( 2) Find the median of X, giving your answer to 3 significant figures. (3)

Transcription

1 S Continuous random variables. The lifetime, X, in tens of hours, of a battery has a cumulative distribution function F() given by F( ) = 9 ( + ) <.5 >.5 (a) Find the median of X, giving your answer to significant figures. () (b) Find, in full, the probability density function of the random variable X. () (c) Find P(X.) () A camping lantern runs on batteries, all of which must be working. Four new batteries are put into the lantern. (d) Find the probability that the lantern will still be working after hours. () (Total marks). The random variable y has probability density function f(y) given by ky f ( y ) = ( a y) y otherwise where k and a are positive constants. (a) (i) Eplain why a (ii) Show that k = 9 a ( ) (6) Edecel Internal Review

2 S Continuous random variables Given that E(Y) =.75 (b) show that a = and write down the value of k. (6) For these values of a and k, (c) sketch the probability density function, () (d) write down the mode of Y. () (Total 5 marks). A continuous random variable has cumulative distribution function F() =, +, 6, < > (a) Find P(X < ). () (b) Find the probability density function f() of X. () (c) Write down the name of the distribution of X. () (d) Find the mean and the variance of X. () Edecel Internal Review

3 S Continuous random variables (e) Write down the value of P(X = ). () (Total marks). The continuous random variable has probability density function f() given by f() = where k is a constant. k( + k ) < < otherwise (a) Show that k = 9 () (b) Find the cumulative distribution function F(). (6) (c) Find the mean of X. () (d) Show that the median of X lies between =.6 and =.7 () (Total 7 marks) Edecel Internal Review

4 S Continuous random variables 5. The diagram above shows a sketch of the probability density function f() of the random variable X. The part of the sketch from = to = consists of an isosceles triangle with maimum at (,.5). (a) Write down E(X). () The probability density function f() can be written in the following form. f() = a b a otherwise (b) Find the values of the constants a and b. () (c) Show that σ, the standard deviation of X, is.86 to decimal places. (7) (d) Find the lower quartile of X. () (e) State, giving a reason, whether P( σ < X < + σ) is more or less than.5 () (Total 5 marks) Edecel Internal Review

5 S Continuous random variables 6. The length of a telephone call made to a company is denoted by the continuous random variable T. It is modelled by the probability density function kt f ( t ) = t otherwise (a) Show that the value of k is. 5 () (b) Find P(T > 6). () (c) Calculate an eact value for E(T) and for Var(T). (5) (d) Write down the mode of the distribution of T. () It is suggested that the probability density function, f(t), is not a good model for T. (e) Sketch the graph of a more suitable probability density function for T. () (Total marks) 7. A random variable X has probability density function given by f ( ) = otherwise (a) Show that the cumulative distribution function F() can be written in the form a + b + c, for where a, b and c are constants. () (b) Define fully the cumulative distribution function F(). () Edecel Internal Review 5

6 S Continuous random variables (c) Show that the upper quartile of X is.5 and find the lower quartile. (6) Given that the median of X is.88 (d) describe the skewness of the distribution. Give a reason for your answer. () (Total marks) 8. A random variable X has probability density function given by f ( ) = k otherwise where k is a constant. (a) Show that k = 5 () (b) Calculate the mean of X. () (c) Specify fully the cumulative distribution function F(). (7) (d) Find the median of X. () (e) Comment on the skewness of the distribution of X. () (Total marks) Edecel Internal Review 6

7 S Continuous random variables 9. The continuous random variable Y has cumulative distribution function F(y) given by F(y ) = k( y + y ) y < y y > (a) Show that k =. 8 () (b) Find P(Y >.5). () (c) Specify fully the probability density function f(y). () (Total 7 marks). The continuous random variable X has probability density function f() given by ( ) f ( ) = otherwise (a) Sketch f() for all values of. () (b) Write down the mode of X. () Find (c) E(X), () (d) the median of X. () Edecel Internal Review 7

8 S Continuous random variables (e) Comment on the skewness of this distribution. Give a reason for your answer. () (Total marks). The continuous random variable X has probability density function given by f ( ) 6 = < < < otherwise (a) Sketch the probability density function of X. () (b) Find the mode of X. () (c) Specify fully the cumulative distribution function of X. (7) (d) Using your answer to part (c), find the median of X. () (Total marks). The continuous random variable X is uniformly distributed over the interval α < < β. (a) Write down the probability density function of X, for all. () (b) Given that E(X) = and P(X < ) = 8 5 find the value of α and the value of β. () Edecel Internal Review 8

9 S Continuous random variables A gardener has wire cutters and a piece of wire 5 cm long which has a ring attached at one end. The gardener cuts the wire, at a randomly chosen point, into pieces. The length, in cm, of the piece of wire with the ring on it is represented by the random variable X. Find (c) E(X), () (d) the standard deviation of X, () (e) the probability that the shorter piece of wire is at most cm long. () (Total marks). The continuous random variable X has cumulative distribution function, F() =,, < >. (a) Find P(X >.). () (b) Verify that the median value of X lies between =.59 and =.6. () (c) Find the probability density function f(). () (d) Evaluate E(X). () (e) Find the mode of X. () Edecel Internal Review 9

