  32.7,  43.5,  52.0,  70 (minutes)
   The two ends from the curve's endpoints, the three middle numbers from the readings already taken at 20, 40 and 60 on the cumulative frequency axis. A box plot needs nothing else.

3. Fig. 13.10

   Reuse the horizontal scale of the curve. Apart from making the plot quick to draw, it lets you check each mark against the curve you read it from.
4. Left whisker: 32.7 − 10 = 22.7 min.  Right whisker: 70 − 52.0 = 18.0 min.  Inside the box: 43.5 − 32.7 = 10.8 min below the median, 52.0 − 43.5 = 8.5 min above it.
   Shape questions are answered by measuring the four sections, not by looking. Each section holds 20 of the 80 students, so the widths are directly comparable.
5. Both sections below the median are wider than the matching sections above it, so the faster half of the students is more spread out: the times bunch up towards the slow end of the distribution.
   Same 20 students in each section, so a wider section means the same number of people spread over more minutes. Describe it in students and minutes — "the plot is skewed" is a label, not an interpretation.
6. The box plot shows only the five-number summary. The mean needs every value (or every frequency and mid-value), and those have been thrown away, so the mean cannot be found from the box plot.
   This is the honest limitation of the picture: a question that asks it is testing whether you know what a representation keeps and what it discards. The mean could still be estimated, but from the original frequency table using mid-values.

Walkthrough 13 — standard deviation of a plain list Basic

1. $\sum x$ = 14 + 9 + 17 + 12 + 20 + 11 + 15 = 98   and   $\sum x^2$ = 196 + 81 + 289 + 144 + 400 + 121 + 225 = 1456
   Two totals, and that is all the formula needs. Each value occurs once, so this is the f = 1 case and the frequency column would be a column of 1s — there is no point writing it.
2. Mean $\overline{x}=\frac{\sum x}{n}=\frac{98}{7}$ = 14 chin-ups
   The mean comes first because the standard deviation formula needs it. Here it works out exactly, which is a small mercy — when it does not, keep the full decimal on your calculator rather than rounding it now.
3. s.d. = $\sqrt{\frac{\sum x^2}{n}-{\overline{x}}^2}$ = $\sqrt{\frac{1456}{7}-{14}^2}$
   Quote the formula, then substitute: that line is worth a mark on its own. The $\sum x^2$ goes on top of n; the mean is squared and subtracted.
4. = $\sqrt{208-196}$ = $\sqrt{12}$ = 3.4641…
   208 is the mean of the squares and 196 is the square of the mean; the gap between them is what the spread costs. It is positive, as it must be — a negative would mean the two were subtracted the wrong way round.
5. = 3.46 chin-ups (to 3 s.f.)
   Round once, at the end, to the accuracy asked for. The standard deviation carries the same units as the data, so it is 3.46 chin-ups — a typical distance from the mean of 14.

Walkthrough 14 — standard deviation from a frequency table Basic

1. Siblings (x)	Students (f)	fx	fx²
   0	6	0	0
   1	14	14	14
   2	12	24	48
   3	6	18	54
   4	2	8	32
   Sum	$\sum f$ = 40	$\sum fx$ = 64	$\sum f{x}^2$ = 148

   Four columns, three totals. The last column is f × x²: for x = 3 that is 6 × 9 = 54, not 18². Check $\sum f$ = 40 against the question before going on.
2. Mean $\overline{x}=\frac{\sum fx}{\sum f}=\frac{64}{40}$ = 1.6 siblings
   Straight from the formula sheet. Dividing by 40 (the number of students) and not by 5 (the number of rows) is the point of the $\sum f$ on the bottom.
3. s.d. = $\sqrt{\frac{\sum f{x}^2}{\sum f}-{\overline{x}}^2}$ = $\sqrt{\frac{148}{40}-{1.6}^2}$
   The formula, then the substitution, on separate lines. Both totals come straight out of the table, which is exactly why the table was built.
4. = $\sqrt{3.7-2.56}$ = $\sqrt{1.14}$ = 1.0677…
   3.7 is the mean of the squares, 2.56 is the square of the mean. Keep several decimal places off the calculator; the rounding is one step, at the end.
5. Mean = 1.6 siblings,  standard deviation = 1.07 siblings (to 3 s.f.)
   The mean is exact, so it is not rounded; the standard deviation is not, so it goes to 3 s.f. Answer both parts explicitly — a solution that stops at 1.0677 has not stated the mean it was also asked for.

