Further Statistics 1 FS1 - Pearson
omndNEW FOR2017ChapterSam1va Disria crbl etees raple11 - 19 PROGRESSIONEdexcel AS and A level Further MathematicsFurther Statistics 1FS1
1ObjectivesA er completing this chapter you should be able to: Find the expected value of a discrete random variable X pages 2–5 Find the expected value of X 2 pages 3–5 Find the variance of a random variable pages 5–7 Use the expected value and variance of a function of X pages 11–14T Solve problems involving random variables pages 7–11AFSample materialDiscrete randomvariablesDREdexcel AS and A level Further MathematicsPrior knowledge check1a P(X 2)b P(X 2)c P(8 X 11)2Discrete random variables are animportant tool in probability. Banksand stockmarket traders use randomvariables to model their risks oninvestments that have an elementof randomness. By calculating theexpected value of their profits, they canbe confident of making money in thelong term.The random variable X B(12, 16 ). Find:Statistics and Mechanics Year 1, Section 6.3The discrete random variable Y hasprobability mass function P(Y y) ky2,y 1, 2, 4, 5, 10 a Find the value of k. Statistics and Mechanicsb Find P(Y is prime). Year 1, Section 6.13Solve simultaneously:3x 2y z 52x y 8x z 3 Pure Year 1, Chapter 31
Chapter 1Discrete random variablesLinksThe probabilities of any discrete randomvariable add up to 1. For a discrete randomvariable, X, you write P(X x) 1. Statistics and Mechanics Year 1, Chapter 6If you take a set of observations from a discrete randomvariable, you can find the mean of those observations.As the number of observations increases, this value willget closer and closer to the expected value of thediscrete random variable. The expected value of the discreteTThe expected value is sometimesreferred to as the mean,mean, and is sometimesdenoted by µµ.AF1The expected value is atheoretical quantity, and gives informationabout the probability distribution of arandom variable.Notationrandom variable X is denoted E(X )and defined as E(X ) xP(X x)ExampleWatch outb 2p 4q 2p 2q so 2q q p 10.80.20.10.4 q0.4 0.10.3Multiply (1) by 2.Subtract bottom line from top line.xP(x x)123456111111666666b The expected value of X is:6E(X) xP(X x)) 61 26 6721 6 2 3.53A discrete random variable X has a probability distribution.xSince the dice is fair, each side is equally likelyto end facing up, so the probability of any face1ending up as the uppermost is 6DRDRaP(X x)12341264315252525a Write down the probability distribution for X 2.b Find E(X 2).Substitute values from the probabilitydistribution into the formula then simplify.a The distribution for X 2 isIf you know the probability distribution of X then you can calculate the expected value. Notice that inExample 1 the expected value is 3.5, but P(X 3.5) 0. The expected value of a random variable doesnot have to be a value that the random variable can actually take. Instead this tells us that in the longrun, we would expect the average of several rolls to get close to 3.5.x1x2P(X 2 x2)234149161264325252525b E(X 2) x2P(X x2)Example46312 1 4 9 16 252525252The random variable X has a probability distributionas shown in the table.a Given that E(X ) 3, write down two equationsinvolving p and q.xp(x)E(X ) xP(X x)If X is a discrete random variable, then X 2 is also a discrete random variable. You can use this rule todetermine the expected value of X 2.Links Any function of a random variable is E(X 2) x2P(X x)also a random variable. Section 1.3Exampleb Use the probability distribution of X to calculate E(XE(X ).Remember that the probabilities must add upto 1. You will o en have to use