Question: A cumulative frequency curve shows the marks of $100$ students. Reading the curve: cumulative frequency $50$ ↔ $62$ marks, $25$ ↔ $47$ marks, $75$ ↔ $74$ marks; also $30$ marks ↔ cf $6$ and $40$ marks ↔ cf $16$. Find the median, the interquartile range, and the number of students scoring $30 \le x < 40$.

[diagram in original sheet: S-shaped cumulative frequency curve, 0–100 marks]

Step 1. On a CF curve, positions use $\dfrac{1}{2}n$, $\dfrac{1}{4}n$, $\dfrac{3}{4}n$ (no $+1$):

$$\text{Median position} = \dfrac{100}{2} = 50 \;\Rightarrow\; \text{Median} = 62 \text{ marks}$$

Step 2. Quartiles the same way:

$$Q_1 \text{ (cf } 25) = 47, \qquad Q_3 \text{ (cf } 75) = 74$$

Step 3. Interquartile range:

$$\text{IQR} = 74 - 47 = 27 \text{ marks}$$

Step 4. A frequency between two marks is a difference of two readings:

$$\#(30 \le x < 40) = \text{cf}(40) - \text{cf}(30) = 16 - 6 = 10 \text{ students}$$

⚠ Watch out: Read the curve in the right direction — quartiles start from the cumulative-frequency axis and move across to the curve. And a single cf reading is "how many scored less than this mark", so band counts always need a subtraction.

Question: A stem-and-leaf diagram shows the Maths test scores of 12 students:

Stem $5$: leaves $3, 8$ — Stem $6$: leaves $0, 1, 4$ — Stem $7$: leaves $2, 2, 6, 9$ — Stem $8$: leaves $1, 1, 2$.

Find the mean, the median and the mode.

Step 1. Read the diagram: stem = tens digit, leaf = units digit. The scores are

$$53, 58, 60, 61, 64, 72, 72, 76, 79, 81, 81, 82$$

Step 2. Mean — add all twelve, divide by 12:

$$\bar{x} = \dfrac{839}{12} = 69\dfrac{11}{12} \approx 69.9 \quad (3 \text{ s.f.})$$

Step 3. Median — the leaves are already in order. Position $\dfrac{12+1}{2} = 6.5$, so average the 6th and 7th scores:

$$\text{Median} = \dfrac{72+72}{2} = 72$$

Step 4. Mode — the most frequent score. Both $72$ and $81$ appear twice:

$$\text{Mode} = 72 \text{ and } 81$$

Yes — a data set can have two modes.

⚠ Watch out: A stem-and-leaf diagram needs a key (e.g. $5\,|\,3$ means $53$) when you draw one, and every leaf must be listed in ascending order. The built-in ordering is the whole point: the median position can be counted straight off the diagram.

Question: A box-and-whisker plot of students' travel times (minutes) has whiskers ending at $5$ and $50$, box edges at $22$ and $32$, and the line inside the box at $25$. Find the median, the range and the interquartile range.

[diagram in original sheet: box-and-whisker plot on a 0–60 min scale]

Step 1. Name the five landmarks first:

minimum $= 5$, $\;Q_1 = 22$, $\;$median $= 25$, $\;Q_3 = 32$, $\;$maximum $= 50$.

Step 2. Median — the line inside the box:

$$\text{Median} = 25 \text{ min}$$

Step 3. Range — whisker end to whisker end:

$$\text{Range} = 50 - 5 = 45 \text{ min}$$

Step 4. Interquartile range — the width of the box:

$$\text{IQR} = Q_3 - Q_1 = 32 - 22 = 10 \text{ min}$$

Each section of the plot (whisker–box–box–whisker) holds $25\%$ of the data — so, for instance, $25\%$ of students took more than $32$ minutes.

⚠ Watch out: The box edges are the quartiles, not the extremes — range uses the whisker tips, IQR uses the box. And the median line is usually not in the middle of the box; an off-centre line just means the middle 50% is skewed.

Question: The ages of 100 buyers fall in the intervals $37$–$46$ (19 people), $47$–$56$ (26), $57$–$66$ (27), $67$–$76$ (28). Estimate the mean and standard deviation, and state the modal class.