Walkthrough 15 — grouped data, using mid-values Intermediate

1. Mid-values:   $\frac{0+4}{2}$ = 2,  6,  10,  14,  18,  22
   Grouped data does not tell you the individual values, so each class is represented by its mid-value — the mean of the two class boundaries. That substitution is the reason every answer here is an estimate.

2. Class	Mid-value (x)	Frequency (f)	fx	fx²
   0 < d ≤ 4	2	8	16	32
   4 < d ≤ 8	6	17	102	612
   8 < d ≤ 12	10	31	310	3100
   12 < d ≤ 16	14	26	364	5096
   16 < d ≤ 20	18	13	234	4212
   20 < d ≤ 24	22	5	110	2420
   Sum		$\sum f$ = 100	$\sum fx$ = 1136	$\sum f{x}^2$ = 15 472

   The same four columns as before; only the meaning of x has changed. For the third class, fx² = 31 × 10² = 3100 — square the mid-value, then multiply by the frequency. $\sum f$ = 100 agrees with the question.
3. Estimated mean = $\frac{1136}{100}$ = 11.36 GB
   "Estimate" is not politeness — the mean of the mid-values is not the mean of the real amounts, and the word belongs in the answer. Keep the full 11.36 for the next line rather than rounding it to 11.4.
4. s.d. = $\sqrt{\frac{\sum f{x}^2}{\sum f}-{\overline{x}}^2}$ = $\sqrt{\frac{15472}{100}-{11.36}^2}$
   Formula, then substitution. Substituting the unrounded mean matters here: using 11.4 instead of 11.36 would change the answer in the third significant figure, which is exactly where it will be marked.
5. = $\sqrt{154.72-129.0496}$ = $\sqrt{25.6704}$ = 5.0665…
   Both numbers under the root are large and close together, so carry all the decimals — chopping 129.0496 to 129 would move the answer. This is the step where an early rounding does the most damage.
6. Estimated mean = 11.4 GB (3 s.f.),  estimated standard deviation = 5.07 GB (3 s.f.)
   Round both, once, at the very end. Both are estimates and both carry the units of the data. A standard deviation of about 5 GB against a mean of about 11 GB says the usage varies a great deal from student to student.

Walkthrough 16 — a scale that reads low Advanced

1. Every recorded mass is 15 g too small, so the correct masses are (recorded mass + 15) g.
   Read the error carefully: it is the same 15 g on every loaf, not a percentage and not a random amount. That single fact is what makes the question answerable without knowing any of the seven masses.
2. Correct mean = 248 + 15 = 263 g
   Adding 15 to all seven masses adds 7 × 15 to the total, so it adds 15 to the mean. The whole distribution slides 15 g to the right and its centre slides with it.
3. Each loaf's distance from the mean is unchanged: both the value and the mean have gone up by the same 15 g.
   This is the reason for the answer that follows. Since (x + 15) − ($\overline{x}$ + 15) = x − $\overline{x}$, every deviation in the formula is exactly what it was before.
4. Correct standard deviation = 12.6 g — unchanged.
   Every squared deviation is unchanged, so their mean is unchanged and so is its square root. Say "unchanged" out loud in the answer: a marker needs to see that this was a decision, not a value copied by accident.

Walkthrough 17 — putting two groups together Advanced ⓘ Extension

1. $\sum x$ for the boys = 18 × 168 = 3024   and   $\sum x$ for the girls = 12 × 160 = 1920
   Rearranging $\overline{x}=\frac{\sum x}{n}$ gives $\sum x=n\overline{x}$. Totals can be added together; means cannot, because the two groups are different sizes.
2. Combined mean = $\frac{3024+1920}{30}$ = $\frac{4944}{30}$ = 164.8 cm
   One total height divided by one total number of students. Note it is not the average of 168 and 160 — that would be 164, and it is wrong because there are more boys than girls pulling the mean upwards.
3. From $\text{s.d.}=\sqrt{\frac{\sum x^2}{n}-{\overline{x}}^2}$, rearranged: $\sum x^2=n({\text{s.d.}}^2+{\overline{x}}^2)$
   Standard deviations cannot be added either, but $\sum x^2$ can. So the plan is to recover each group's $\sum x^2$ from the summary it was given as, add the two, and put the total back into the same formula.
4. Boys: 18(6² + 168²) = 18 × 28 260 = 508 680
   Girls: 12(5² + 160²) = 12 × 25 625 = 307 500
   Total $\sum x^2$ = 816 180
   Two substitutions into the rearranged formula, then one addition. The numbers are large — that is normal here, because every height is being squared — so use the calculator's memory rather than copying digits by hand.
5. Combined s.d. = $\sqrt{\frac{816180}{30}-{164.8}^2}$ = $\sqrt{27206-27159.04}$ = $\sqrt{46.96}$ = 6.8527…
   The same formula as always, now with the totals for all 30 students and the combined mean of 164.8. The two numbers under the root differ by only 47 out of 27 000, so keep every digit of both.
6. Mean = 164.8 cm,  standard deviation = 6.85 cm (to 3 s.f.)
   The combined spread, 6.85 cm, is larger than either group's own (6 cm and 5 cm) — as it should be, since the two groups have different means, so the gap between boys and girls adds to the variation inside each group.

