Cumulative frequency

The cumulative frequency is obtained by adding up the frequencies as you go along, to give a 'running total'.

Drawing a cumulative frequency diagram

The table shows the lengths (in cm) of 32 cucumbers.

Before drawing the cumulative frequency diagram, we need to work out the cumulative frequencies. This is done by adding the frequencies in turn.

 

Length	Frequency	Cumulative Frequency
21-24	3	3
25-28	7	10 (= 3 + 7)
29-32	12	22 (= 3 + 7 + 12)
33-36	6	28 (= 3 + 7 + 12 + 6)
37-40	4	32 (= 3 + 7 + 12 + 6 + 4)

The points are plotted at the upper class boundary. In this example, the upper class boundaries are 24.5, 28.5, 32.5, 36.5 and 40.5. Cumulative frequency is plotted on the vertical axis.

There are no values below 20.5cm.

Cumulative frequency graphs are always plotted using the highest value in each group of data, (because the table gives you the total that are less than the upper boundary) and the cumulative frequency is always plotted up a graph, as frequency is plotted upwards.

Cumulative frequency diagrams usually have this characteristic S-shape, called an ogive.

Finding the median and quartiles

When looking at a cumulative frequency curve, you will need to know how to find its median, lower and upper quartiles, and the interquartile range.

By drawing horizontal lines to represent ¹/₄ of the total frequency,¹/₂ of the total frequency and ³/₄ of the total frequency, we can read estimates of the lower quartile, median and upper quartile from the horizontal axis.

Quartiles are associated with quarters. The interquartile range is the difference between the lower and upper quartile.

From these values, we can also estimate the interquartile range: 33 - 28 = 5.

Drawing a pie chart

Look at this record of traffic travelling down a particular road.

Traffic Survey 31 January 2008

Type of vehicle	Number of vehicles
Cars	140
Motorbikes	70
Vans	55
Buses	5
Total vehicles	270

To draw a pie chart, we need to represent each part of the data as a proportion of 360, because there are 360 degrees in a circle.

For example, if 55 out of 270 vehicles are vans, we will represent this on the circle as a segment with an angle of: (⁵⁵/₂₇₀) x 360 = 73 degrees.

This will give the following results:

Traffic Survey 31 January 2008

Type of vehicle	Number of vehicles	Calculation	Degrees of a circle
Cars	140	(140/270) x 360	= 187
Motorbikes	70	(70/270) x 360	= 93
Vans	55	(55/270) x 360	= 73
Buses	5	(5/270) x 360	= 7

This data is represented on the pie chart below.

Cumulative increase and decrease

Compound Interest

Here the interest is added to the principal at the end of each year. So the next year the interest is worked out on a larger amount of money than what was originally borrowed.

This means paying interest on the interest of previous years (unlike simple interest, where you only pay interest on the original amount).

This is how it is calculated:

£400 is borrowed for 3 years at 5% compound interest.

Principal at the start = £400

Interest in the 1st year = ⁵/₁₀₀ × 400 = £20

Principal after 1 year = £420

Interest in the 2nd year = ⁵/₁₀₀ × 420 = £21

Principal after 2 years = £441

Interest in the 3rd year = ⁵/₁₀₀ × 441 = £22.05

Principal after 3 years = £463.05

The total interest charged under compound interest will be £63.05.

This is different to the simple interest worked out above.

Four types of question

In percentage questions, read the question carefully and decide what you are being asked to do. You may need to:

• Find a given percentage of an amount.
• Work out a percentage when given 2 amounts.
• Work backwards from a percentage increase or decrease (reverse percentages).
• Find a cumulative change.

Mean, mode and median

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Measures of average in grouped and continuous data

The mean

We already know how to find the mean from a frequency table. Finding the mean for grouped or continuous data is very similar.

The grouped frequency table shows the number of CDs bought by a class of children in the past year.

 

Number of CDs	Frequency (f)
0-4	10
5-9	12
10-14	6
15-19	2
>19	0

• We know that 10 children have bought either 0, 1, 2, 3 or 4 CDs, but we do not know exactly how many each child bought.
• If we assumed that each child bought 4 CDs, it is likely that our estimate of the mean would be too big.
• If we assumed that each child bought 0 CDs, it is likely that our estimate would be too small.
• It therefore seems sensible to use the mid-point of the group and assume that each child bought 2.

Finding the mid-points of the other groups, we get:

 

Number of CDs	f	Mid-point, x	fx
0-4	10	2	20
5-9	12	7	84
10-14	6	12	72
15-19	2	17	34
>19	0	-	0

The mean is 

Remember: This is only an estimate of the mean.

The median

As explained previously, the median is the middle value when the values are arranged in order of size.

As the data has been grouped, we cannot find an exact value for the median, but we can find the class which contains the median.

 

Number of CDs	Frequency (f)
0-4	10
5-9	12
10-14	6
15-19	2
>19	0

There are 30 children, so we are looking for the class which contains the (30 + 1) ÷ 2 = 15¹₂th value. The median is therefore within the 5-9 class.

The mode

The mode is the most common value.

We cannot find an exact value for the mode, and therefore give themodal class. The modal class is 5-9.

Finding the nth term

Sometimes, rather than finding the next number in a linear sequence, you want to find the 41^st number, or 110^th number, say.

Writing out 41 or 110 numbers takes a long time, so you can use a general rule.

To find the value of any term in a sequence, use the n^th term rule.

Question

What is the n^th term of this sequence?

