A level Maths Year 2 Student Book: Chapter 13

Sample Chapter This is draft content and subject to change

Edexcel A-level Mathematics Year 2

This Student Book will: • support you through the course with detailed explanations, clear worked examples and plenty of practice on each topic, with full worked solutions available to teachers

• help you take control of your learning with prior knowledge checks to assess readiness and end-of-chapter reviews that test understanding • prepare you for linear assessment with clear links between topics embedded in each chapter and exam-style practice questions that help build synoptic understanding • build your confidence in the key A-level skills of problem solving, modelling, communicating mathematically and working with proofs • demonstrate how technology, such as graphing software and scientific calculators, can aid and deepen the study of A-level Maths with ‘Technology’ boxes throughout • prepare you for further study and working life by setting maths in real-world contexts that emphasise practical applications.

Year 1 and and AS Student Book

Edexcel A-level Mathematics Year 1 and AS Student Book 978-0-00-820495-2

Maths_ALevel_Covers_v02.indd 2

Edexcel A-level Mathematics Year 2 Student Book Helen Ball Kath Hipkiss Michael Kent Chris Pearce

Edexcel A-level Mathematics Helen Ball Kath Hipkiss Michael Kent Chris Pearce

Edexcel A-level Mathematics Year 2 Student Book

Written by our expert authors for the 2017 Edexcel specification, this Student Book covers all of the content needed for A-level Year 2 Mathematics.

It combines comprehensive and supportive explanations with plenty of practice to prepare you for the demands of A-level and beyond.

ISBN 978-0-00-820496-9

9 780008 204969

09/01/2017 18:45

CONTENTS Introduction v 1 – Algebra and functions 1: Functions

1

1.1 Definition of a function 2 1.2 Composite functions 8 1.3 Inverse functions 12 1.4 The modulus of a linear function 15 1.5 Transformations involving the modulus function 21 1.6 Functions in modelling 24 Summary of key points 28 Exam-style questions 28

2 – ALGEBRA AND FUNCTIONS 2: PARTIAL FRACTIONS 31 2.1 Simplifying algebraic fractions 2.2 Partial fractions without repeated terms 2.3 Partial fractions with repeated terms 2.4 Improper fractions 2.5 Using partial fractions in differentiation, integration and series expansion Summary of key points Exam-style questions

32 33 36 39 41 43 43

3 – Coordinate Geometry 1: Parametric equations 45 3.1 Parametric equations of curves 46 3.2 Converting between Cartesian and parametric forms 50 3.3 Problems involving parametric equations 53 Summary of key points 59 Exam-style questions 59

4 – Sequences and series 4.1 Types of sequences 4.2 Sigma notation 4.3 Arithmetic sequences and series 4.4 Geometric sequences and series 4.5 Binomial expansions Summary of key points Exam-style questions

5 – Trigonometry 5.1 Radians 5.2 Trigonometric ratios

61 62 66 68 71 75 78 78

80 81 84

5.3 Sketching graphs of trigonometric functions using radians 86 5.4 Practical problems 88 5.5 Small angle approximations 91 5.6 Addition and subtraction formulae 93 5.7 Expressions of the form a cos ø + b sin ø 97 5.8 More trigonometric functions 99 5.9 Inverse functions 103 Summary of key points 105 Exam-style questions 105

6 – Differentiation

107

6.1 Turning points 108 6.2 The chain rule 110 6.3 Differentiating ekx 112 6.4 Differentiating in ax 114 6.5 Differentiating sin x and cos x from first principles 116 Summary of key points 119 Exam-style questions 119

7 – Further differentiation 7.1 The product rule 7.2 The quotient rule 7.3 Differentiating trigonometric functions 7.4 Differentiating parametric equations 7.5 Implicit equations 7.6 Constructing differential equations Summary of key points Exam-style questions

8 – Integration 8.1 Recognising integrals 8.2 Integration with trigonometrical functions 8.3 Definite integrals 8.4 Integration by substitution 8.5 Integration by parts 8.6 Integrating algebraic fractions 8.7 Solving differential equations Summary of key points Exam-style questions

9 – Numerical methods 9.1 Finding roots 9.2 How change of sign methods can fail 9.3 Iterative methods 9.4 Newton–Raphson

121 122 125 127 130 133 137 140 140

143 144 147 149 153 156 160 163 169 169

171 172 175 177 185

iii

CONTENTS

9.5 How iterative methods can vary 187 9.6 Numerical integration 190 9.7 Using numerical methods to solve problems 193 Summary of key points 196 Exam-style questions 196

10 – Three-dimensional vectors 10.1 Vectors in three dimensions 10.2 Vectors and shapes Summary of key points Exam-style questions

11 – Proof 11.1 Proof by contradiction Summary of key points Exam-style questions

12 – Probability

198 199 203 207 207

209 210 214 214

216

12.1 Set notation to describe events and outcomes 217 12.2 Conditional probability 222 Summary of key points 234 Exam-style questions 234

13 – Statistical distributions

238

13.1 The normal distribution 239 13.2 Using the normal distribution 242 13.3 Non-standardised variables 249 13.4 Using the normal distribution to approximate the binomial distribution 257 Summary of key points 263 Exam-style questions 263

14 – Statistical hypothesis testing

265

14.1 Correlation coefficients 266 14.2 Hypothesis tests for the mean of a normal distribution 275 14.3 Non-linear regression 280 Summary of key points 285 Exam-style questions 285

15 – Kinematics

287

15.1 Equations of constant acceleration 288 15.2 Velocity vectors 291 15.3 Equations of constant acceleration using vectors 298 15.4 Vectors with calculus 301 15.5 Projectiles 305 Summary of key points 318 Exam-style questions 318

iv

16 – Forces 16.1 Resolving forces 16.2 Adding forces 16.3 Coefficient of friction 16.4 Connected particles Summary of key points Exam-style questions

17 – Moments

321 322 332 335 342 352 352

355

17.1 Turning moments 17.2 Horizontal rods 17.3 Equilibrium of rigid bodies Summary of key points Exam-style questions

356 365 373 383 383

Exam-style extension questions

xxx

Answers xxx Formulae xxx Glossary xxx Index xxx

13

STATISTIcAl DISTRIBUTIonS

Data such as the scores of students in an exam will have a mean and a standard deviation showing the spread of the data. The majority of students taking the exam will get within two standard deviations of the mean.

You can use this information to predict other information about the data and the real-life situation. If you are told that the bottom 5% of students will fail the course, you could work out the lowest mark that a student can have and still be awarded a passing grade, or the probability of a student’s score being in the top 10%.