Step 1. The exact ages are unknown, so represent each interval by its mid-value:

$$\dfrac{37+46}{2} = 41.5, \qquad 51.5, \qquad 61.5, \qquad 71.5$$

Step 2. Estimated mean — each mid-value weighted by its frequency:

$$\bar{x} = \dfrac{19(41.5) + 26(51.5) + 27(61.5) + 28(71.5)}{100} = \dfrac{5790}{100} = 57.9$$

Step 3. Standard deviation, with $\sum x^2 = 346945$ from the calculator's STAT mode:

$$\text{SD} = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2} = \sqrt{\dfrac{346945}{100} - 57.9^2} = \sqrt{117.04} \approx 10.8$$

Step 4. Modal class — the interval with the highest frequency (28):

$$67\text{–}76$$

⚠ Watch out: Divide by the total frequency ($100$), never by the number of intervals ($4$). The answer is an estimate — say so — because mid-values stand in for unknown exact ages. And it is the modal class for grouped data, not a single mode.

Question: The times spent by 200 customers at a café are: $40 \le t < 60$ (36 customers), $60 \le t < 80$ (57), $80 \le t < 100$ (49), $100 \le t < 120$ (58), drawn as a histogram. Estimate the mean time.

[diagram in original sheet: histogram with equal-width bars, no gaps]

Step 1. A histogram shows grouped continuous data — bars touch, and each bar covers its whole interval. To estimate the mean, take each class's mid-value:

$$\dfrac{40+60}{2} = 50, \qquad 70, \qquad 90, \qquad 110$$

Step 2. Weight each mid-value by its frequency:

$$\sum fx = 36(50) + 57(70) + 49(90) + 58(110)$$ $$= 1800 + 3990 + 4410 + 6380 = 16580$$

Step 3. Divide by the total frequency:

$$\bar{t} = \dfrac{16580}{200} = 82.9 \text{ minutes}$$

Step 4. Sense-check against the histogram: the two tallest bars sit at $60$–$80$ and $100$–$120$, so a mean in the low $80$s — between them — is plausible. ✓

⚠ Watch out: Bars in a histogram must touch — gaps belong to bar charts of discrete data. Frequency is read off the bar height only when the class widths are equal, as here; and the mean from grouped data is always an estimate.

Question: Box-and-whisker plots show the speeds of 100 vehicles before and after a speed camera was installed. Before: median $82$ km/h, $Q_1 = 76$, $Q_3 = 86$. After: median $76$ km/h, $Q_1 = 71$, $Q_3 = 80$. Has the camera been effective? Compare the two distributions.

Step 1. Compute the comparison numbers — one measure of centre, one of spread:

Before: median $= 82$, IQR $= 86 - 76 = 10$.

After: median $= 76$, IQR $= 80 - 71 = 9$.

Step 2. Sentence 1 — compare centres, with figures:

The camera has been effective: the median speed after ($76$ km/h) is lower than the median speed before ($82$ km/h).

Step 3. Sentence 2 — compare spreads, interpreted in context:

The speeds after are also more consistent, as the IQR after ($9$ km/h) is smaller than before ($10$ km/h).

Step 4. That two-sentence shape — centre with figures, then spread with figures, each tied to the context — is exactly what the 2-mark comparison question wants.

⚠ Watch out: "Compare" needs both a centre (median/mean) and a spread (IQR/SD), each with its numbers quoted — "B did better" alone scores nothing. And interpret spread as consistency in context, not just "the IQR is smaller".

Question: The heights of 9 seedlings, in cm, are $13, 18, 13, 14, 13, 16, 14, 21, 13$. Find the mean, the median and the mode.

Step 1. Mean — total divided by how many:

$$\bar{x} = \dfrac{13+18+13+14+13+16+14+21+13}{9} = \dfrac{135}{9} = 15 \text{ cm}$$

Step 2. Median — arrange in ascending order first:

$$13,\ 13,\ 13,\ 13,\ 14,\ 14,\ 16,\ 18,\ 21$$

The centre position is $\dfrac{n+1}{2} = \dfrac{9+1}{2} = 5$, so the median is the 5th value:

$$\text{Median} = 14 \text{ cm}$$

Step 3. Mode — the value appearing most often. $13$ appears four times:

$$\text{Mode} = 13 \text{ cm}$$

Step 4. Sense-check the three: here mode $(13)$ < median $(14)$ < mean $(15)$ — the single large value $21$ pulls the mean up but leaves the median untouched. That is why the median is preferred when there is an outlier.

