Chapter 4: Apply statistics and probability

Applied statistics and probability are essential tools for agricultural engineers in Kenya, enabling them to analyze data related to crop yields, soil properties, machinery performance, and environmental factors. Accurate data presentation helps in making informed decisions, optimizing resource use, and improving agricultural productivity. This chapter focuses on the mathematical treatment of different types of data encountered in agricultural engineering, emphasizing clear and effective presentation methods.

4.1 Data presentation

Data presentation turns a list of recorded values into a table or a graph that can be read at a glance. The graph chosen depends on the type of variable and on whether the values are listed one by one or sorted into classes. A continuous variable is measured and can take any value in a range, such as soil temperature or the height of a maize plant. A discrete variable is counted and takes separate values, such as the number of breakdowns of a tractor in a year. Data listed value by value are ungrouped, and data sorted into class intervals are grouped. Example: moisture readings from a field are continuous and are usually grouped, while the number of trips made by a farm trailer each day is discrete and can be left ungrouped.

Every graph needs a title, labelled axes with units, an even scale and a clear key where more than one series is shown.

4.1.1 Continuous variables

Continuous variables are measured, not counted, so any value in a range is possible: soil moisture content, fuel consumption in litres per hour, temperature and plant height. Because exact repeats are rare, the measurements are sorted into classes with a lower boundary that belongs to the class and an upper boundary that belongs to the next class. A value of exactly 110 therefore belongs to the class 110 to under 120. Continuous data are shown with a histogram, which shows how the values are spread, and with a line graph, which shows how a measurement changes with time.

4.1.1.1 Histogram

A histogram shows the frequency of each class of a continuous variable by a bar. The bars touch because one class begins where the previous class ends, and the horizontal axis is a number line with the class boundaries marked on it. When all classes have the same width, the height of each bar is the frequency. When the widths differ, the height must be the frequency density, which is shown in the Histogram under Grouped data. Example: heights of maize plants in classes of 10 cm give bars of equal width, and the tallest bar marks the most common height range.

To draw a histogram from a table with equal class widths:

1. Write the title and decide the unit of the variable.
2. Draw the horizontal axis as a number line and mark the class boundaries at equal spacing.
3. Draw the vertical axis from 0 and choose a scale that fits the largest frequency.
4. Label both axes with their names and units.
5. Draw a bar above each class, from its lower boundary to its upper boundary, with height equal to the frequency.
6. Leave no gaps between neighbouring bars.
7. Read the graph: the tallest bar is the modal class, and adding bar heights gives the number of values above or below a boundary.

Three worked examples follow.

Example 1: The heights of 50 maize plants were measured and sorted into classes of 10 cm: 100 to under 110 (7 plants), 110 to under 120 (15), 120 to under 130 (18) and 130 to under 140 (10). Describe the histogram and find the percentage of plants below 120 cm.

Given: classes of equal width 10 cm, total 7 + 15 + 18 + 10 = 50 plants

Bars: boundaries 100, 110, 120, 130, 140 with heights 7, 15, 18, 10

Tallest bar: 120 to under 130 cm (18 plants)

Working: plants below 120 cm = 7 + 15 = 22

Result: 22 ÷ 50 × 100 = 44%

Answer: the modal class is 120 to under 130 cm, and 44% of the plants are below 120 cm

Example 2: The moisture content of 40 bags of maize grain was recorded in classes of 1 percentage point: 12 to under 13 (4 bags), 13 to under 14 (9), 14 to under 15 (15), 15 to under 16 (8) and 16 to under 17 (4). Find the percentage of bags at 15% or above.

Given: total 4 + 9 + 15 + 8 + 4 = 40 bags

Working: bags at 15% or above = 8 + 4 = 12

Result: 12 ÷ 40 × 100 = 30%

Answer: the modal class is 14 to under 15%, and 30% of the bags are at 15% or above

Example 3: The electricity use of 30 grain dryer batches was recorded in classes of 2 kWh: 8 to under 10 (3), 10 to under 12 (11), 12 to under 14 (12) and 14 to under 16 (4). Find the number and percentage of batches using 12 kWh or more.