P(X x) 1when solving problems involving discreterandom variables.From (2).A fair six-sided dice is rolled. The number on the uppermost face is modelled by the randomvariable X.a Write down the probability distribution of X.X.Problem-solvingTRecall that a random variable is a variable whosevalue depends on a random event. The randomvariable is discrete if it can only take certainnumerical values.a p q 0.1 0.3 0.2 1p q 1 0.6p q 0.4(1)(1 0.1) 2p (3 0.3) 4q (5 0.2) 32p 4q 3 (0.1 0.9 1)2p 4q 1(2)AF1.1 Expected value of a discrete random variable123450.1p0.3q0.2120 25 4.8X can take values 1, 2, 3, 4, so X 2 can takevalues 12, 22, 32, 42.Note that because X takes only positivevalues, P(X 2 x 2) P(X x).Watch outE(X 2) is, in general, not equal toIn this example E(X ) 1.92 and1.922 4.8.(E(X ))2.b Find the value of p and the value of q.23
Chapter 1E/P1Aa(1 x)P(X x) {b b1 For each of the following probability distributions write out the distribution of X 2 and calculateboth E(X ) and E(X 2).a xP(X x)24680.30.30.20.2 2 1120.10.40.10.4b xP(X x)E/PWatch outNote that, for example,P(X 2 4) P(X 2) P(X 2).34560.10.10.10.20.40.1Find E(X ) andE(X 2).8 A biased six-sided dice has a 8 chance of landing on any of the numbers 1, 2, 3 or 4.The probabilities of landing on 5 or 6 are unknown. The outcome is modelled as a randomvariable, X. Given that E(X ) 4.1,1a find the probability distribution of X.b Calculate the probability that the dice lands on 6 at least 3 times.By modelling the amount paid out in prize money as adiscrete random variable, determine the maximum valueof P in order for Jorge to make a profit on his game.AFx 2, 3, 6b Work out E(X ) and E(X 2).Challengec State whether or not (E(X ))2 E(E(XX 2).The expected profit from thegame is the cost of playing the gameminus the expected value of theamount paid out in prize money.DRDR2 x x 1, 2, 3, 4P(X x) { 442x 5a Construct a table giving the probability distribution of X.b Calculate E(X ) and E(XE(X 2).1.2 Variance of a discrete random variablec State whether or not (E(X(E(X ))2 E(E(X 2).If you take a set of observations from a discrete random variable, you can find the variance of thoseobservations. As the number of observations increases, this value will get closer and closer to thevariance of the discrete random variable.5 The random variable X has the following probability distribution.xP(X x)123450.1ab0.20.1 The variance of X is usually written as Var(X ) and is defined asVar(X ) E((X E(X ))2)(5 marks)Given that E(X) 2.9 find the value of a and the value of b.6 The random variable X has the following probabilitydistribution.x 2 112P(X x)0.1abcGiven that E(X ) 0.3 and4HintThree fair six-sided dice are rolled. The discrete random variable X isdefined as the largest value of the three values shown. Find E(XE( ).4 The random variable X has a probability function given byE/P(3 marks)9 Jorge has designed a game for his school fete. Students can pay 1 to roll a fair six-sided dice.If they score a 6 they win a prize of 5. If they score a 4 or a 5 they win a smaller prize of P.a Construct tables giving the probability distributions of X and X 2.E/P(5 marks)The dice is rolled 10 times.E3 The random variable X has a probability function1P(X x) x(6 marks)T2TP(X x)1x 2, 1, 0x x5 5Given that E(X ) 1.2, find the value of a and the value of b.2 The score on a biased dice is modelled by a random variable X with probability distributionx7 The discrete random variable X has probability functionAFExerciseDiscrete random variablesE(X 2) 1.9, find a, b and c.