10 S Continuous random variables (f) Comment on the skewness of X. Justify your answer. () (Total marks). The continuous random variable X has probability density function f ( ) +, = k, otherwise. (a) Show that k =. () (b) Specify fully the cumulative distribution function of X. (5) (c) Calculate E(X). () (d) Find the value of the median. () (e) (f) Write down the mode. Eplain why the distribution is negatively skewed. () () (Total 6 marks) Edecel Internal Review

11 S Continuous random variables 5. A continuous random variable X has probability density function f() where k( ), f ( ) =,, Otherwise where k is a positive constant. (a) Show that k =. () Find (b) E(X), () (c) the cumulative distribution function F(). (6) (d) Show that the median value of X lies between.7 and.75. () (Total 5 marks) 6. A continuous random variable X has probability density function f() where f() = k( ),,, otherwise, where k is a positive constant. (a) Show that k =. () Edecel Internal Review

12 S Continuous random variables Find (b) E(X), () (c) the mode of X, () (d) the median of X. () (e) Comment on the skewness of the distribution. () (f) Sketch f(). () (Total 8 marks) 7. The random variable X has probability density function f() = k(, + 5 ),, otherwise. (a) Show that k = 9. () Find (b) E(X), () (c) the mode of X. () Edecel Internal Review

13 S Continuous random variables (d) the cumulative distribution function F() for all. (5) (e) Evaluate P(X.5). () (f) Deduce the value of the median and comment on the shape of the distribution. () (Total 7 marks) 8. A random variable X has probability density function given by f() =, 8 5,,,, otherwise. (a) Calculate the mean of X. (5) (b) Specify fully the cumulative distribution function F(). (7) (c) Find the median of X. () (d) Comment on the skewness of the distribution of X. () (Total 7 marks) Edecel Internal Review

14 S Continuous random variables 9. The continuous random variable X has probability density function f() = k( 5 ),, otherwise, where k is a constant. (a) Show that k = 56. () (b) Find the cumulative distribution function F() for all values of. () (c) Evaluate E(X). () (d) Find the modal value of X. () (e) Verify that the median value of X lies between. and.5. () (f) Comment on the skewness of X. Justify your answer. () (Total 8 marks). A continuous random variable X has probability density function f() where f() = k( + + ),, otherwise where k is a positive integer. Edecel Internal Review

15 S Continuous random variables (a) Show that k =. () Find (b) E(X), () (c) the cumulative distribution function F(), () (d) P(. < X <.). () (Total 5 marks). The continuous random variable X has cumulative distribution function, F( ) =, ( ), <,, >. (a) Find P(X >.7). (b) Find the probability density function f() of X. (c) Calculate E(X) and show that, to decimal places, Var (X) =.57. () () (6) One measure of skewness is Mean Mode. Standard deviation (d) Evaluate the skewness of the distribution of X. () (Total marks) Edecel Internal Review 5

16 S Continuous random variables. The lifetime, in tens of hours, of a certain delicate electrical component can be modelled by the random variable X with probability density function, ( ) =, f 7 < 6, 6, otherwise. (a) Sketch f() for all values of. () (b) Find the probability that a component lasts at least 5 hours. () A particular device requires two of these components and it will not operate if one or more of the components fail. The device has just been fitted with two new components and the lifetimes of these two components are independent. (c) Find the probability that the device breaks down within the net 5 hours. () (Total 9 marks). The continuous random variable T represents the time in hours that students spend on homework. The cumulative distribution function of T is F( t ) =, k(t, t ) t <, t.5, t >.5. where k is a positive constant. (a) Show that k = 6 7. () (b) Find the proportion of students who spend more than hour on homework. () (c) Find the probability density function f(t) of T. () Edecel Internal Review 6

17 S Continuous random variables (d) Show that E(T ) =.9. (e) Show that F(E(T )) =.75. () () A student is selected at random. Given that the student spent more than the mean amount of time on homework, (f ) find the probability that this student spent more than hour on homework. () (Total marks) Edecel Internal Review 7

18 S Continuous random variables 9 m + m =.5 M m + m.5 =. (a) ( ) ± m = M m =.6,.6 (median =).6 A Note M putting F() =.5 M using correct quadratic formula. If use calc need to get.6 (8...) A cao.6 must reject the other root. If they use Trial and improvement they have to get the correct answer to gain the second M mark. (b) Differentiating d ( + ) 9 d 9 = ( + ) M A 8 ( + ).5 f( ) = 9 Bft otherwise Note M attempt to differentiate. At least one n n A correct differentiation B must have both parts- follow through their F () Condone < (c) P(X.) = F(.) M =.7 7 =, awrt.67 A.67 Note 8 + d d or 9. 5 " their f ( )"d.condone missing d..5 M finding/writing F(.) may use/write ( ). or ( ) A awrt.67 (d) (.667) =.5 awrt.5 or.55 M A Note M (c) If epressions are not given you need to check the calculation Edecel Internal Review 8