Given: total 3 + 11 + 12 + 4 = 30 batches

Working: batches using 12 kWh or more = 12 + 4 = 16

Result: 16 ÷ 30 × 100 = 53.3% (to 1 decimal place)

Answer: 16 batches, which is 53.3% of the batches

4.1.1.2 Line

A line graph of a continuous variable plots each measurement against the time at which it was taken and joins the points with straight lines. It shows trends, highs and lows, and the rate of change. Because the variable changes smoothly with time, a value between two readings can be estimated from the joining line. Example: soil temperature measured every three hours shows the daily rise and fall, and a point on the line between two readings estimates the temperature at a time that was not measured.

To draw a line graph:

1. Write the title and choose a scale for time on the horizontal axis and for the measurement on the vertical axis.
2. Label both axes with names and units.
3. Plot each pair of readings as a point.
4. Join the points in order of time with straight lines.
5. Read values between points from the line, treating them as estimates.
6. Do not extend the line far beyond the last reading, because the trend may not continue.

Three worked examples follow.

Example 1: Soil temperature at 10 cm depth was recorded every three hours: 06:00 17.5 °C, 09:00 20.0 °C, 12:00 25.5 °C, 15:00 27.0 °C, 18:00 22.5 °C and 21:00 19.0 °C. Find the highest reading, the average rate of rise from 06:00 to 12:00 and the estimated temperature at 10:30.

Given: six readings plotted against time

Highest reading: 27.0 °C at 15:00

Working: rise from 06:00 to 12:00 = 25.5 − 17.5 = 8.0 °C over 6 hours

Rate: 8.0 ÷ 6 = 1.33 °C per hour

Working: between 09:00 and 12:00 the rate is (25.5 − 20.0) ÷ 3 = 1.8333 °C per hour, and 10:30 is 1.5 hours after 09:00

Result: 20.0 + 1.5 × 1.8333 = 22.75 °C

Answer: the highest reading is 27.0 °C, the average rate is 1.33 °C per hour and the estimate at 10:30 is 22.75 °C

Example 2: The water level in a farm tank was read each hour: 0 h 120 cm, 1 h 108 cm, 2 h 97 cm, 3 h 85 cm and 4 h 73 cm. Find the fall each hour and the average rate of fall.

Given: five readings at hourly intervals

Working: falls are 120 − 108 = 12, 108 − 97 = 11, 97 − 85 = 12 and 85 − 73 = 12 cm

Result: total fall = 120 − 73 = 47 cm over 4 hours, so 47 ÷ 4 = 11.75 cm per hour

Answer: the level falls by about 11.75 cm per hour, so a straight line fits well; continuing the line to 5 hours gives about 61.25 cm, which is an estimate only

Example 3: The height of a maize crop was measured weekly: week 1 12 cm, week 2 25 cm, week 3 41 cm, week 4 60 cm and week 5 82 cm. Describe the growth from the graph.

Given: five weekly readings

Working: weekly growth is 25 − 12 = 13, 41 − 25 = 16, 60 − 41 = 19 and 82 − 60 = 22 cm

Answer: the weekly growth increases each week (13, 16, 19, 22 cm), so the line becomes steeper and the crop is growing faster as the weeks pass

4.1.2 Discrete variable

A discrete variable is counted, so it takes separate whole-number values: the number of breakdowns or the number of sprinkler heads on a field. Counts of named categories, such as the number of ploughs, harrows and planters in a workshop, are shown in the same way. There is nothing between one value and the next, so each value is shown by its own bar or its own sector. Discrete data are shown with a bar graph or a pie graph. Example: the number of delivery trips made by a farm trailer each day is 2, 3, 4, 5 or 6, and a value of 3.5 trips does not exist.