Reveal

For example, to find the 10^th term, work out 5 × 10 = 50. To find the 7^th term, work out 5 × 7 = 35

So the 41^st term is 5 × 41 = 205 and the 110^th term is 5 × 110 = 550

Parallel lines will never meet, no matter how far they are extended. Here are some interesting facts about angles and parallel lines.

Parallel lines

In addition to the basic angle facts, you need to memorise the rules of angles and parallel lines, and apply these to problems.

Small arrows are used to show that 2 lines are parallel to each other.

Click on the button below to see examples of these different types of angle.

• Vertically opposite angles are equal.
• Corresponding angles are equal.
• Alternate angles are equal.
• Co-interior angles add up to 180°.

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Question

Find the angles a, b and c. Give reasons for your answers.

Question

Find the angle B A D. Give a reason for your answer.

When answering questions on angles, remember the following points:

• Look out for parallel lines (squares, rectangles, parallelograms and marked parallel lines) and see which of the above rules apply.
• Look out for lines of equal length. Remember that an isosceles triangle has two sides of equal length and that the angles which are directly opposite these sides are also equal.
• Diagrams are unlikely to be drawn to scale, so do not be tempted to measure!
• Do not assume things which may turn out to be untrue (for example, an angle is not necessarily a right angle just because it looks as though it should be!).
• It is likely that you will not be able to find the required angle immediately. Do not worry - just write as much information as possible onto the diagram. It should eventually lead you to the answer. If not, you might still get some marks for your working!

Solving simultaneous equations using graphs

Solving simultaneous equations using a graph is easier than you might think. First, you need to draw the lines of the equations. The points where the lines cross is the solution.

Linear equations

The graphs of linear equations will give straight lines.

Example

• Solve these simultaneous equations by drawing graphs:
• 2x + 3y = 6
• 4x - 6y = - 4

For example, to draw the line 2x + 3y = 6 pick two easy numbers to plot. One when x = 0 and one where y= 0

• When x = 0 in the equation 2x + 3y = 6
• This means 3y = 6 so y = 2
• So one point on the line is (0, 2)

• When y = 0
• 2x = 6 so x = 3
• So another point on the line is (3 ,0)

In an exam, only use this method if you are prompted to by a question. It is usually quicker to use algebra if you are not asked to use graphs.

Linear and quadratic equation

Example

• Solve the simultaneous equations by drawing graphs.
• y - 2x = 1
• y = x² - 2

How to convert a recurring decimal into a fraction

Example 1

Write 0.77..$\mathrm{0.77..}$ as a fraction in its lowest terms.

Our first step is to form a simple equation where x=0.77..$x=\mathrm{0.77..}$. By multiplying both sides by 10$10$ we can obtain another equation with 10x=7.77..$10x=\mathrm{7.77..}$. Now we eliminate the recurring part of the decimal by subtracting x$x$ from 10x$10x$.

x10x9xx=0.77..=7.77..=7=79$$\begin{array}{rl}x& =\mathrm{0.77..}\\ 10x& =\mathrm{7.77..}\\ 9x& =7\\ x& =\frac{7}{9}\end{array}$$

So we have our answer 0.77...=79$\mathrm{0.77...}=\frac{7}{9}$.
The important part to remember is to get two equations in x$x$ where the recurring part after the decimal point is exactly the same.

Example 2

Write 0.277..$\mathrm{0.277..}$ as a fraction in its lowest terms.

Again our first step is write x=0.277..$x=\mathrm{0.277..}$. Now multiplying by 10$10$ gives us 10x=2.77..$10x=\mathrm{2.77..}$. This time we need another equation to match the recurring part of the equation. So multiplying by 10$10$ again gives 100x=27.77..$100x=\mathrm{27.77..}$. Now we have two equations with the same recurring part we subtract one from the other as before.

x10x100x90xxx=0.277..=2.77..=27.77..=25=2590=518$$\begin{array}{rl}x& =\mathrm{0.277..}\\ 10x& =\mathrm{2.77..}\\ 100x& =\mathrm{27.77..}\\ 90x& =25\\ x& =\frac{25}{90}\\ x& =\frac{5}{18}\end{array}$$

So in its lowest terms 0.277..=518$\mathrm{0.277..}=\frac{5}{18}$.

Example 3

Write 0.0855.. as a fraction in its lowest terms.

As always our first step it to write x=0.0855..$x=\mathrm{0.0855..}$. This time to get two equations with the same recurring part after the decimal point we need to multiply by 100$100$ and then 1000$1000$. This gives us:

100x1000x900xx=8.55..=85.55..=77=77900$$\begin{array}{rl}100x& =\mathrm{8.55..}\\ 1000x& =\mathrm{85.55..}\\ 900x& =77\\ x& =\frac{77}{900}\end{array}$$

This cannot be simplified any further so 0.855..=77900$\mathrm{0.855..}=\frac{77}{900}$.

Example 4

Write 0.0234234.. as a fraction in its lowest terms.

This time the '234$234$' part is the important bit. Now to get two multiples of x with 234 as the recurring part we need to multiply first by 10$10$ and then another 1000$1000$ to move the digits 3 places to the right.

x=0.0234234...10x=0.234234..10000x=234.2349990x=234xx=2349990x=13555$$\begin{array}{r}x=\mathrm{0.0234234...}\\ 10x=\mathrm{0.234234..}\\ 10000x=234.234\\ 9990x=234x\\ x=\frac{234}{9990}\\ x=\frac{13}{555}\end{array}$$

So 0.0234=13555$0.0234=\frac{13}{555}$ as a fraction in its lowest terms.