Walkthrough 18 — comparing with the mean and standard deviation Intermediate

1. $\sum x$ = 4000   and   $\sum x^2$ = 2 000 078,  with n = 8
   The two totals the formula needs. The squares are large — around 250 000 each — so keep every digit; the whole calculation turns on a difference of about 10 between two numbers near 250 000.
2. Mean = $\frac{4000}{8}$ = 500 g
   Exactly the labelled mass. That is worth noticing straight away, because it means the comparison in part (b) cannot be settled by the average — the two machines are level on it.
3. s.d. = $\sqrt{\frac{2000078}{8}-{500}^2}$ = $\sqrt{250009.75-250000}$ = $\sqrt{9.75}$ = 3.12 g (to 3 s.f.)
   This is why the digits mattered: 250 009.75 − 250 000 = 9.75, and rounding either number to the nearest whole would have changed the answer completely. Round only the final square root.
4. The mean mass is 500 g for both machines, so on average both machines fill the packets to the labelled mass.
   Comment one, about the average. "The same" is a perfectly good comparison — and here it is the important one, because it says neither machine is systematically over- or under-filling.
5. Machine P has the smaller standard deviation (3.12 g against 4.2 g), so the masses of P's packets are less spread out about 500 g — P is the more consistent machine.
   Comment two, about the spread, and a genuinely different point from the first. Both figures are quoted, the direction is stated, and it is then translated into what it means about packets of rice.
6. Machine P, because with the same mean mass its standard deviation is smaller, so fewer of its packets will be noticeably underweight or overweight.
   The averages tie, so the decision has to rest on the spread. Naming the machine is not enough on its own — the justification is what is being marked, and it has to be about packets, not about numbers.

Walkthrough 19 — comparing two cumulative frequency curves Advanced

1. n = 80 for each school, so read across at $\frac{80}{4}$ = 20, $\frac{80}{2}$ = 40 and $\frac{3\times 80}{4}$ = 60 — the same three lines serve both curves.
   The two schools have the same number of students, so one set of horizontal lines gives six readings. If the two sets were different sizes, each curve would need its own $\frac{n}{4}$, $\frac{n}{2}$ and $\frac{3n}{4}$.
2. School A: $Q_1$ = 32.7, median = 43.5 min, $Q_3$ = 52.0, so IQR = 52.0 − 32.7 = 19.3 min
   Three readings and a subtraction, exactly as in Walkthrough 6. Read to 1 decimal place — the grid will not carry more — and keep the construction lines on the graph.
3. School B: $Q_1$ = 28.2, median = 35.5 min, $Q_3$ = 43.0, so IQR = 43.0 − 28.2 = 14.8 min
   The same three lines, stopped at the dashed curve instead of the solid one. School B's curve lies to the left of School A's throughout, so every one of its readings is smaller.
4. School B has the lower median (35.5 min against 43.5 min), so on average the students at School B finished the paper about 8 minutes faster than those at School A.
   Comment one, about the average. Both medians quoted, direction stated, and translated into students and minutes. Note that here lower is better — for a time taken, small is fast.
5. School B has the smaller interquartile range (14.8 min against 19.3 min), so the times at School B are less spread out — School B's students were more consistent in how long they took.
   Comment two, about the spread, using the IQR because the averages being compared are medians. Median with IQR, mean with standard deviation — the pairs are not interchangeable.

6. Fig. 13.13

   Each box plot is built from its curve: the two class boundaries at the ends, and the three quartiles just read off. Drawing them on one scale is what makes the two distributions comparable at a glance.
7. School B's box is both further left and narrower — the two comparisons above, seen in one picture. The plots also show that the two whisker ends are identical, which is a limitation of the data rather than a real similarity: grouped into classes of width 10, neither school's fastest or slowest time can be estimated any better than by the class boundaries 10 and 70.
   Say what the picture adds and what it cannot say. The equal whiskers are an artefact of the grouping, so treating them as a real finding — "both schools had the same range" — reads more into the data than it holds.