In this chapter you will learn how to use the normal distribution to calculate missing information such as in the example above and to find the probability of certain data items occurring.

lEarninG obJEctiVES You will learn how to:

› › › › › ›

understand and use the normal distribution find probabilities using the normal distribution using a calculator analyse the shape, symmetry and points of inflection of the normal distribution link the normal distribution to the binomial distribution apply continuity corrections select an appropriate probability distribution for a particular context.

toPic linKS You will need the skills and techniques that you learnt in Book 1, Chapter 12 Data presentation and interpretation to interpret histograms and calculate the mean and standard deviation. The content from Book 1, Chapter 13 Probability and statistical distributions will help you calculate probabilities. From that same chapter, the ability to use binomial expansions will help you calculate exact and cumulative probabilities.

Prior KnoWlEdGE You should already know how to:

› ›

interpret the mean and standard deviation and draw histograms

›

apply statistics to describe a population.

infer properties of populations or distributions from a sample, while knowing the limitations of sampling

The normal distribution

13.1

You should be able to complete the following questions correctly: 1 The number of calls received by the emergency services in 20 consecutive minutes was: 6, 1, 3, 4, 0, 4, 2, 3, 9, 4, 4, 3, 5, 0, 4, 6, 6, 4, 2, 5 Calculate : a the mean

b the standard deviation.

In the following 5 minutes the following amounts of calls were made: 0, 0, 2, 5, 1 c Calculate the new mean and standard deviation.

13.1 The normal distribution If you measured the heights of all the people at a concert on any particular night and plotted the results, the graph would look something like this histogram.

Probability density

50 40 30 20 10

0

1.0

1.2

1.4

1.6

1.8

2.0

Heights (m) By plotting the continuous variable of height, you can represent all the possible probabilities of any person being any height.

Continuous distributions are distributions for continuous variables and are more commonly known as probability densities.

A probability density shaped like a bell like this is called the normal distribution.

Each end of the bell curve extends asymptotically. The y-axis in the normal distribution represents the probability density. For example, in the diagram above, the area under the curve represents one. Using this you can see the probability of a person being about 1.2 m is about half of the probability of a person being about 1.4 m.

Normal distributions deal with continuous variables.

‘Asymptotically’ means that the curve never touches the axis.

You do not need to understand the concept of probability density in detail as it is above the scope of the course. However, you should learn the following features of a curve that describes a continuous distribution like the normal distribution.

›

The area under the curve equals 1.

Using the example of heights at a concert, you would conclude that the probability of having a height is 1!

›

The probability of any exact value of X is 0.

3

13 StatiStical diStribUtionS

In the concert example, the probability that a particular person will have height of exactly 1.3 m is essentially zero.

›

You can make the probability as close as you like to zero by making the height measurement more and more precise.

The area under the curve and bounded between two given points on the x-axis is the probability that a number chosen at random will fall between the two points.

In the concert example, suppose that the probability of measuring 1 . Then the continuous distribution between 1.3 m and 1.4 m is 10 for possible measurements would have a shape that places 10% of the area below the curve in the region bounded by 1.3 m and 1.4 m on the x-axis.

Stop and think

What would the histograms look like if you had separated the heights of males and females?

›

Normal distributions are defined by two parameters, the mean (μ) and the standard deviation (s).

›

68% of the area of a normal distribution is within 1 standard deviation of the mean.

›

Approximately 95% of the area of a normal distribution is within two standard deviations of the mean.

›

99.75% of the data lies within ±3 standard deviations of the mean.

›

The mean, μ, and standard deviation, s, determine the shape of the normal curve.

›

The mean gives the central location of the data, which is the line of symmetry.

›

The smaller the standard deviation, the less spread out the data. The larger the standard deviation, the more spread out the data.

0

1

2

3

4

5

6

7

8

9

10

In the diagram above, the red and green curves have the same shape and width. This means the standard deviations are the same. The green and blue curves have the same mean. The blue curve is wider, so has the larger standard deviation, so its data is more spread out.

4

68% –3

–2

–1

0

1

2

3

2

3

Standard deviations

95% –3

–2

–1

0

1

Standard deviations

The normal distribution

A point of inflection is where the point on the curve stops increasing and starts decreasing.

The inflection points are the between the mean, μ, and the standard deviation, s. This means that the inflection points are between μ − s and μ + s.

13.1

inflection points

total area = 1

Look at these two distributions.

μ – 3σ μ – 2σ μ – σ

μ

μ+σ

μ + 2σ

μ + 3σ

Both distributions in the diagram have the same mean. The red curve has a smaller standard deviation, meaning its distribution is less spread out on either side of the mean. The blue curve has a larger standard deviation, meaning its distribution is more spread out on either side of the mean.

Example 1 The marks in an examination are normally distributed. a What percentage of the marks would be above the mean weight? b What percentage of the marks would be between −1 and +1 standard deviation? c What percentage of the marks would be between the mean and +2 standard deviations? d What percentage of the marks would be between −2 standard deviation and +1 standard deviations?

Solution a As the normal distribution is symmetrical either side of the mean, you would expect 50% of the marks to be above the mean. b You would expect that between −1 and +1 standard deviations either side of the mean to be 68% of the marks. c Between the mean and +2 standard deviations would be 47.5% of the marks because there is 95% of the data between −2 and +2 standard deviations either side of the mean. d Between −2 and +2 standard deviations is 95% of the data and between −1 and +1 standard deviations is 68% of the data.

So from −2 standard deviations to the mean is 47.5% of the data and from the mean to +1 standard deviation is 34% of the data. Combining these gives you 81.5% of the data.

5

13 StatiStical diStribUtionS

exercise 13.1A 1

page xx

Correct the following list of properties of a normal distribution.

a The distribution is symmetrical about the standard deviation. b The mode, median and mean are all different. c The total area above the curve is 1. d The distribution is defined by three parameters: the mean, the median and the standard deviation.

CM

2

PS

3 The heights of Year 13 students are normally distributed.

Comment on the features of the two distributions in the diagram.

a What percentage of the Year 13 students would be above the mean height? b What fraction of the Year 13 students would have heights between the mean and +1 standard deviation?

c What percentage of the Year 13 students would have heights between the mean and +2 standard deviations?

d What fraction of the Year 13 students would have heights between −3 standard deviation and +2 standard deviations?

4

Describe each of the curves in this diagram, using the terms mean and standard deviation.

y Percent

CM

40

A

30 20

B

10 6

9

12

15

18

21

x

13.2 Using the normal distribution You can calculate the area under each section of a normal distribution curve using integration.