(7 marks)HintYou can use the giveninformation to write downsimultaneous equations for a, b and cwhich can be solved using the matrixinverse operation on your calculator. Core Pure Book 1, Section 6.6 Sometimes it is easier to calculate the variance usingthe formula Var(X) E(X2) (E(X ))2The random variable (X E(X ))2 is the squared deviationfrom the expected value of X. It is large when X takes valuesthat are very different to E(X ).NotationThe variance issometimes denoted by σ 2, whereσ is the standard deviation.From the definition you can see that Var(X ) 0 for any random variable X. The larger Var(X ) themore variable X is. In other words, the more likely it is to take values very different to its expectedvalue.5
Chapter 1Discrete random variables3 Given that Y is the score when a single unbiased six-sided dice is rolled, find E(Y ) and Var(Y ).4A fair six-sided dice is rolled. The number on the uppermost face is modelled by the randomvariable X.PCalculate the variance using both formulae and check that you get the same answer.123456x214916263611616166161b E(D)222 6.25 6 2.25 6 0.25 6The expected value ofis6DR493591Var(X ) E(X 2) (E(X )) 2 6 4 12Exercise1B1 The random variable X has a probability distribution given byxP(X x) 101231111155555a Write down E(X ).b Find Var(X ).2 Find the expected value and variance of the random variable X with probability distributiongiven bya xP(X x)b xP(X x)cxP(X x)6t123P(T t)111244b Find the expected value and variance of T.T.91E(X 2) x 2 P(X x) 61 (1 4 36) 6So using the alternative formula6 A fair coin is tossed repeatedly until a head appears or three tosses have been made. The randomvariable T represents the number of tosses of the coin.a Show that the distribution of T isSubstitute values into the formula forvariance.AF351 (6.25 2.25 0.25) 3 12X2c Var(D).1ESo the variance isVar(X ) (x E(X )) 2 P(X x)d the standard deviation.a the distribution of D and show that P(D 3) 86.25 2.25 0.25 0.25 2.25 6.25P(X x)c Var(S )5 Two fair tetrahedral (four-sided) dice are rolled and D is the difference between their scores.Find:This was calculated in the first exampleof the previous section.T(x E(X ))2b E(S )Txa the distribution of SAFWe have that E(X ) 3.5The distributions of X, X 2 and (X E(X))2 are given by4 Two fair cubical dice are rolled and S is the sum of their scores. Find:E(3 marks)(6 marks)7 The random variable X has probability distribution given byx12P(X x)ab3aDRExamplewhere a and b are constants.a Write down E(X ).(2 marks)b Given that Var(XVar(X ) 0.75, find the values of a and b.(5 marks)1.3 Expected value and variance of a function of XIf X is a discrete random variable, and g is a function, then g(X ) is also a discrete random variable.You can calculate the expected value of g(X ) using the formula: E(g(X )) g(x)P(X x)1231116This is a more general version of the formula for E(X 2). For simple functions, such as addition andmultiplication by a constant, you can learn the following rules: 101111 If X is a random variable and a and b are constants, then E(aX b) aE(X ) b 2 1121111343223 If X and Y are random variables, then E(X Y ) E(X ) E(Y ).466You can use a similar rule to simplify variance calculations for some functions of random variables: If X is a random variable and a and b are constants then Var(aX b) a2 Var(X ).7
Chapter 1Example5A discrete random variable X has a probability distribution234P(X x)643252525a Find E(X ) and Var(X ).25a Write down the probability distribution for Y where Y 2X 1.b Find E(Y ).1234y31225562574259325P(Y y)b E(Y ) yP(Y y)When x 1, y 2 1 1 3x 2, y 2 2 1 5etc.121 25 4.84DR612 2 c E(X ) xP(X x) 1 25254348 3 4 1.922525 252 1.92 1 4.84ExampleNotice how the probabilities relating to Xare still being used, for example,P(X 3) P(Y 7).AF46312 3 5 7 9 25252525d Var(X 2)a The distribution of X is0x10121P(X x)4E(X ) 10 by symmetry.2014Var (X ) E(X 2) (E(X ))21 102 501 202 1 102 Var(X ) 02 424b E(S) E(X 10) E(X ) 10 10 10 0()1 E(1 10 5 01 X 5 E(T ) EX ) 5 E(X222( )1Var(T ) 2250Var(X ) 12.54d Their total scores should both be approximatelyzero, but Susan’s scores should be more spreadout than Thomas’s.Find:c Var(3X )d Comment on any likely differences or similarities.c Var(S ) Var(X) 50A random variable X has E(X ) 4 and Var (X ) 3.b E(X 2)Susan and Thomas play a game using two 10p coins. The coins are tossed and Susan records herscore using the random variable S and Thomas uses the random variable TT. After a large numberof tosses they compare their scores.If you know or are given E(X ) you can use theformula to find E(Y ) quickly.6a E(3X )1c Find Var(S ) and Var (T ).TxS X 10 and T 2 X 5b Show that E(S ) E(T ).c Compute E(X ) and verify that E(Y ) 2E(X ) 1.a The distribution for Y isThe random variables S and T are defined as follows:T112Two fair 10p coins are tossed. The random variable X represents the total value of the coins thatland heads up.AFx7DRExampleDiscrete random variablesThe distribution of X is symmetricaround 10. More precisely, X has the samedistribution as 10 (X 10) 20 X.Therefore E(E(X) E(20 X ) 20 E(X ),so E(E(X ) 10.Use the formulae for the expected valueof a sum.Subtracting a constant doesn’t changethe variance, so Var(S ) Var(X ).Both random variables have expected value0, so we would expect both Susan andThomas to have a score of approximately0. The random variable S which representsSusan’s score has higher variance, meaningwe should expect it to vary more.e E(X 2)Examplea E(3X ) 3E(X ) 3 4 12The random variable X has the following distributionb E(X 2) E(X ) 2 4 2 2c Var(3X ) 32 Var(X ) 9 3 27d Var (X 2) Var(X ) 3e E(X 2) Var(X ) (E(X ))2 3 42 1988Rearrange Var(X) E(X2) (E(X))2x0 30 60 90 P(X x)0.40.20.10.3Calculate E(sinX ).9
Chapter 1Discrete random variablesThe distribution of sin X is010.40.2a Write down E(X ).120.1They are paid 200 plus 100 times the score on the dice. The amount paid to each player ismodelled as a discrete random variable Y.0.31E(sin X) sin x P(X x) 0 0.4 0.22Using the general formula for E(g(X )).b Write Y in terms of X.2c Find the expected pay out each time a player makes it around the board. 0.1 1 0.38 3 0.48720ExerciseP7 John runs a pizza parlour that sells pizza in three sizes: small (20 cm diameter), medium (30 cmdiameter) and large (40 cm diameter). Each pizza base is 1 cm thick. John has worked out that3 95on average, customers order a small, medium or large pizza with probabilities , and2010 20respectively. Calculate the expected amount of pizza dough needed per customer.E/P8 Two tetrahedral dice are rolled. The random variable X represents the result of subtracting thesmaller score from the larger.1CxP(X x)12340.10.30.20.4T1 The random variable X has distribution given byb Find E(Y ).AFa Write down the probability distribution for Y where Y 22XX 3.c Calculate E(X ) and verify that E(2XE(2X 3) 2E(2E(XX ) 3.2 The random variable X has distribution given by 2 10.10.10120.20.40.2DRxP(X x)b Calculate E(Y ).3 The random variable X has E(XE(X ) 1 and Var(X ) 2. Find:a E(8X )b E(X 3)c Var(X 3)d Var(3X )e Var(1 2X )f E(X 2)4 The random variable X has E(X ) 3 and E(X 2) 10. Find:b E(3 4X )d Var(X )e Var(3X 2)c E(X2 4X )5 The random variable X has a mean µ and standard deviation σ.Find, in terms of µ and σ :10a E(4X)b E(2X 2)d Var(2X 2)e Var(2X 2)(7 marks)44XX 1The random variables Y and Z are defined as Y 2X and Z .2b Show that E(Y ) E(Z ).