4.1.2.1 Bar graph

A bar graph shows each value or category of a discrete variable by a bar whose height is its frequency. The bars have equal widths and equal gaps between them, and the gaps show that the values are separate. Example: counting how many of 20 tractors had 0, 1, 2, 3 or 4 breakdowns in a year gives five bars, one for each count.

To draw a bar graph:

1. Write the title and list the values or categories along the horizontal axis.
2. Draw the vertical axis from 0 with an even scale up to the largest frequency.
3. Label both axes with names and units.
4. Draw one bar of equal width for each value or category.
5. Leave an equal gap between neighbouring bars.
6. Make the height of each bar equal to its frequency.

Three worked examples follow.

Example 1: The number of breakdowns in one year was counted for 20 tractors: 0 breakdowns for 4 tractors, 1 for 7, 2 for 5, 3 for 3 and 4 for 1. Describe the bar graph and find the percentage of tractors with 2 or more breakdowns.

Given: bars of heights 4, 7, 5, 3 and 1 over the values 0, 1, 2, 3 and 4, with total 20 tractors

Tallest bar: 1 breakdown (7 tractors)

Working: tractors with 2 or more breakdowns = 5 + 3 + 1 = 9

Result: 9 ÷ 20 × 100 = 45%

Answer: 45% of the tractors had two or more breakdowns

Example 2: A farm trailer made 2, 3, 4, 5 or 6 delivery trips a day over 25 days: 2 trips on 3 days, 3 trips on 6 days, 4 trips on 9 days, 5 trips on 5 days and 6 trips on 2 days. Find the number of days with at least 5 trips.

Given: bar heights 3, 6, 9, 5, 2 with total 3 + 6 + 9 + 5 + 2 = 25 days

Working: days with at least 5 trips = 5 + 2 = 7

Result: 7 ÷ 25 × 100 = 28%

Answer: 7 days, which is 28% of the days

Example 3: A workshop holds 12 ploughs, 8 harrows, 5 planters, 10 sprayers and 5 trailers. Draw the bar graph and find the share of sprayers.

Given: categories along the horizontal axis with bar heights 12, 8, 5, 10 and 5

Working: total = 12 + 8 + 5 + 10 + 5 = 40 machines

Result: 10 ÷ 40 × 100 = 25%

Answer: the tallest bar is the ploughs, and sprayers are 25% of the machines

4.1.2.2 Pie graph

A pie graph is a circle divided into sectors, where the size of each sector shows the share of one value or category in the whole. It suits a few categories that add up to a total, such as the share of costs. The angle of a sector is its frequency divided by the total, multiplied by 360°. Example: a category that makes up a quarter of the total takes a sector of 90°.

angle of sector = (frequency ÷ total frequency) × 360°

To draw a pie graph:

1. Add the frequencies to find the total.
2. Work out the angle of each sector from the formula.
3. Check that the angles add up to 360°.
4. Draw a circle and a radius, and measure each angle in turn with a protractor.
5. Label each sector with its category and its frequency or percentage.
6. Give the graph a title.

Three worked examples follow.

Example 1: A farm machinery workshop records 300 hours of work in a month: welding 120 hours, machining 60, painting 40, assembly 30 and testing 50. Find the sector angles.

Given: total = 120 + 60 + 40 + 30 + 50 = 300 hours

Working: welding 120 ÷ 300 × 360 = 144°

Working: machining 60 ÷ 300 × 360 = 72°, painting 40 ÷ 300 × 360 = 48°, assembly 30 ÷ 300 × 360 = 36° and testing 50 ÷ 300 × 360 = 60°

Check: 144 + 72 + 48 + 36 + 60 = 360

Answer: the angles are 144°, 72°, 48°, 36° and 60°

Example 2: A pie graph shows 80 bags of seed. The sector for beans is 90°. Find the number of bags of beans.

Given: total = 80 bags, angle = 90°

Formula: frequency = (angle ÷ 360) × total

Substitution: 90 ÷ 360 × 80 = 0.25 × 80

Answer: 20 bags of beans

Example 3: The running costs of a 2 ha maize plot are fertiliser Ksh 18 000, seed Ksh 9 000, labour Ksh 12 000, fuel Ksh 6 000 and repairs Ksh 3 000. Find the percentage and angle of each sector.

