Organise, interpret and display data into appropriate tabular and graphical representations including frequency distribution tables, both ungrouped and grouped using class intervals and class centres, and cumulative frequency

Part (a): the frequencies sum to the $25$ households, so $x = 25 - (3 + 8 + 5 + 2) = 25 - 18 = 7$. One mark for $x = 7$.

Part (b): cumulative frequencies are the running totals $3, 10, 18, 23, 25$. One mark, and markers check the final cumulative frequency equals $n = 25$.

Part (c): "fewer than $3$ children" means $0$, $1$ or $2$, the cumulative frequency at the $2$ row, which is $3 + 7 + 8 = 18$ households. One mark for the answer and one for correctly reading the cumulative column (not including the $3$-children row).

Markers award the value of $x$, a correct cumulative column ending at $25$, and the correct cumulative read for "fewer than $3$".

Part (a): the class centre is $\frac{41 + 60}{2} = 50.5$. One mark.

Part (b): the modal class is $41\text{-}60$ because it has the highest frequency, $28$. One mark.

Part (c): the cumulative frequencies are $4, 14, 42, 68, 80$, ending at the total $80$. One mark, with the final value checked against $n = 80$.

Part (d): "more than $60$" means the $61\text{-}80$ and $81\text{-}100$ classes, so $26 + 12 = 38$ students passed. Equivalently, the cumulative frequency up to $41\text{-}60$ is $42$, and $80 - 42 = 38$. One mark for the method, one for $38$.

Markers reward the class centre, the modal class identified by frequency, a cumulative column ending at $80$, and the correct "more than $60$" count.

Part (a): the second class is $5\text{-}9$, so its upper class limit is $9$ and its class centre is $\frac{5 + 9}{2} = 7$. One mark for both.

Part (b): the four frequencies sum to $6 + 11 + 14 + 9 = 40$, the total number of parcels, so the four classes account for every parcel. The cumulative frequency of the fourth class is this running total, $40$. One mark for the sum equalling $40$ and one for stating the cumulative frequency is $40$ (the total).

Markers reward the correct upper limit and class centre, the demonstration that the frequencies total $40$, and the cumulative frequency stated as $40$.

Part (a) - add the frequency column. The frequency column should total the number of students surveyed:

This matches the $30$ students, so the table is complete (every student is counted exactly once).

Part (b) - "at least $2$" means $2$ or more. Add the frequencies for $2$, $3$ and $4$ pets:

so $13$ students own at least $2$ pets. (Sanity check: $13$ is less than the total $30$, as it must be, and the $17$ students with $0$ or $1$ pet make up the rest, since $6 + 11 = 17$ and $17 + 13 = 30$.)

Set up the rows. List each distinct shoe size from smallest to largest in its own row: $7, 8, 9, 10$.

Tally each value, then count. Work through the list once, adding one stroke per value (grouping in fives with a gate stroke), then write the frequency as the number of strokes:

Check the total. The frequency column must add to the $20$ players:

which matches, so no value has been missed or double-counted.

Part (a) - the class centre is the average of the two endpoints. Add the lower and upper limit of each interval and halve:

so the class centres are $5.5, 15.5, 25.5$ and $35.5$.

Part (b) - locate the value. Since $21 \le 24 \le 30$, the value $24$ falls in the interval $21\text{-}30$. (Note the class centres are evenly spaced $10$ apart, which is the class width - a quick check that they are right.)

Decide the rows from the given intervals. The four classes are $0\text{-}9$, $10\text{-}19$, $20\text{-}29$ and $30\text{-}39$, each $10$ wide.

Tally each value into its class, then count the frequency. Going through the list once and dropping each height into the interval that contains it gives the frequencies below. The class centre of each interval is the average of its endpoints, for example $\dfrac{0 + 9}{2} = 4.5$.

Check the total. The frequencies must add to the $25$ seedlings:

which matches, so every seedling has been placed in exactly one class. (The class centres $4.5, 14.5, 24.5, 34.5$ rise by $10$, the class width, confirming they are evenly spaced.)

Build the cumulative frequency by running totals. Each cumulative frequency is the sum of all frequencies up to and including that row, so add each new frequency to the running total:

Check the final cumulative total. The last cumulative frequency must equal the number of matches, $n = 30$, and it does ($28 + 2 = 30$), so the column is correct.

Read off "at most $2$ goals". "At most $2$" means $0$, $1$ or $2$ goals, which is exactly the cumulative frequency in the $2$ row:

so the team scored at most $2$ goals in $16$ matches.

Part (a) - tally into classes, then add the class-centre and cumulative columns. Each value is placed in the one class that contains it; the class centre is the average of the endpoints (for example $\dfrac{0 + 9}{2} = 4.5$); and each cumulative frequency is the running total of the frequencies.

The frequencies add to $7 + 14 + 15 + 9 + 5 = 50$, and the final cumulative frequency is $50 = n$, so the table is consistent.

Part (b) - modal class. The highest frequency is $15$, so the modal class is $20\text{-}29$ minutes.

Part (c) - "less than $30$ minutes". Times under $30$ fall in the classes $0\text{-}9$, $10\text{-}19$ and $20\text{-}29$, which is the cumulative frequency at the $20\text{-}29$ row:

so $36$ workers commute less than $30$ minutes.

Part (d) - estimate the mean from class centres. With grouped data the mean is estimated by treating every value in a class as its class centre, so multiply each class centre by its frequency, add, and divide by $n$:

so the estimated mean commuting time is about $22.7$ minutes. (It is an estimate, not the exact mean, because the class centres stand in for the real values once the data is grouped.)

Part (a) - class centres. The class centre is the average of the lower and upper limit of each interval:

so the class centres are $104.5, 114.5, 124.5, 134.5$ and $144.5$ grams.

Part (b) - cumulative frequency column. Add the frequencies as a running total:

The final cumulative frequency is $60 = n$, confirming the column.

Part (c) - "large" apples are $130$ g or more. These are the two top classes, $130\text{-}139$ and $140\text{-}149$:

As a percentage of the $60$ apples:

so about $19$ apples (roughly $31.7\%$) are large. (Cross-check with the cumulative column: $41$ apples are under $130$ g, and $60 - 41 = 19$ are $130$ g or more, which agrees.)