Cumulative Frequency: past paper questions & worked solutions

The median is the \(60/2=30\)th value. Draw horizontally from cumulative frequency \(30\) to the curve, then vertically to the weight axis: about \(117\text{ kg}\).

The median splits the 60 observations in half.

Read the lower quartile at cumulative frequency \(60/4=15\), giving \(Q_1\approx112\text{ kg}\). Read the upper quartile at cumulative frequency \(3(60)/4=45\), giving \(Q_3\approx128\text{ kg}\).

Quartiles occur at one-quarter and three-quarters of the total frequency.

Interquartile range is upper quartile minus lower quartile.

Use age (years) on the horizontal axis from \(0\) to \(60\), and cumulative frequency on the vertical axis from \(0\) to \(80\). Plot \((0,0),(10,19),(20,37),(30,52),(40,63),(50,73),(60,80)\).

Each cumulative total is plotted at its upper age boundary.

Join the points in order with an increasing curve or straight line segments. Do not draw bars or a staircase.

This represents the accumulated number up to each age boundary.

Read horizontally from cumulative frequency 40 to the graph, then vertically down to the age axis.

Using straight segments, \[20+\frac{40-37}{52-37}\times10=22\text{ years}.\]

The median is estimated between ages 20 and 30; a smooth graph gives a nearby estimate.

Read vertically from age \(16\) to the graph, then horizontally to the cumulative-frequency axis.

The ordinate is the estimated number aged 16 and under.

Using straight segments, \[19+\frac{16-10}{20-10}(37-19)=29.8\approx30.\]

Round the estimated count to a whole number of people.

Read across from \(\frac{80}{2}=40\): about \(25\) seconds.

At \(50\) seconds the curve reads about \(75\), so \(80-75=5\) adults.

There are \(50\) plants, so read across from \(25\): the curve reaches \(25\) at about \(28.5\) cm.

Read the curve at both ends of the class: about \(44\) at \(40\) cm and \(30\) at \(30\) cm.

At \(35\) cm the curve reads about \(37\), so about \(50-37=13\) plants.

**Part (a) Median:** Total plants = 40. The median position is at cumulative frequency = 40/2 = 20. Reading across from CF = 20 to the curve and down to the x-axis gives:

\(a\) 30 cm

Lower Quartile (Q1) is at CF = 10. Reading the graph ⟹ Height = 20 cm.

Upper Quartile (Q3) is at CF = 30. Reading the graph ⟹ Height = 40 cm.

IQR = *Q*3 − *Q*1 = 40 − 20 = 20 cm

\(b\) 20 cm

**Part (c) Premium plants revenue:** Premium plants have height  \> 40 cm. From the graph, cumulative frequency at 40 cm is 30. This means 30 plants are ≤ 40 cm. Number of premium plants = 40 − 30 = 10 plants Total Money = 10 plants × 30 euros = 300 euros

Read across from \(\frac{40}{2}=20\): about \(27\) cm.

Lower quartile at \(10\): about \(19\) cm. Upper quartile at \(30\): about \(35\) cm.

At \(40\) cm the curve reads about \(34\), so \(40-34=6\) plants are taller than \(40\) cm.

Plot \((35,12),(45,36),(55,64),(65,76),(75,86),(85,90)\), start at \((25,0)\), and join with a smooth curve.

At \(52\) minutes: about \(56\). At \(42\) minutes: about \(29\).

There are \(90\) bags, so the median is at cumulative frequency \(45\).

From the graph,

Large bags sold:

So \(25\) bags have weight greater than \(d\).

bags have weight \(d\) grams or less.

From the graph, cumulative frequency \(65\) gives

There are \(80\) people.

For the interquartile range:

From the graph,

At \(50\) minutes, the cumulative frequency is about \(42\).

No women took \(50\) minutes or less, so these \(42\) people are men.

There are \(60\) people.

The median is at cumulative frequency \(30\).

From the graph,

For the interquartile range:

From the graph,

Hospital B has median \(28\) minutes and IQR \(19\) minutes.

So Hospital A has a shorter median waiting time and a smaller spread.

Plot \((20,9),(30,28),(40,45),(50,63),(60,76),(70,80)\), start at \((0,0)\), and join with a smooth curve.

Read across from \(\frac{80}{2}=40\): about \(37\) years.

Plot \((130,8),(140,35),(150,52),(160,58),(170,60)\), start at \((120,0)\) and join with a smooth curve.

\(\tfrac12\times60=30\): read across from 30 → about \(138\) cm.

\(\tfrac14\times60=15\): read across from 15 → about \(133\) cm.

Up from 155 cm the curve reads about \(55\), so \(60-55=5\) plants.

Plot \((20,12),(30,27),(40,47),(50,65),(60,74),(70,80)\), start at \((10,0)\), and join with a smooth curve.

Read up from \(46\): about \(58\) people.

\(80-58=22\) people.

(About \(26\%\) to \(30\%\) is accepted, depending on the curve.)

Plot \((10,8),(20,21),(30,33),(40,50),(50,57),(60,60)\), start at \((0,0)\), and join with a smooth curve.

Read across from \(\frac{60}{2}=30\): about \(27.5\) minutes.