The diagram below shows a normal distribution with a mean of 24 and a standard deviation of 1. The shaded area between 23 and 25 contains 68% of the distribution – so the probability of getting a value between 23 and 25 would be 0.68. Similarly, 95% of the data 6

Using the normal distribution

13.2

lies between 22 and 26 and anything below 21 or above 27 would be classed as an outlier.

The area under a normal curve shows probabilities between two values. The probability of the variable taking a value between two limits is the area under the curve between those limits. Not all probabilities will be as straightforward as 1 or 2 standard deviations, so you will need to use a formula.

20

21

22

23

24

25

26

27

28

If a variable X has a normal distribution then you can write this using mathematical notation: X ∼ N(μ, s 2) where

› › › › › ›

X is the random variable (this can be any letter) ∼ is short hand for ‘is distributed’ N tells you that it has a normal distribution μ is the mean s is the standard deviation s 2 is the square of the standard deviation and is called the variance.

For example, if X ∼ N(25, 12), this tells you that

› › ›

the random variable X has a normal distribution with a mean μ = 25 and a standard deviation s = 1.

This means that 68% of the values lie between 24 (= 25 − 1) and 26 (= 25 + 1).

Don’t forget that the second parameter is the variance.

The standard deviation is the square root of this number.

Example 2 The heights of males of females in a town were recorded and had results with normal distributions as follows: Males: X ∼ N(178, 72) Females: Y ∼ N(167, 64) Write down mean, variance and standard deviation for each gender.

Solution Using X ∼ N(μ, σ2), you know the first parameter is the mean and the second is the variance. With that in mind: Males: X ∼ N(178, 72) has a mean of 178 cm, variance of 49 (or 72) and standard deviation of 7.

Females: Y ∼ N(167, 64) has a mean of 167 cm, variance of 64 (or 82) and standard deviation of 8.

7

13 StatiStical diStribUtionS

exercise 13.2A 1

page xx

During a kickboxing tournament, two different fighters had scores with normal distributions as follows: Fighter A: mean = 3, variance = 4 Fighter B: mean = 2, standard deviation = 3 Use the correct mathematical notation to describe the distribution of each of the fighters’ scores.

2

A school took part in a Maths Challenge over two different years. The scores with normal distributions as follows: Year 1: μ = 0, s 2 = 5 Year 2: μ = −1, s = 2 Use the correct mathematical notation to describe the distribution of the scores for each year.

CM

3

A mathematics student was asked to write down the mathematical notation for the scores on an app which have a normal distribution with mean μ = 7 and standard deviation s = 11.

The student wrote N ∼ X(11, 49).

Write down everything that is wrong with the student’s notation, then write down the correct notation.

4 Two mathematics students sat a set of tests and had results with normal distributions as follows:

Student A: X ∼ N(5, 72) Student B: Y ∼ N(6, 81) a Write down the mean, variance and standard deviation of the students’ scores. b Sketch this as a graph. the standard normal distribution It is not necessarily the case that normal distributions have the same means and standard deviations. A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution.

The standard normal distribution can be used to work out probabilities for variables with a normal distribution. The standard normal distribution is defined as follows: Z ∼ N(0, 1) This tells you that:

› › › ›

the standard normal random variable is denoted by Z Z has a normal distribution with a mean μ = 0 and a standard deviation s = 1.

The diagram shows the standard normal distribution.

8

–1

0

1

Using the normal distribution

Any normal distribution can be made to fit the standard normal distribution. The process is called standardising. Since the distribution has a mean of 0 and a standard deviation of 1, the value of Z is equal to the number of standard deviations below (or above) the mean. This is often referred to as the z-score. For example, a z-score of −2.5 represents a value 2.5 standard deviations below the mean.

Below is a graph of the standard normal distribution. It is symmetrical about the mean, 0. The shaded area under the curve shows probabilities that Z takes that are less than or equal to z. To calculate the probability of this, you would need to find the area.

An uppercase Z signifies the random variable which is being standardised. A lowercase z signifies the z-score.

The z-scores can all be calculated on your calculator. The calculator shows values of the normal cumulative distribution function, F (z), which correspond to the area under the curve to the left of Z for different values of z. That is, the probability that Z ⩽ z.

P(Z < z)

There are three different cases to consider.

› › ›

13.2

0

Z is less than value a: (Z < a)

z

Z is greater than value a: (Z > a)

Key InFoRMATIon

Z is between values a and b: (a < Z < b).

If you are given Z ∼ N(0, 1), you can say that the probability that Z is less than a value a (where a > 0) is shown by P(Z < a)

›

F (z) = P(Z ⩽ z)

y

where P is the probability.

You know that the mean of the standard normal distribution is 0 so the diagram will look like this: If you are given Z ∼ N(0, 1), you can say that the probability that Z is greater than a value a (where a > 0) is shown by P(Z > a)

0

a

x

where P is the probability.

You know that the mean of the standard normal distribution is 0 so the diagram will look like this:

Technology Your calculator should have a distribution mode with various settings for the normal distribution. Make sure that you know how to find values of the cumulative distribution function.

y

0

a

x 9

13 StatiStical diStribUtionS

If you are given Z ∼ N(0, 1), you can say that the probability that Z is between the value a and b (where a, b > 0 and b > a) is shown by P(a < Z < b) where P is the probability.

You know that the mean of the standard normal distribution is 0 so the diagram will look like this:

0

a

exercise 13.2B 1

b

page xx

Given that Z ∼ N(0, 1), draw a clear sketch diagram for:

a P(Z > a) where a < 0 b P(Z < a) where a < 0 c P(a < Z < b) where a < 0, b > 0 d P(a < Z < b) where a, b < 0, and b > a e P(0 < Z < a) where a > 0.

CM

2

A mathematics student was asked to write down the mathematical notation for the standard normal distribution.

The student wrote N ∼ Z(1, 02).

List everything that is wrong with what the student wrote and then write down the correct notation.

Using a calculator for the normal distribution Cumulative probabilities associated with the z-score can all be calculated using your calculator.

If you are given Z ∼ N(0, 1) and asked to find, for example, P(Z < 0.43), begin by drawing a clear diagram.

The expression for the probability is: P(Z < 0.43) = F (0.43) where F represents the area to the left of the value in brackets – that is, any given value of Z. On your calculator, set the mode to normal cumulative probability as you want to know the probability up to and including 0.43. Set the lower bound as a low negative number (for example –999, as the curve is an asymptote) and the upper bound as 0.43. The standard deviation should be 1 as this is a standard normal distribution.

So P(Z < 0.43) = 0.666 402.