(3 marks)c Find Var(Z ).(2 marks)HintChallengeShow that E((X E(X ))2) E(E(XX 2) (E((E(XX ))2.a Write down the probability distribution for Y where Y X 3.a E(2X )a Find E(X ) and Var(X ).T 3AFP(X x)2 3DRsin x6 In a board game, players roll a fair six-sided dice each time they make it around the board.The score on the dice is modelled as a discrete random variable X.Remember that for two randomvariables X and Y we haveE(X Y ) E(X ) E(Y )1.4 Solving problems involving random variablesSuppose we have two random variables X and Y g(X ). If g is one-to-one, and we know the meanand variance of Y, then it is possible to deduce the mean and variance of X.Example9X 150X is a discrete random variable. The discrete random variable Y is defined as Y 50Given that E(Y ) 5.1 and Var(Y ) 2.5, find:a E(X )b Var(X ).c E(2X 2)11
Chapter 1aDiscrete random variablesX 150Y 50X 50Y 150E(X ) E(50Y 150) 50E(Y ) 150 255 150 405b Var(X ) In matrix1 1 1 1( 1 1Rearrange to get an expression for X in terms of Y.form this is0.451 a2 b 0.1( ) (4) c0.6 )By inverting the matrix on a calculator (orby hand) we find values for a, b and c.So, by inverting the matrix we find2 3a1b 63(c ) 6 ( 2 0Use your expression for X in terms of Y.Remember that the ‘ 150’ does not affect thevariance, and that you have to multiply Var(Y ) by502 to get Var(X ).Var(50Y 150)502Var(Y )502 2.562500.4510.2 30.1 0.22 )( 0.6 ) (0.05)So a 0.2, b 0.2 and c 0.05.Use the expression for Y to write everythingin terms of X only.c P(X Y) P(X 3X 1) P(1 2X 0)So P(X Y ) 0.3 0.2 0.25 0.75.1 2X 0 when X 2, 1, 0. SoP(X Y ) P(P(XX 2)2) P(P(X 1) P(X 0).Example 10012P(X x)0.3a0.25bc1 X is a discrete random variable. The random variable Y is defined by Y 44XX 6. Given thatE(Y ) 2 and Var(Y ) 32, find:a E(X )The discrete random variable Y is defined as Y 33XX 1.b Var(X )Given that E(Y ) 2.5 and Var(Y ) 13.95, finda E(X ) and E(X 2)b the values of a, b and cDRc P(X Y ).Y 1a We have X 31 (E(Y 1 Y ) 1) 0.50.5E(X) E ((E(Y3 ) 31 Var(Y 1 Y ) 1.55Var(X) Var (Var(Y3 ) 9SoE(X 2) Var(X ) (E(X ))2 1.55 0.25 1.8.b We havea b c 1 0.3 0.25 0.45E(X ) 2 0.3 1 a 0 0.25 1 b 2 c 0.5So a b 2c 0.5 0.6 0.1E(X2) 4 0.3 1 a 0 0.25 1 b 4 c 1.8Soa b 4c 1.8 1.2 0.64 3X2 X is a discrete random variable. The random variable Y is defined by Y 21 and Var(YVar(Y ) 9, find:Given that E(Y ) 1Rearrange the formula Y 3X 1 toget it in terms of X.a E(X )b Var(X )c E(X 2)Adding a constant does not changevariance, so Var(Y 1) Var(Y ).Var(aX b) a2 Var(X ).P3 The discrete random variable X has probability distribution given byxP(X x)The probabilities must sum to 1.12340.3ab0.2The random variable Y is defined by Y 2X 3. Given that E(Y ) 8, find the values of a and b.We know that E(X ) 0.5 from part a.E/P4 The discrete random variable X has probability distribution given byxP(X x)90 180 270 ab0.3The random variable Y is defined as Y sin X .We know that E(X 2) 1.8 from part a.12c the standard deviation.AF 1DR 2AFx1DTTExerciseThe discrete random variable X has probability distribution given bya Find the range of possible values of E(Y ).(5 marks)b Given that E(Y ) 0.2, write down the values of a and b.(2 marks)13