At \(45\) minutes the curve reads about \(54\), so about \(60-54=6\) people. (Answers from about \(5\) to \(8\) are accepted, depending on the curve.)

Cumulative frequencies:

The median is the \(30\)th value. Interpolating in the \(20\) to \(30\) interval gives about \(27.5\) minutes.

For more than \(45\) minutes, the estimate is about \(7\) people.

Plot \((1,4),(2,19),(3,39),(4,50),(5,56),(6,60)\), start at \((0,0)\), and join with a smooth curve.

Read across from \(\frac{60}{2}=30\): about \(2.55\) kg.

At \(3.7\) kg the curve reads about \(47\), so about \(60-47=13\) parcels.

Plot \((60,15),(70,31),(80,52),(90,66),(100,74),(110,80)\), start at \((50,0)\), and join with a smooth curve.

Read across from \(40\): about \(74\) minutes.

Lower quartile at \(20\): about \(63\) minutes. Upper quartile at \(60\): about \(86\) minutes.

Complete the cumulative frequencies:

The median is at cumulative frequency \(40\).

This is between \((20,33)\) and \((30,57)\):

So the median is about \(23\) minutes.

At \(42\) minutes:

Number taking more than \(42\) minutes:

Complete the cumulative frequencies:

The median is at cumulative frequency \(60\).

This lies between \((15,40)\) and \((20,68)\):

So the median is about \(18.6\) minutes.

At \(23\) minutes:

Number who waited longer than \(23\) minutes:

\(a\) Median(b) IQR(c) % days  \> 75 km(d) More consistent?

Median at the $\\frac{60}{2} = 30$th value — read from graph at cf  = 30:

Median  ≈ 62 km

*Q*₁ at cf  = 15, *Q*₃ at cf  = 45. Read from graph: *Q*₁ ≈ 56 km,   *Q*₃ ≈ 74 km IQR = *Q*₃ − *Q*₁ = 69 − 55 = 14 km

IQR  ≈ 14 km

Graph-reading answers are accepted within  ± 2 km of the model answer.

From the graph, read the cumulative frequency at *d* = 75 km: cf at 75 km ≈ 52

Number of days with distance  \> 75 km: 60 − 52 = 8 days

Percentage: $$\\frac{8}{60}\\times100 = 13.3\\%$$

13.3% of days Ehrick drives more than 75 km

Plot \((5,6),(10,17),(15,27),(20,42),(25,53),(30,60)\), start at \((0,0)\), and join with a smooth curve.

Lower quartile at \(\frac{60}{4}=15\): about \(9\) hours. Upper quartile at \(45\): about \(21.4\) hours.

(Answers from about \(11\) to \(14\) are accepted, depending on the curve.)

So \(60-25=35\) workers spent \(W\) hours or less. Reading across from \(35\): \(W\approx17.7\).

\(17-6=11\) workers, so

Plot \((10,7),(20,24),(30,42),(40,56),(50,66),(60,70)\), start at \((0,0)\), and join with a smooth curve.

Lower quartile at \(\frac{70}{4}=17.5\): about \(16\) km. Upper quartile at \(\frac34\times70=52.5\): about \(37.5\) km.

(Answers from about \(19\) to \(23\) km are accepted, depending on the curve.)

At \(46\) km the curve reads about \(62\), so

The median is at cumulative frequency \(30\).

This is between \((20,13)\) and \((30,36)\):

Median age is about \(27.4\) years.

For the interquartile range:

For people older than \(55\):

Complete the cumulative frequencies:

The median is at cumulative frequency \(40\).

This is between \((30,32)\) and \((40,52)\):

So the median age is about \(34\) years.

Estimate the number aged between \(18\) and \(65\):

60% are going to work:

Plot \((5,17),(10,32),(15,57),(20,70),(25,76),(30,80)\), start at \((0,0)\), and join with a smooth curve.

Read across from \(40\): about \(11.6\) km.

Lower quartile at \(20\): about \(6\) km. Upper quartile at \(60\): about \(16.2\) km.

1. Office A has a lower median (about \(11.6\) km against \(15\) km), so on average workers at Office A travel a shorter distance.

2. Office A has a larger IQR (about \(10\) km against \(5\) km), so the distances at Office A are more varied (less consistent).

Plot the cumulative frequency points and join with a smooth increasing curve:

There are \(60\) people, so the median is read at cumulative frequency \(30\).

Using linear interpolation between \((20,13)\) and \((30,36)\):

So the median is about \(27\) years.

For the interquartile range, read \(Q_1\) at cumulative frequency \(15\) and \(Q_3\) at cumulative frequency \(45\):

So the IQR is about \(25\) years.

At age \(55\), the cumulative frequency is about \(49.5\), so the number older than \(55\) is

This gives about \(11\) people.

There are \(60\) students.

The median is at cumulative frequency \(30\). From the graph,

At \(58\) marks, the cumulative frequency is about \(45\), so about \(45\) students gained \(58\) marks or less.

For the interquartile range:

From the graph,

The geometry IQR is \(9\), so Luis is correct because the algebra IQR is larger.

For grade A, use mark \(64\). From the graph, about \(50\) students gained \(64\) or less, so about

students gained more than \(64\).

Probability both selected students were grade A: