The cumulative frequency at \(m \le 15\) is \(25\), so \(25\) parcels do not have a mass of more than 15 kg.

The total number of parcels is the last cumulative frequency, \(54\).

The number with a mass of more than 15 kg is \(54 - 25 = 29\).

Add up the frequencies of every class up to \(20\) minutes:

\(3 + 6 = 9\) students

That running total is what the graph is plotted from — the point above \(20\) is at a height of \(9\).

There are \(30\) students altogether, so the median is read at half of that: \(30 \div 2 = 15\).

Go across from \(15\) on the cumulative frequency axis to the graph, then straight down — the dashed lines on the diagram.

Median \(\approx 25\) minutes, give or take a small square of the grid: anything from \(23\) to \(27\) is a fair reading.

It is an estimate because the graph only records which class each time fell in, not its exact value.

Each cumulative total is the running total of the frequencies so far:

\(6 = 6\)

\(6 + 10 = 16\)

\(16 + 12 = 28\)

\(28 + 11 = 39\)

A cumulative frequency point is always plotted at a class's upper boundary, \(m = 10\), not its midpoint.

The cumulative total of \(6\) is not reached until every one of the dogs in that class has been counted, and that is only true right up to the top of the class.

Each cumulative frequency is plotted at the upper boundary of its class, because it counts everything up to that value.

So the points are \((10, 2)\), \((20, 12)\), \((30, 18)\), \((40, 26)\) and \((50, 28)\).

Nothing is below \(0\) minutes, so the graph can start from \((0, 0)\). Join the points with a smooth curve.

Plotting at the middle of each class instead is the usual slip: it moves every point half a class to the left.

Each cumulative frequency adds the next frequency to the total so far.

The first is the first frequency, \(12\). The second is \(12 + 16 = 28\), the third is \(28 + 16 = 44\), and so on to the last class.

The cumulative frequencies are 12, 28, 44, 56, 64. The last one is the total number of students, \(64\).

Each cumulative frequency is plotted at the upper boundary of its class, because it counts everything up to that value: the heights are 12, 28, 44, 56, 64 at 150, 160, 170, 180, 190.

Nothing is counted below the lower boundary of the first class, so the graph starts from \((140, 0)\). That point is not in the table.

Join the points with a smooth curve or with straight lines.

There are \(64\) students, so the median is read at half of that: \(64 \div 2 = 32\).

The graph reaches \(32\) between the points \((160, 28)\) and \((170, 44)\). Along that stretch it rises \(16\) over a width of \(10\), and it has to rise \(32 - 28 = 4\) to reach \(32\), so the median is \(160 + 10 \times \frac{ 4 }{ 16 } \approx 162.5\).

Go across from \(32\) on the cumulative frequency axis to the graph, then straight down to the height axis.

Median \(\approx \frac{325}{2}\) cm, give or take a small square of the grid: anything from \(161\) to \(164\) is a fair reading.

There are \(80\) plants, so the median is read at \(80 \div 2 = 40\).

Across from \(40\) to the graph and straight down:

Median \(\approx 70\) cm. A reading a small square either side, from \(65\) to \(75\), is just as fair.

The lower quartile is read at a quarter of the total, \(80 \div 4 = 20\), which gives about \(51\) cm.

The upper quartile is read at three quarters of the total, \(3 \times 20 = 60\), which gives about \(89\) cm.

Interquartile range \(\approx 89 - 51 = 38\) cm

Each quartile can fairly be read a small square either way, so any interquartile range from \(28\) to \(48\) cm comes from a fair pair of readings.

Go up from \(22\) on the time axis to the graph, then straight across to the cumulative frequency axis — the dashed lines on the diagram.

About \(19\) of the students had a time of less than \(22\) minutes. The line across does not have to land exactly on a ruled line; anything from \(18\) to \(20\) is a fair reading.

The graph counts upwards, so it only ever answers "how many were less than". Everything else is the rest of the students.

\(40 - 19 = 21\) students, or about that — it is \(40\) take away whatever was read in part (a).

There are \(40\) donations, so the median is the \(20\)th value.

The running totals are \(3\), \(7\), \(33\), \(37\), \(40\). The first one to reach \(20\) is \(33\), at the end of the class \(20 < a \le 40\).

Go across from \(20\) on the cumulative frequency axis to the graph, then straight down to the amount axis.

Median \(\approx 30\) pounds. A reading a small square either side, from \(28\) to \(32\), is just as fair.

The unequal classes change nothing about the reading: the curve is still plotted at the top end of each class, so the wide class simply takes longer to climb.

Each frequency is the difference between consecutive cumulative totals:

\(17 - 8 = 9\)

\(35 - 21 = 14\)

With the missing frequencies filled in, the five frequencies are \(8\), \(9\), \(4\), \(14\) and \(4\).

The largest of these is \(14\), for the class \(40 < m \le 50\) — not the last class in the table.

Read across from \(80\) g on the curve: \(88\) of the onions have a mass of \(80\) g or less, so they cost \(35\)p each.

The rest are heavier than \(80\) g: \(220 - 88 = 132\) onions at \(46\)p each.

\(88 \times 35 = 3080\)p and \(132 \times 46 = 6072\)p

\(3080 + 6072 = 9152\)p, which is \(£91.52\).