⚠ Watch out: Sorting comes before finding the median — the middle of the unsorted list is meaningless. And with an even $n$ the position $\frac{n+1}{2}$ falls between two values: average them.

Question: For the 9 seedling heights $13, 13, 13, 13, 14, 14, 16, 18, 21$ (already sorted, in cm), find $Q_1$, $Q_3$ and the interquartile range.

Step 1. Locate the quartiles by position (data listed, not grouped):

$$Q_1 \text{ position} = \dfrac{1}{4}(n+1) = \dfrac{10}{4} = 2.5$$

Step 2. Position $2.5$ sits between the 2nd and 3rd values ($13$ and $13$):

$$Q_1 = \dfrac{13+13}{2} = 13$$

Step 3. Same for the upper quartile:

$$Q_3 \text{ position} = \dfrac{3}{4}(n+1) = 7.5 \;\Rightarrow\; Q_3 = \dfrac{16+18}{2} = 17$$

Step 4. The interquartile range is the spread of the middle 50%:

$$\text{IQR} = Q_3 - Q_1 = 17 - 13 = 4 \text{ cm}$$

(Compare: the full range is $21 - 13 = 8$ cm — the IQR ignores the extreme values, so the outlier $21$ doesn't inflate it.)

⚠ Watch out: For a listed data set the quartile positions use $\frac{1}{4}(n+1)$ and $\frac{3}{4}(n+1)$; for a cumulative frequency curve they switch to $\frac{1}{4}n$ and $\frac{3}{4}n$. Mixing up the two conventions is the classic statistics slip.

Question: The marks of 10 students are $19, 17, 22, 24, 24, 24, 19, 27, 18, 25$. Find the standard deviation.

Step 1. Standard deviation measures how far the data spreads from the mean. The exam formula is

$$\text{SD} = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}$$

Step 2. Get the three ingredients (calculator STAT mode, or by hand):

$$\sum x = 219 \;\Rightarrow\; \bar{x} = \dfrac{219}{10} = 21.9$$ $$\sum x^2 = 19^2 + 17^2 + \cdots + 25^2 = 4901, \qquad n = 10$$

Step 3. Substitute:

$$\text{SD} = \sqrt{\dfrac{4901}{10} - 21.9^2} = \sqrt{490.1 - 479.61} = \sqrt{10.49}$$

Step 4. Evaluate and round:

$$\text{SD} = 3.238\ldots \approx 3.24 \quad (3 \text{ s.f.})$$

Check. The marks span $17$ to $27$ — a spread of about $\pm 5$ around the mean — so an SD of about $3$ is the right size. An SD of $30$ or $0.3$ would signal a slip.

⚠ Watch out: $\sum x^2$ means "square each value, then add" — it is NOT $(\sum x)^2 = 219^2$. Subtract $\bar{x}^2$ inside the square root, and keep $\bar{x}$ exact ($21.9$) rather than a rounded copy.

Question: A dot diagram shows the number of story books read by 8 students in a month: one dot each at $2, 5, 6$; three dots at $3$; two dots at $4$. Find the median and the mode, and describe the distribution.

[diagram in original sheet: number line 2–6 with stacked dots]

Step 1. Read the data off the diagram, in order — each dot is one student:

$$2,\ 3,\ 3,\ 3,\ 4,\ 4,\ 5,\ 6$$

Step 2. Median — with $n = 8$, position $\dfrac{8+1}{2} = 4.5$, between the 4th and 5th values:

$$\text{Median} = \dfrac{3+4}{2} = 3.5 \text{ books}$$

Step 3. Mode — the tallest stack of dots:

$$\text{Mode} = 3 \text{ books}$$

Step 4. Describe the shape in words: the data clusters around $3$–$4$ books, with a slight tail towards the higher values ($5$ and $6$) and no extreme outliers.

A dot diagram keeps every individual value visible, so clusters, gaps and outliers can be seen at a glance.

⚠ Watch out: "Describe the distribution" wants features — cluster, spread, tail/skew, outliers — with numbers attached, not a recalculated mean. And when the median position lands between two dots, average them; don't round the position to a whole number.