Given: total = 18 000 + 9 000 + 12 000 + 6 000 + 3 000 = Ksh 48 000

Working: fertiliser 37.5% and 135°, seed 18.75% and 67.5°, labour 25% and 90°, fuel 12.5% and 45°, repairs 6.25% and 22.5°

Check: 135 + 67.5 + 90 + 45 + 22.5 = 360

Answer: fertiliser takes the largest sector at 135°, and labour takes a quarter of the circle

4.1.3 Grouped data

Grouped data are values sorted into class intervals, which is the usual way to handle many measurements. The terms used are the class limits (the smallest and largest values the class can hold), the class boundaries (the points halfway between the upper limit of one class and the lower limit of the next), the class width (upper boundary minus lower boundary) and the class midpoint (the average of the two boundaries). Example: the daily water use in cubic metres of 30 irrigated plots was 12, 17, 23, 14, 28, 19, 21, 25, 16, 31, 18, 22, 27, 13, 20, 24, 29, 15, 26, 33, 19, 22, 17, 25, 30, 21, 18, 23, 28 and 20. Sorting them into classes of width 5, with the lower boundary included in the class, gives:

The frequencies add up to 30, the number of values. For a discrete count, classes are written with limits that leave no gaps between whole numbers, such as 0 to 9 and 10 to 19.

4.1.3.1 Histogram

A histogram of grouped data is drawn from the grouped frequency table. When the classes have equal width, the bar heights are the frequencies. Grouped tables often have classes of unequal width, and then the bar heights must be the frequency density, so that the area of each bar, and not its height, shows the frequency. Example: a class twice as wide as its neighbour holds twice as many values for the same density, so plotting frequency would make it look too tall.

frequency density = frequency ÷ class width

The area of a bar = class width × frequency density = frequency.

Three worked examples follow.

Example 1: Draw the histogram for the water-use table of 30 plots given above and name the modal class.

Given: equal class widths of 5 m³ and frequencies 3, 8, 9, 7 and 3

Bars: boundaries 10, 15, 20, 25, 30, 35 with heights 3, 8, 9, 7 and 3

Answer: the modal class is 20 to under 25 m³ with 9 plots

Example 2: The lengths of 50 offcuts of irrigation pipe were grouped as 0 to under 1 m (6), 1 to under 2 m (14), 2 to under 4 m (18) and 4 to under 8 m (12). Find the bar heights.

Given: class widths 1, 1, 2 and 4 m

Formula: frequency density = frequency ÷ class width

Working: 6 ÷ 1 = 6, 14 ÷ 1 = 14, 18 ÷ 2 = 9 and 12 ÷ 4 = 3

Check: areas are 1 × 6 = 6, 1 × 14 = 14, 2 × 9 = 18 and 4 × 3 = 12, which equal the frequencies

Answer: the bar heights are 6, 14, 9 and 3 per metre, so the class 1 to under 2 m has the tallest bar although the class 2 to under 4 m holds more offcuts

Example 3: A greenhouse sensor recorded 20 temperatures: 22 to under 24 °C (3 readings), 24 to under 27 °C (5), 27 to under 30 °C (7) and 30 to under 33 °C (5). Find the frequency densities.

Given: class widths 2, 3, 3 and 3 °C

Formula: frequency density = frequency ÷ class width

Working: 3 ÷ 2 = 1.5, 5 ÷ 3 = 1.67, 7 ÷ 3 = 2.33 and 5 ÷ 3 = 1.67

Answer: the frequency densities are 1.5, 1.67, 2.33 and 1.67 readings per °C

4.1.3.2 Bar

A bar graph of grouped data is used when the variable is a count that has been sorted into groups of whole numbers, such as 0 to 9 and 10 to 19. Each group is a separate bar with gaps between the bars, because the counts do not run continuously from one group to the next. The bar height is the frequency of the group. Example: the number of bags harvested from each plot, grouped in tens, is shown as separate bars, which differs from the touching bars of a histogram used for measured values.