This means the probability of Z having a value < 0.43 is 0.666 402.

How do you find P(Z > 0.43)? The calculator only gives the probabilities to the left of a value – i.e. less than the value.

10

0

0.43

F is the capital Greek letter F called phi, pronounced ‘fi’.

Using the normal distribution

However, the total area under the curve is 1, so you can calculate the difference.

So P(Z > 0.43) = 1 − F (0.43) = 1 − 0.666 402 = 0.333 598 So the probability of Z having a value > 0.43 is 0.333 598.

13.2

Technology This time you can select the lower bound as 0.43 and the upper bound as a large number (for example 999, as the curve is an asymptote).

Note that P(Z < 0.43) + P(Z > 0.43) = 1.

How do you find P(Z < −0.43)?

You know that the normal distribution is symmetrical about the mean, so: P( Z < −0.43) = P( Z > 0.43) = 1 − P(Z < 0.43)

–0.43

= 1 − F (0.43) = 1 − 0.666 402

Technology

= 0.333 598

Select the lower bound to a small number and upper bound to −0.43 and standard deviation to 1.

This means the probability of Z having a value < 0.43 is 0.333 598.

Note this is the same as P(Z > 0.43).

Finally, how do you find P(0.1 < Z < 0.31)?

To find the probability (area) between 0.31 and 0.1 you need to find the probability less than 0.31 and subtract the probability less than 0.1.

P(0.1 < Z < 0.31) = P(Z < 0.31) − P(Z < 0.1) = F (0.31) − F (0.1) = 0.621 72 − 0.539 83 = 0.081 89 What is the probability that Z will have a value greater than the mean but less than 1.5?

To find the probability (area) between the mean, 0, and 1.5, you need to find the probability that Z will have a value less than 1.5 and subtract the probability that Z will have a value less than 0.

0

0.1

0.31

It is important to show stages in your workings and not rely solely on writing down the answer from your calculator.

Technology On your calculator you can set the upper bound as 0.31 and the lower bound as 0.1 and the standard deviation as 1. This will get the same result.

0

1.5

P(0 < Z < 1.5) = P(Z < 1.5) − P(Z < 0) 11

13 StatiStical diStribUtionS

The probability that Z is less than 0 is 0.5 since 0 is the mean and half of a normal distribution lies below the mean and half above it.

P(0 < Z < 1.5) = P(Z < 1.5) − P(Z < 0)

Technology On your calculator, set the lower bound to 0 and the upper bound to 1.5 and the standard deviation to 1.

= F (1.5) − 0.5 = 0.933 19 − 0.5 = 0.433 19

Example 3 Find the following probabilities using your calculator: a P(Z ⩽ 0.34)

b P(Z < 0.2)

c P(Z > 0.6)

Solution a Select a small lower bound and upper bound as 0.34 and a standard deviation of 1.

P(Z ⩽ 0.34) = 0.633 07 b Select a small lower bound and upper bound as 0.2 and a standard deviation of 1.

P(Z < 0.2) = 0.579 26 c Select a lower bound of 0.6 and a large upper bound and a standard deviation of 1.

P(Z > 0.6) = 0.274 25

exercise 13.2c For this exercise, use the correct notation and draw a diagram in each case.

1

Find the probability that Z will be less than 1.4.

2

Find the probability that Z will be less than 0.87.

3

Find the probability that Z will be greater than 1.06.

4

Find the probability that Z will be less than −3.

5

Find the probability that Z will be greater than −1.32.

6

Find the probability that Z will be between 1.1 and 2.1.

7

Find the probability that Z will be between −1.32 and 1.21.

8

Find the probability that Z will be between −2.65 and −1.43.

12

page xx

Non-standardised variables

13.3

13.3 Non-standardised variables You know that the area under a normally distributed curve (that is, with mean 0 and standard deviation 1) is equal to one. To find the probability that a random variable x is in any interval within the curve, you need to calculate the area of that interval. To find the area of any interval under any normal curve, convert to a z-score.

Use the following formula to find a z-score: Value - Mean Standard deviation x−µ = . σ

z=

The horizontal scale of the curve of the standard normal distribution corresponds to z-scores.

If the random variables are converted into z-scores, the result will be the standard normal distribution. Following this conversion, the area that falls in the interval under the non-standard normal curve is the same as the area under the standard normal curve within the corresponding boundaries.

Example 4 Given the random variable X ∼ N(15, 42), find: a P(X < 25)

b P(X > 25)

c P(10 < X < 25)

Solution a Draw a clear diagram.

15

(

25

25 − 15 4 = P(Z < 2.5)

P(X < 25) = P Z <

)

= F (2.5) = 0.993 79 So, the probability of X having a value < 25 is 0.993 79.

The variance is 42, so the standard deviation is 4.

Although your calculator can do this in a single calculation, make sure you show your working and how to standardise your variable.

You can use the single-step calculation as a way to double-check your answer.

13

13 StatiStical diStribUtionS

b Draw a clear diagram.

15

Technology

25

P(X > 25) = 1 – P(X < 25)

(

P ( X > 25) = 1 − P Z <

25 − 15 4

)

On your calculator, set the lower bound to a small number and the upper bound to 2.5 not 25 – the z-score is 2.5, as you are modelling using the standard normal distribution. This means you will use a standard deviation of 1, not 4.

= 1 − P( Z < 2.5) = 1 − F (2.5) = 1 − 0.993 79 = 0.006 21 So, the probability of X having a value > 25 is 0.006 21. c Draw a clear diagram.

10

25

P (10 < X < 25) = P

(10 −4 15 < Z < 25 4− 15 )

= P(−1.25 < Z < 2.5) = P(Z < 2.5) − P(Z < −1.25)

Technology You calculator can also do this in a single calculation by inputting the mean as 15 and the standard deviation as 4, the lower bound as a small number and the upper bound as 25.

Technology Set the lower bound to 2.5 and the upper bound to a large number and the standard deviation to 1.

= F (2.5) − F (−1.25) = 0.993 79 − 0.105 65 = 0.888 14 So, the probability of X having a value between 10 and 25 is 0.888 14.

14

Technology Set the lower bound to −1.25 and the upper bound to 2.5 and standard deviation to 1.

Non-standardised variables

Using the large data set 1

13.3

Assuming that daily mean wind speed follows a normal distribution, and given the mean wind speed is 9.4 mph and standard deviation is 3.3 mph, find the probability that the daily mean wind speed in Cambourne in 2015 is: a above 14 mph b below 9 mph c

between 9 and 14 mph.

exercise 13.3A

page xx

For this exercise, present your working as shown in the examples above and use the correct notation.

1

Find the area of the indicated region under the standard normal curve.

0

–0.5 0

2

PS

3

4

1.5

–2.25

Z

Z

0

0

2

Z

Given that X ∼ N(30, 100), find the following probabilities.

a P(X < 35)

b P(X > 38.6)

c P(X > 20)

d P(35 < X < 40)

e P(15 < X < 32)

f P(17 < X < 19)

The distribution of heights of 20-year-old men is modelled using the normal distribution with a mean of 177 cm and a standard deviation of 7 cm. Find the probability that the height of a randomly selected 20-year-old man is:

a under 172 cm PS

Z

1.2

b over 180 cm

c

between 172 cm and 180 cm.

When Tim makes a hot chocolate he uses instant powder and uses a spoon to add it to into a mug. The weight of powder in grams may be modelled by the normal distribution with a mean of 6 g and a standard deviation of 1 g. Tim has found that if he uses more than 8 g Jenni says it’s too strong and if he uses less than 4.5 g she says it’s too weak. Find the probability that he makes the hot chocolate:

a too weak

b too strong

c just right.

15

13 StatiStical diStribUtionS

5

A particular breed of hens produces eggs with a mean mass of 60 g with a standard deviation of 4 g and mass is found to be normally distributed.

The eggs are classified as small, medium, large or extra large depending on their weight, as follows. classiďŹ cation

Weight

small

less than 55 g

medium

between 55 g and 65 g

large

between 65 g and 70 g

extra large

greater than 70 g

a Find the proportion of eggs that are classified as small. b Find the proportion of eggs that are classified as medium. c Find the proportion of eggs that are classified as large or extra large. d If the classification for medium eggs is changed to between 55 g and 68 g, by how much does the proportion of eggs classified as medium increase?

6

The police regularly monitor the speeds of cars on a section of motorway and it is found that the speeds are normally distributed. The mean speed is 68.5 mph with a standard deviation of 5 mph.

a Find the proportion of motorists that break the speed limit (70 mph). b A motorist believes that people arenâ&#x20AC;&#x2122;t fined unless they are 10% over the speed limit.

What proportion of motorists will be fined on this basis?

c A motorist is given a fixed penalty fine for doing 85 mph.

What percentage of motorists exceed this speed?

Finding values of variables from known probabilities Sometimes you may know the probability of an event and then need to work backwards to find the corresponding value X.

Example 5 A random variable X is normally distributed with a mean 9 and standard deviation 0.27. Find the value of x so that P(X < x) = 0.751 75.

Solution First, write down the distribution in the question as a standardised normal distribution.

(

P Z <

)

xâ&#x2C6;&#x2019;9 = 0.75175 0.27

On your calculator, set the area (the probability up to and including the value you are trying to find) as 0.751 75. The standard deviation is 0.27 and the mean is 9. x = 9.1836 Probability of x = 9.1836 with a mean of 9. Standard deviation of 0.27 is 0.751 75.

16

Technology Your calculator now needs to be in inverse normal distribution mode.

Non-standardised variables

Using the large data set 2

13.3

Assume that in Hurn in 2015 the data for the daily maximum gust is normally distributed. a Find the mean. b Find the standard deviation. c

Find the value of x so that P(X < x) = 0.850.

exercise 13.3B

page xx

1 The lifetime of an electrical component may be modelled by a normal distribution with mean 150 hours and variance 52 hour2.

a Find the probability that a component lasts i more than 150 hours iii between 100 and 150 hours.

ii less than 100 hours

b How long would you expect i 15% of the components to last? ii 30% of the components to last? iii 90% of the components to last?

2

A factory produces bags of sugar. The actual weights may be modelled by a normal distribution with mean 498.7 g and standard deviation 7.3 g. Bags are rejected if they are 15% underweight or 90% overweight. What weights are these?

3

Assume a test is normally distributed with a mean of 19 and a standard deviation of 2.4. Let X be the distribution of test scores. Find a and b where:

a P(X > a) = 0.432 4

PS

5

b P(X < b) = 0.205

Given that X â&#x2C6;ź N(30, 100), find:

P(X < x) = 0.99

P(X < x) = 0.979 82

P(X > x) = 0.194 89

P(X < x) = 0.75

P(X ) > 0.99

P(X > x) = 0.05

A manufacturer produces a new fluorescent light bulb and claims that it has an average lifetime of 4000 hours with a standard deviation of 375 hours.

Find the lifetime such that 90% of light bulbs will last for less than this duration.

PS

6

A double bed is the right length for 92.9% of men.

The heights of men in the UK are normally distributed with mean height 1.753 m and standard deviation 0.1 m.

What is the greatest height a man can be to fit into a double bed?

PS

7

The police regularly monitor the speeds of cars on a section of motorway and it is found that the speeds are normally distributed. The mean speed is 68.5 mph with a standard deviation of 5 mph.

Find the range within which 80% of the speeds lie.

17

13 StatiStical diStribUtionS

PS

8

An examination has a mean mark of 80 and a standard deviation of 12. The marks can be assumed to be normally distributed.

a What is the least mark needed to be in the top 25% of the students taking this examination? b Between which two marks will the middle 95% of students lie? c 200 students take this examination. Calculate the number of students likely to score 90 or more. determining the mean or standard deviation There are instances where a random variable is known to be normally distributed but the value of the mean or the standard deviation (or both) is not known. When you have information about the probabilities, it is possible to determine the unknown parameters.

Example 6 A farm which produces potatoes models the weights (in g) of a particular brand of potato by a random variable, J ∼ N(μ, 25).

The processing plant observes that 2% of the potatoes weigh more than 350.7 g. Determine the mean, μ.

Solution Draw a sketch of the given information:

μ

350.7g

You need to find the area of the shaded region. The question says 2% of the potatoes have a weight of more than 350.7 g, therefore the area is 2% to the right.

This gives P(Z > 2.05375) = 0.02 The standardised value of 350.7 satisfies the equation: 350.7 − µ = 2.0537 5 Multiply by 5 to give 350.7 − μ = 10.2685 Solving this for μ = 340.4315 = 340 g

Example 7 A new process is introduced at the farm. This changes the variance so that the weight of the potatoes is now modelled by the random variable M ∼ N(340, s 2). The processing plant reveals that 90% of them have a weight above 322.3 g.

Determine the variance of the random variable.

18

Technology On your calculator, set the area to 0.98 (remembering that the calculator is set to read from the left of the graph) and as it is modelled using the normal distribution the mean will be 0 and the standard deviation 1.