Chapter 13 A discrete random variable X has the probability distribution shown in the table below.5 The discrete random variable X has probability distribution given byxP(X x) 2 1012abcbax0P(X x)1b55 ba Find the value of b.b Show that E(X ) 1.3.a show that:2a 2b c 110a 4b c 2.4a b 0.4c Find the exact value of Var(X ).d Find the exact value of P(X 1.5).4 The discrete random variable X has a probability functionk(1 x)P(X x) k(x 1){0c Find P(2X 3 Y ).a Show that k 41(1 mark)(2 marks)DRd Find P(X 5 Y ).xP(X x)188b E(E(XX)c E(3XE(3X 1)e E(log(XE(log(X 1))12340.40.20.10.3b E(XE( )3 Xd E( )2e E( X )c Var(X )f E(2 x)7 A discrete random variable is such that each of its values is assumed to be equally likely.Find:a Write the name of the distribution.d Var(X )b Give an example of such a distribution.A discrete random variable X as defined above can take values 0, 1, 2, 3, and 4.Find:2 The discrete random variable X has the probability distribution in the table below.x 2 10123P(X x)0.10.20.3t0.10.1c E(X )Pb P( 1 X , 2)c E(2X 3)d Var(2X 3)d Var(X )e the standard deviation.8 The random variable X has a probability distribution.xFind:a r2a P(3 X 2 10)a Construct a table giving the probability distribution of X.f E(X 3)3
Statistics and Mechanics b Find P(Y is prime). Year 1, Section 6.1 3 Solve simultaneously: 3x 2y z 5 2x y 8 x z 3 Pure Year 1, Chapter 3 AFT Prior knowledge check 01_Chapter_1_001-018.indd 1 19/06/2017 12:16 Sample material Edexcel AS and A level Further Mathematics
Oakland (Live) Tuesday 9 a.m. FS1 AMA Supercross Series Oakland (Replay) SOCCER English League Soccer Friday 2 p.m. FS1 Manchester United at Cambridge United (Live) Saturday 9 a.m. FS1 Middlesbrough at Manchester City (Live) International Wednesday 5 p.m. FS1 United States at Chile (Live) TENNIS Australian Open
Pearson Education LTD. Pearson Education Australia PTY, Limited. Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia, Ltd. Pearson Education Canada, Ltd. Pearson Educación de Mexico, S.A. de C.V. Pearson Education—Japan Pearson Education Malaysia, Pte. Ltd. The Libra
Pearson Education LTD. Pearson Education Australia PTY, Limited. Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia, Ltd. Pearson Education Canada, Ltd. Pearson Educatión de Mexico, S.A. de C.V. Pearson Education—Japan Pearson Education Malaysia, Pte. Ltd. Library of Co
Pearson (UK) 80 Strand, London WC2R 0RL, UK T 44 (0)20 7010 2000 F 44 (0)20 7010 6060 firstname.lastname@pearson.com www.pearson.com Pearson (US) 1330 Avenue of the Americas, New York City, NY 10019, USA T 1 212 641 2400 F 1 212 641 2500 firstname.lastname@pearson-inc.com www.pearson.com Pearson Education One Lake Street, Upper Saddle River,
Pearson BTEC Level 4 HNC The Pearson BTEC Level 4 HNC in Business is a qualification with a minimum of 120 credits of which 60 are mandatory core. The Pearson BTEC Level 4 HNC programme must contain a minimum of 65 credits at level 4. Pearson BTEC Level 5 HND The Pearson BTEC Lev
7 p.m. TNT Chicago Bulls at Toronto Raptors (Live) 9:30 p.m. TNT Brooklyn Nets at Golden State Warriors (Live) BOXING Professional Tuesday 8 p.m. FS1 Golden Boy Promotions: Rene Alvarado vs Rocky Juarez from Fort Bliss Arena in El Paso, Texas (Replay) Thursday 9 p.m. FS1 FOX Fight Nigh
The FortiGate/FortiWiFi-60C series represent a new generation of desktop network security appliances from Fortinet, and include the first Fortinet System-on-a-chip (SoC), the FS1. Integrating FortiASIC acceleration logic together with a RISC-based main processor and other system components, the FS1 SoC simplifies appliance
‘Stars’ can allow a business to be a market leader ‘Problem Child’ products give businesses opportunity to invest ‘Dogs’ should be divested Increased profits can ari se f rom selling different products Newer products can replace thos e at the end of the life cycle A range of pro ducts increases brand awareness Easier to launch new products with larg e existing portfolio 5 Award 1 .