Three worked examples follow.

Example 1: The bags harvested from 40 plots were grouped as 0 to 9 (6 plots), 10 to 19 (15), 20 to 29 (12) and 30 to 39 (7). Describe the bar graph and find the percentage of plots with 20 or more bags.

Given: four separate bars with heights 6, 15, 12 and 7, total 40 plots

Working: plots with 20 or more bags = 12 + 7 = 19

Result: 19 ÷ 40 × 100 = 47.5%

Answer: the tallest bar is the group 10 to 19, and 47.5% of the plots gave 20 or more bags

Example 2: The number of sprinkler heads on each of 24 fields was grouped as 1 to 5 (4 fields), 6 to 10 (9), 11 to 15 (7) and 16 to 20 (4). Find the percentage of fields with more than 10 sprinkler heads.

Given: bar heights 4, 9, 7 and 4, total 24 fields

Working: fields with more than 10 heads = 7 + 4 = 11

Result: 11 ÷ 24 × 100 = 45.8% (to 1 decimal place)

Answer: 45.8% of the fields

Example 3: Express the plot data of Example 1 as percentages so that a bar graph can use a percentage scale.

Given: frequencies 6, 15, 12 and 7, total 40

Working: 6 ÷ 40 = 15%, 15 ÷ 40 = 37.5%, 12 ÷ 40 = 30% and 7 ÷ 40 = 17.5%

Check: 15 + 37.5 + 30 + 17.5 = 100

Answer: the bar heights on the percentage scale are 15%, 37.5%, 30% and 17.5%

4.1.3.3 Cumulative frequency

The cumulative frequency of a class is the total of its frequency and the frequencies of all earlier classes. It tells how many values lie below the upper boundary of that class. The table is built by adding down the frequency column, and the last cumulative frequency equals the total number of values. Example: if 4 tractors fall in the first class and 9 in the second, the cumulative frequency at the end of the second class is 13.

Three worked examples follow.

Example 1: The seasonal hours of use of 50 tractors were grouped as 0 to under 200 (4), 200 to under 400 (9), 400 to under 600 (17), 600 to under 800 (14) and 800 to under 1 000 (6). Build the cumulative frequency table and find the number of tractors used for fewer than 600 hours.

Given: five classes of width 200 hours, total 50

Working: cumulative frequencies at the upper boundaries are 4, 4 + 9 = 13, 13 + 17 = 30, 30 + 14 = 44 and 44 + 6 = 50

Answer: 30 tractors were used for fewer than 600 hours, which is 60%

Example 2: A cumulative frequency table for the diameters of 60 fence posts reads 5, 17, 38, 52 and 60 at the upper boundaries of five classes. Find the class frequencies.

Given: cumulative frequencies 5, 17, 38, 52, 60

Working: 5, 17 − 5 = 12, 38 − 17 = 21, 52 − 38 = 14 and 60 − 52 = 8

Check: 5 + 12 + 21 + 14 + 8 = 60

Answer: the class frequencies are 5, 12, 21, 14 and 8

Example 3: For the tractor data of Example 1, find the number of tractors used between 400 and 800 hours and express the cumulative frequencies as percentages.

Working: cumulative frequency at 800 hours is 44 and at 400 hours is 13, so 44 − 13 = 31 tractors

Result: 31 ÷ 50 × 100 = 62%

Working: cumulative percentages are 4 ÷ 50 = 8%, 13 ÷ 50 = 26%, 30 ÷ 50 = 60%, 44 ÷ 50 = 88% and 50 ÷ 50 = 100%

Answer: 31 tractors, which is 62%

4.1.3.4 Ogive

An ogive is the graph of cumulative frequency. Each cumulative frequency is plotted against the upper class boundary of its class, a starting point of zero frequency is plotted at the lower boundary of the first class, and the points are joined by a smooth curve (or by straight lines). The curve rises from left to right and levels out at the total number of values. Medians, quartiles and the number of values below or above any chosen value can be read from it. Example: with 40 values, the value on the horizontal axis at a height of 20 on the curve is the median.