Non-standardised variables

Solution

Technology

Draw a sketch:

90% 322.3

13.3

340

You now need to find the area of the shaded region. The question says 90% of the potatoes have a weight of more than 322.3 g, therefore the area is 90% to the right.

On your calculator, set the area to 0.1 (remembering that the calculator is set to read from the left of the graph) and as it is modelled using the normal distribution the mean will be 0 and the standard deviation 1.

This gives P(Z > −1.2816) = 0.90 The standardised value of 322.3 satisfies the equation: 322.3 − 340 = −1.2816 σ Multiply by σ to give 322.3 − 340 = −1.2816s −17.7 = −1.2816s Solving this for s = 13.81 So the variance, s 2 = 190.72 g

Example 8 The weight of potatoes put into bags by the processing plant is modelled by a normal distribution with a mean of μ g and a standard deviation of s g.

The processing plant observe that in the bags 83% contain less than 354.8 g and 8.5% contain less than 343.1 g. Determine μ and s .

Solution As there are two pieces of information, two sketches are required.

μ

354.8

343.1

μ

You need to find the area of the shaded regions.

Area 1: The question says 83% of the potatoes have a weight of less than 354.8 g, therefore the area is 83%.

This gives P(Z < 0.954) = 0.83 19

13 StatiStical diStribUtionS

The standardised value of 354.8 satisfies the equation: 354.8 − µ = 0.954 σ Area 2: The question says 8.5% of the potatoes have a weight of less than 343.1 g, therefore the area is 8.5%.

This gives P(Z < −1.372) = 0.085 The standardised value of 343.1 satisfies the equation: 343.1 − µ = −1.372 σ You can now rearrange both of these equations to set up a pair of simultaneous equations in μ and s .

354.8 − μ = 0.954s 343.1 − μ = −1.372s Subtract the equations to eliminate μ.

11.7 = 2.326s s = 5.03 Therefore, substituting back into one of the two derived equations: 354.8 − μ = 0.954s 354.8 − μ = 4.799 μ = 350

Using the large data set 3

The daily maximum gust in Leeming in 1987 is modelled by a normal distribution with a mean of μ mph and a standard deviation of s mph.

Meteorologists observe that 6.8% of the daily gusts are above 30 mph and 60.1% of the daily gusts are below 40 mph. Determine μ and s.

exercise 13.3c PS

1

page xx

Assume a test is normally distributed with a mean of 43 and a standard deviation of σ.

Let X be the distribution of test scores.

Find σ if the probability of getting a score above 48 is 0.2.

PS

2

The length of a bike frame may be modelled by a normal distribution with standard deviation 13 mm. 11% of the components are longer than 47 cm.

Calculate the mean length of a frame.

PS

3 The volume of the contents of a can of fizzy drink may be modelled by a normal distribution.

18% have a volume of more than 325.42 ml and 72% have a volume of more than 332.91 ml.

Find the mean and standard deviation of the volume.

20

Using the normal distribution to approximate the binomial distribution

PS

4

The quartiles of a normal distribution are known to be 9.92 and 12.24. Find the mean and standard deviation of the distribution.

PS

5

Engineers make certain components for aeroplanes. The company believes that the time taken to make this component may be modelled by the normal distribution with a mean 90 minutes and a standard deviation 3 minutes.

13.4

Assuming the company’s beliefs to be true, find the probability that the time taken to make one of these components, selected at random, was:

a over 95 minutes

b under 87 minutes

c between 87 and 95 minutes.

Stan believes that the company is allowing too long for the job and times himself manufacturing the component. He finds that only 10% of the components take him over 85 minutes, and that 20% take him less than 65 minutes.

Estimate Stan’s mean and standard deviation.

6

A power lifter knows from experience that she can deadlift at least 140 kg once in every 5 attempts. She also knows that she can deadlift at least 130 kg on 8 out of 10 attempts. Find the mean and standard deviation of the weights the powerlifter can lift.

13.4 Using the normal distribution to approximate the binomial distribution In Book 1, Chapter 13 Probability and statistical distributions you learnt about the binomial distribution. You are now going to use binomials with the normal distribution.

A fair coin has the same probability of landing on heads as it does tails. If you want to know the probability of getting exactly 9 heads out of 15 flips you could use the binomial distribution.

› › ›

The mean is μ = np = 15 × 0.5 = 7.5.

The variance is σ2 = np(1 − p) = 15 × 0.5 × 0.5 = 3.75.

The standard deviation is σ = √3.75 = 1.9364.

To find the exact probability of flipping 9 heads out of 15 you would use X ∼B(n, p) on your calculator or 15C9 × 0.59 × 0.56 = 0.152 74 So P(exactly 9 heads) = 0.152 74 Remembering to find the z-score: z= =

Value − Mean Standard deviation x−µ . σ

− 3.75) = 2.712 standard The z-score for exactly 9 heads is (91.9364 deviations above the mean of the distribution.

21

13 Statistical distributions

You already know the probability of any one specific point is 0, as the normal distribution is a continuous distribution. However, the binomial distribution is a discrete probability distribution, so you need to use something called a continuity correction.

To find P(exactly 9 heads) you need you to consider any value from 8.5 to 9.5. Once you are considering a range, you will have an interval, so you can work out the area under a normal curve from 8.5 to 9.5.

The area of the interval is 0.151 94, which is the approximation of the binomial probability. For these parameters, the approximation is very accurate.

Using the normal distribution as an approximation for the binomial You may use the normal distribution as an approximation for the binomial B(n, p) (where n is the number of trials each having probability p of success) when:

›› n is large ›› p is close to 0.5.

This ensures that the distribution is reasonably symmetrical and not skewed at either end.

The parameters for the normal distribution are then:

›› mean = μ = np ›› variance = σ2 = np(1 − p) ›› standard deviation = np(1 – p).

If np ⩾ 5 and n(1 − p) ⩾ 5 then the binomial random variable x is approximately normally distributed, with mean μ = np and standard deviation σ = np(1 – p).

Example 9 In two surveys, respondents answer ‘yes’ or ‘no’. In each case, decide whether you can use the normal distribution to approximate x, the number of people who answer ‘yes’. If you can, find the mean and standard deviation. If you cannot, explain why.

22

Technology Using the normal cumulative probability on your calculator, set the lower bound to 8.5 and the upper bound to 9.5, the mean is 7.5 and the standard deviation 1.9364.

Using the normal distribution to approximate the binomial distribution

13.4

a 34 per cent of people in Lincoln say that they are likely to make a New Year’s resolution. You randomly select 15 people in Lincoln and ask each if they are likely to make a New Year’s resolution. b 6 per cent of people in Lincoln who made a New Year’s resolution resolved to exercise more. You randomly select 65 people in Lincoln who made a resolution and ask each if they have resolved to exercise more.