To draw and read an ogive:

1. Build the cumulative frequency table.
2. Mark the upper class boundaries on the horizontal axis and the cumulative frequency on the vertical axis.
3. Plot (lower boundary of the first class, 0) as the starting point.
4. Plot each cumulative frequency above the upper boundary of its class.
5. Join the points with a smooth curve.
6. For the median, go across from n ÷ 2 on the vertical axis to the curve and then down to the horizontal axis.
7. For the lower quartile use n ÷ 4 and for the upper quartile use 3n ÷ 4 in the same way.

Three worked examples follow.

Example 1: The flow rate of 40 spray nozzles was tested and grouped as 4 to under 6 litres per minute (3 nozzles), 6 to under 8 (8), 8 to under 10 (15), 10 to under 12 (10) and 12 to under 14 (4). List the points to plot.

Given: total 40 nozzles

Working: cumulative frequencies are 3, 3 + 8 = 11, 11 + 15 = 26, 26 + 10 = 36 and 36 + 4 = 40

Points: (4, 0), (6, 3), (8, 11), (10, 26), (12, 36) and (14, 40)

Answer: the ogive rises from (4, 0) to (14, 40)

Example 2: Read the median and the quartiles from the ogive of Example 1.

Given: n = 40, so the median is at 20, the lower quartile at 10 and the upper quartile at 30

Method: the plotted points are joined by straight lines, so a reading between two points is found by proportion along the line. A smooth curve drawn by eye gives a slightly different estimate.

Working: the median lies between (8, 11) and (10, 26), so 8 + (20 − 11) ÷ (26 − 11) × 2 = 8 + 1.2 = 9.2

Working: the lower quartile lies between (6, 3) and (8, 11), so 6 + (10 − 3) ÷ 8 × 2 = 7.75

Working: the upper quartile lies between (10, 26) and (12, 36), so 10 + (30 − 26) ÷ 10 × 2 = 10.8

Answer: median 9.2, lower quartile 7.75 and upper quartile 10.8 litres per minute, so the middle half of the nozzles lies in a range of 10.8 − 7.75 = 3.05 litres per minute

Example 3: Estimate from the ogive of Example 1 the number of nozzles delivering less than 9 litres per minute and more than 11 litres per minute.

Working: at 9, the curve lies halfway between (8, 11) and (10, 26), so 11 + 0.5 × 15 = 18.5 nozzles

Result: 18.5 ÷ 40 × 100 = 46.25%

Working: at 11, the curve lies halfway between (10, 26) and (12, 36), so 26 + 0.5 × 10 = 31, and 40 − 31 = 9 nozzles are above 11

Answer: about 18.5 nozzles (46%) deliver less than 9 litres per minute, and about 9 nozzles (22.5%) deliver more than 11

4.1.4 Ungrouped data

Ungrouped data are the individual observations, listed one by one or as each value with its frequency, without class intervals. This suits small sets of values, such as monthly totals or counts with only a few different values. Example: the number of fertiliser bags used on each of 20 fields takes only the values 1 to 5, so a table of each value with its frequency is enough and no classes are needed.

4.1.4.1 Line

A line graph of ungrouped data plots each separate observation, such as a monthly total, against its position in the sequence and joins the points. It shows the trend over the sequence. The points are joined only to show the pattern, because there are no measurements between two months, so the line is not used to read in-between values. The same graph plotted with each value on the horizontal axis and its frequency on the vertical axis shows how often the values occur. Example: rainfall totals for six months give six points joined in order of the months.

Three worked examples follow.

Example 1: Monthly rainfall from January to June was 45, 38, 92, 140, 118 and 60 mm. Find the wettest month, the months above 100 mm and the largest rise between two months.

Given: six plotted points

Wettest month: April, with 140 mm

Months above 100 mm: April and May

Working: changes from month to month are February 38 − 45 = −7, March 92 − 38 = 54, April 140 − 92 = 48, May 118 − 140 = −22 and June 60 − 118 = −58 mm

Answer: the largest rise is 54 mm from February to March, and the largest fall is 58 mm from May to June

Example 2: The number of faulty bolts in each of 25 boxes was 0 in 4 boxes, 1 in 7, 2 in 8, 3 in 4 and 4 in 2. Describe the line graph of frequency against the number of faulty bolts.

Given: points (0, 4), (1, 7), (2, 8), (3, 4) and (4, 2), with total 4 + 7 + 8 + 4 + 2 = 25 boxes

Answer: the line rises to a peak at 2 faulty bolts (8 boxes) and then falls, so 2 is the most frequent value

Example 3: Two milling machines in a maize mill used electricity in kWh from January to April: machine A 410, 395, 430 and 450, and machine B 380, 400, 390 and 420. Compare them on one graph.

Given: two lines drawn on the same axes, with a key

Working: the monthly differences A − B are 30, −5, 40 and 30 kWh

Answer: machine A used more electricity in three of the four months, the gap is largest in March at 40 kWh, and machine B used 5 kWh more in February

4.1.4.2 Cumulative frequency

The cumulative frequency of ungrouped data is the running total of the frequencies of the values, taken in order from the smallest value. It shows how many observations are at or below each value. Because the values are separate, the graph is a step graph, with a horizontal line at each cumulative frequency and a jump at each new value. Example: if 2 fields used 1 bag and 5 fields used 2 bags, then 7 fields used at most 2 bags.

Three worked examples follow.

Example 1: The number of fertiliser bags used on each of 20 fields was 1 bag on 2 fields, 2 bags on 5, 3 bags on 6, 4 bags on 4 and 5 bags on 3. Find the cumulative frequencies and the number of fields using at most 3 bags.

Given: total 2 + 5 + 6 + 4 + 3 = 20 fields

Working: cumulative frequencies are 2, 2 + 5 = 7, 7 + 6 = 13, 13 + 4 = 17 and 17 + 3 = 20

Result: 13 ÷ 20 × 100 = 65%

Answer: 13 fields used at most 3 bags, which is 65%, and 20 − 13 = 7 fields used more than 3 bags

Example 2: The number of maize shellers in use on each of 30 days was 3 on 2 days, 4 on 6 days, 5 on 10 days, 6 on 8 days and 7 on 4 days. Find the number of days with fewer than 6 shellers in use.

Given: total 2 + 6 + 10 + 8 + 4 = 30 days

Working: cumulative frequencies are 2, 8, 18, 26 and 30

Answer: 18 days had fewer than 6 shellers in use, which is 60% of the days

Example 3: Express the cumulative frequencies of Example 2 as percentages.

Working: 2 ÷ 30 = 6.7%, 8 ÷ 30 = 26.7%, 18 ÷ 30 = 60%, 26 ÷ 30 = 86.7% and 30 ÷ 30 = 100%

Answer: the cumulative percentages are 6.7%, 26.7%, 60%, 86.7% and 100%

Chapter Summary

This chapter explored various methods for presenting data, including the use of histograms and line graphs for continuous variables, bar and pie charts for discrete variables, and specialized techniques such as cumulative frequency and ogive graphs for grouped data. It then examined measures of central tendency, detailing how to calculate the mean, mode, and median for both grouped and ungrouped data sets. Following this, the chapter addressed measures of dispersion, focusing on standard deviation and variance, again differentiating between grouped and ungrouped data calculations. The concept of probability was introduced with examples of scenarios involving replacement and without replacement, highlighting how these affect outcomes. Probability distribution functions were discussed with emphasis on the binomial and Poisson distributions, explaining their applications and characteristics. Finally, the chapter concluded with the normal distribution, describing its properties and significance in statistical analysis. Together, these topics provide a comprehensive foundation for applying statistics and probability in practical contexts.