Solution a In this binomial experiment, n = 15, p = 0.34, and (1 − p) = 0.66. So, np = (15)(0.34) = 5.1 and n(1 − p) = (15) (0.66) = 9.9.

Because np and n(1 − p) are greater than 5, you can use the normal distribution with μ = 5.1 and σ = √np(1 − p) = 1.83 b In this binomial experiment, n = 65, p = 0.06, and (1 − p) = 0.94. So, np = (65)(0.06) = 3.9 and n(1 − p) = (65)(0.94) = 61.1. You should notice here that np < 5, this means you cannot use the normal distribution to approximate the distribution of x.

Example 10 Use a continuity correction to convert each of the following binomial intervals to a normal distribution interval. a The probability of getting between 170 and 290 successes. b The probability of at least 58 successes. c The probability of getting less than 23 successes.

Solution a The discrete midpoint values are 170, 171 up to 290 so for the continuous normal distribution is 169.5 < x < 290.5. b The discrete midpoint values are 58, 59, 60 and so on.

Therefore for the continuous normal distribution is x > 57.5. c The discrete midpoint values are up to including 20, 21, 22 so for the continuous normal distribution is x < 22.5

Example 11 You are told that 38 per cent of people in the UK admit that they look at other people’s mobile phones. You randomly select 200 people in the UK and ask each if they look at other people’s mobile phones. If the statement is true, what is the probability that at least 70 will say yes?

23

13 StatiStical diStribUtionS

Solution Because np = 200 × 0.38 = 76 and n(1 − p) = 200 × 0.62 = 124, the binomial variable x is approximately normally distributed with μ = np = 76

Technology

σ = √(200 × 0.38 × 0.62) = 6.86

This can also be done on your calculator. Set the lower bound to −0.9475 the upper bound to a large number, the mean to 0 and the standard deviation to 1.

Using a continuity correction, you can rewrite the discrete probability P(x > 70) as the continuous probability function P(x > 69.5). The graph shows a normal curve with μ = 76 and σ = 6.86 and a shaded area to the right of 69.5.

Technology x > 69.5

μ = 76

The z-score that corresponds to 69.5 is z = (69.5 − 76) = −0.9475.

6.86 So, the probability that at least 70 will say yes is P(x > 69.5) = P(z > −0.9475) = 1 − P(z < −0.9475) = 1 − 0.171 69 = 0.8283

Your calculator may also let you change the mean and standard deviation, in which case you can check your calculation in one step using the lower bound as 69.5, upper bound as a large number, mean as 76 and standard deviation as 6.86.

Example 12 A survey reports that 95% of teenagers eat a burger each week.

You randomly select 200 teenagers and ask each whether they have eaten a burger this week. What is the probability that exactly 194 will say yes?

Solution Because np = 200 × 0.95 = 190 and n(1 − p) = 200 × 0.05 = 10, the binomial variable x is approximately normally distributed with μ = np = 190 σ = √(200 × 0.95 × 0.05) = 3.0822 Using a continuity correction, you can rewrite the discrete probability P(x = 194) as the continuous probability function P(193.5 < x < 194.5).

The first z-score that corresponds to 193.5 and 194.5 would be z = 193.5 − 190 = 1.135 55 3.0822 The second z-score would be z = 194.5 − 190 = 1.459 996 3.0822 24

Technology This can also be done on your calculator. Set the lower bound to 1.135 55, the upper bound to a 1.459 996, the mean to 0 and the standard deviation to 1.

Using the normal distribution to approximate the binomial distribution

So, the probability that exactly 194 teenagers will say they have eaten a burger is P(193.5 < x < 194.5) = P(1.135 55< z < 1.459 996) = P(x < 1.459 996) â&#x2C6;&#x2019; P(z < 1.135 55) = 0.9279 â&#x2C6;&#x2019; 0.871 93 = 0.0560 There is a probability of about 0.06 that exactly 194 of teenagers will say they have eaten a burger this week.

Technology Your calculator may also let you change the mean and standard deviation, in which case you can check your calculation in one step using the lower bound to 193.5, the upper bound to 194.5, mean as 190 and standard deviation 3.0822.

exercise 13.4A PS

13.4

page xx

1 You decide to use the normal distribution to approximate the binomial distribution. You want to know the probability of getting exactly 16 tails out of 20 coin flips.

a Calculate the mean number of tails in 20 coin flips. b Calculate the variance and standard deviation. c Calculate the probability using the normal distribution of 16 tails out of 20 coin flips.

2

IQ scores are given as an integer value, X. Tests are designed so that the distribution of scores has a mean of 95 and a standard deviation of 11.

Determine the probabilities:

a P(X < 100)

b

P(85 < X < 105)

3 The discrete random variable X has a binomial distribution with n = 200 and p = 0.32.

Determine, by using a suitable approximation, the probabilities:

M

4

a P(X < 150)

b

P(X > 30)

c P(X > 145)

d

P(75 < X < 150)

a State the conditions under which a binomial probability model can be well approximated by a normal model.

T is a random variable with the distribution B(15, 0.4).

b Rahman uses the binomial distribution to calculate the probability that T < 5 and gives his answer to 4 significant figures. What answer does he get?

c Cynthia uses a normal distribution to calculate an approximation for the probability that T < 5 and gives her answer to 4 significant figures. What answer does she get?

d Assuming Cynthia has worked everything out correctly, calculate the percentage error in her calculation.

PS CM

5

A popular brand of moisturiser claims that 72% of people using it see noticeable effects after just two weeks.

25

13 StatiStical diStribUtionS

Assuming the manufacturerâ&#x20AC;&#x2122;s claim is correct for the population using the moisturiser, calculate the probability that at least 13 of a random sample of 15 people using the moisturiser can see noticeable effects:

a using the binomial distribution b using the normal approximation to the binomial. c Comment on the agreement or disagreement between your two values. d Would the agreement be better or worse if the proportion had been 85% instead of 72%.

PS

6

A multiple-choice theory test consists of 35 questions, for each of which the candidate is required to tick one of five possible answers. Exactly one answer to each question is correct. A correct answer gains one mark and an incorrect answer gains no marks.

One candidate guesses every answer without looking at the question first. What is the probability that they get a particular answer correct? Calculate the mean and variance of the number of questions they answer correctly.

The panel of examiners want to ensure no more than 1% of guessers pass the examination in this way. Use the normal approximation to the binomial, working to 3 decimal places, to establish the pass mark that meets this requirement.

26

Key points and Exam-style questions

13

SUMMARy oF Key PoInTS

› › › › ›

A random variable X that is normally distributed, with mean μ and variance σ2, is notated X ∼ N(μ, σ2).

The standard normal distribution is defined as Z ∼ N(0, 1) as the mean is 0 and the standard deviation is 1.

To standardise the variable, use z = x − µ σ The normal distribution may be used to approximate discrete distributions but continuity corrections are required.

The binomial distribution B(n, p) may be approximated by N(np, np(p − 1)) provided n is large and p is close to 0.5.

eXAM-STyle QUeSTIonS 13

page xx

1 The mass of sugar in a 1 kg bag may be assumed to have a normal distribution with mean 1005 g and standard deviation 2 g. Find the probability that:

2

a a 1 kg bag will contain less than 1000 g of sugar

[2 marks]

b a 1 kg bag will contain more than 1007 g of sugar

[2 marks]

c a 1 kg bag will contain between 1000 g and 1007 g of sugar.

[3 marks]

The heights of the UK adult female population are normally distributed with mean 164.5 cm and standard deviation 8.75 cm.

a Find the probability that a randomly chosen adult female is taller than 160 cm.

Nancy is a student in Year 7. She is at the 45th percentile for her height.

[2 marks]

b Assuming that Nancy remains at the 45th percentile, estimate her height as an adult.

[2 marks]

3 The weights of people using a lift are normally distributed with mean 72 kg and standard deviation 10 kg. The lift has a maximum allowed load of 320 kg. If 4 persons from this population are in the lift, determine the probability that the maximum load is [3 marks] exceeded.

PS

4

A university laboratory is lit by a large number of fluorescent tube light bulbs with lifetimes modelled by a normal distribution with a mean 1000 hours and standard deviation 110 hours.

The bulbs remain on continuously.

a What proportion of bulbs have lifetimes that exceed 900 hours?

[1 mark]

b What proportion of bulbs have lifetimes that exceed 1200 hours?

[1 mark]

c Given that a bulb has already lasted 900 hours, what is the probability it will last another 100 hours?

[3 marks]

The university replace the bulbs periodically after a fixed interval.

d To the nearest day, how long should this interval be if, on average, 1% of the bulbs are to burn out between successive replacement times?

[3 marks] 27

13 StatiStical diStribUtionS

PS

5

A factory produces blades for wind turbines. The lengths of these blades are normally distributed with 33% of them measuring 212.6 ft or more and 12% of them measuring 211.8 ft or less.

Write down simultaneous equations for the mean and standard deviation of the distribution and solve to find the values. Hence estimate the proportion of blades, which measure 212 ft or more.

The blades are acceptable if they measure between 211.8 ft and 212.8 ft. What percentage is rejected as being outside of the acceptable range?

[6 marks]

PS

6

A machine is used to fill oil into gearboxes with a nominal value of 3.54 litres. Suppose the machine delivers a quantity of oil which is normally distributed with a mean of 3.58 litres and a standard deviation of 0.13 litres.

a Find the probability that a randomly selected gear box contains less than the nominal volume.

[2 marks]

It is required by law that no more than 4% of gearboxes contain less than the nominal volume.

b Find the least value of μ which will comply with the law when σ = 0.13.

[2 marks]

c Find the greatest value of σ which will comply with the law when μ = 3.58 litres.

[3 marks]

7 A preserve manufacturer produces a pack consisting of 8 assorted pots of differing flavours. The actual weight of each pot may be taken to have an independent normal distribution with mean 416 g and standard deviation σ g. Find the value of σ such that 99% of packs weigh above 400 g.

[3 marks]

8

The weight, X g, of beans put in a tin by machine K is normally distributed with a mean of 300 g and a standard deviation of 11 g.

A tin is selected at random.

a Find the probability that this tin contains more than 308 g The weight stated on the beans is w g.

[2 marks]

b Find w such that P(W < w) = 0.02

[4 marks]

The weight, F g, of beans put into a cardboard carton is trialled by machine J and is normally distributed with a mean μ g and a standard deviation s g.

c Given that P(F < 290) = 0.95 and P(F < 295) = 0.97, find the values of μ and s.

9

[6 marks]

A restaurant monitors how long diners spend in the restaurant. The manager suggests that the length of a visit can be modelled by a normal distribution with mean 100 minutes. Only 15% of diners stay for more than 115 minutes. a Find the standard deviation of the normal distribution.

[4 marks]

b Find the probability that a visit lasts less than 40 minutes.

[3 marks]

The restaurant increases it opening times so shift workers can also dine out. It introduces a closing time of 11:00 pm. Alison arrives at the restaurant at 10:00 pm. c Explain whether or not this normal distribution is still a suitable model for the length of her visit.

[2 marks]

10 A computer game takes a mean of 120 hours to complete, with a standard deviation of 13 hours. The completion time is generalised and may be assumed to be normally distributed.

Find the time, t hours, of 1 gamer in 10 playing longer than t.

28

[6 marks]

Sample Chapter This is draft content and subject to change

Edexcel A-level Mathematics Year 2

This Student Book will: • support you through the course with detailed explanations, clear worked examples and plenty of practice on each topic, with full worked solutions available to teachers

• help you take control of your learning with prior knowledge checks to assess readiness and end-of-chapter reviews that test understanding • prepare you for linear assessment with clear links between topics embedded in each chapter and exam-style practice questions that help build synoptic understanding • build your confidence in the key A-level skills of problem solving, modelling, communicating mathematically and working with proofs • demonstrate how technology, such as graphing software and scientific calculators, can aid and deepen the study of A-level Maths with ‘Technology’ boxes throughout • prepare you for further study and working life by setting maths in real-world contexts that emphasise practical applications.

Year 1 and and AS Student Book

Edexcel A-level Mathematics Year 1 and AS Student Book 978-0-00-820495-2

Maths_ALevel_Covers_v02.indd 2

Edexcel A-level Mathematics Year 2 Student Book Helen Ball Kath Hipkiss Michael Kent Chris Pearce

Edexcel A-level Mathematics Helen Ball Kath Hipkiss Michael Kent Chris Pearce

Edexcel A-level Mathematics Year 2 Student Book

Written by our expert authors for the 2017 Edexcel specification, this Student Book covers all of the content needed for A-level Year 2 Mathematics.

It combines comprehensive and supportive explanations with plenty of practice to prepare you for the demands of A-level and beyond.

ISBN 978-0-00-820496-9

9 780008 204969

09/01/2017 18:45