19 S Continuous random variables is correct to sf. A awrt.5 or.55 []. (a) (i) f(y) > or f() > M ky ( a y) > or k(a ) > or (a y) > or (a ) > a > A cso Note M for putting f(y) or f() or ky (a y) or k(a ) or (a y) or (a ) or state in words the probability can not be negative o.e. A need one of ky(a y) or k(a ) or (a y) or (a ) and a (ii) k ( y ) = integration M ay y k = answer correct A 9a k 9 = answer = M 9a 8 k = k = 9 a * A cso 6 ( ) Note M attempting to integrate (at least one y n y n + ) (ignore limits) A Correct integration. Limits not needed. And equals not needed. M dependent on the previous M being awarded. Putting equal to and have the correct limits. Limits do not need to be substituted. A cso (b) k ( ay y )dy =. 75 Int f ( ) M k ay y =.75 Correct integration A f ( ) =.75 and limits, Mdep Edecel Internal Review 9

20 S Continuous random variables 8 k 9a = a = 5.75( a ) subst k Mdep 8.5a =.5 + a = * Acso k = 9 B 6 Note M for attempting to find y f (y) dy (at least one yn y n + ) (ignore limits) A correct Integration M y f (y) =.75 and limits, dependent on previous M being awarded M subst in for k. dependent on previous M being awarded A cso B cao /9 (c) Note B correct shape. No straight lines. No need for patios. B completely correct graph. Needs to go through origin and the curve ends at. Special case: If draw full parabola from to get B B Allow full marks if the portion between = and = is dotted and the rest of the curve solid. B B (d) mode = B Note Edecel Internal Review

21 S Continuous random variables B cao [5]. (a) P(X < ) = F() M = = A 6 Note M for attempting to find F() by a correct method eg subst into F() or d 6 + Do NOT award M for d or 6 both of which give the correct answer by using F() as the pdf A / o.e or awrt. Correct answer only with no incorrect working gets M A (b) f() = ( ) df d f ( ) = 6 A otherwise M Note M A B for attempting to differentiate F(). (for attempt it must have no s in) for the first line. Condone < signs for the second line. They must have < and < only. B (c) Continuous Uniform (Rectangular) distribution B Note B must have continuous and uniform or Rectangular Edecel Internal Review

22 S Continuous random variables (d) Mean = B Variance is ( ) = M A Note B for mean = [ ± ( b a) ] M for attempt to use, they must subst in values and not just quote the formula, or using (their f()) (their mean), including limits. Must get when they integrate. A cao. (e) P(X = ) = B Note B cao []. (a) ( + ) d + kd = k M + + k k [ ] ( = ) or k + + ( = ) k A 9k = M dep k = * * given * * 9 cso A Note st M attempting to integrate at least one part (at least one n n+ ) (ignore limits) st A Correct integration. Limits not needed. nd M dependent on the previous M being awarded. Adding the two answers together, putting equal to and have the correct limits. nd A cso Edecel Internal Review

23 S Continuous random variables (b) For >, F() ( t t ) For <, F() F() = 9 + dt M = + 9 A = k dt + M = A ( + 6) 7 < = < > B ft B 6 Note st M Att to integrate ( t t + ) 9 (at least one n n+ ). Ignore limits for method mark st A + allow use of t. 9 Must have used/implied use of limit of. This must be on its own without anything else added nd M attempting to find k +... (must get kt or k) and they must use the correct limits and add ( t t + ) or or use + C and use 9 F() = nd A must be correct st B middle pair followed through from their answers. condone them using < or < incorrectly they do not need to match up nd B end pairs. condone them using < or. They do not need to match up NB if they show no working and just write down the Edecel Internal Review

24 S Continuous random variables distribution. If it is correct they get full marks. If it is incorrect then they cannot get marks for any incorrect part. So if < is correct they can get M A otherwise M A. If < is correct they can get M A otherwise M A. you cannot award Bft if they show no working unless the middle parts are correct. (c) E(X) = ( + ) dt + d 9 M = A 9 = or.6 or awrt. A Note st M attempting to use integral of f() on one part st A Correct Integration for both parts added together. Ignore limits. nd A cao or awrt. (d) F(m) =.5 M F(.6) = ( ) = awrt. 8 7 M F(.7) = ( ) awrt = A Hence median lies between.6 and.7 A da Note st M for using F(X) =.5. This may be implied by subst into F(X) and comparing answers with.5. nd M for substituting both.6 and.7 into their F(X).5 or their F(X) st A awrt.8 and.5 if using their F(X) and awrt. and. or if using their F(X).5 Other values possible. You may need to check their values for their correct equation NB these last two marks are B B on epen but mark as M A nd A for conclusion but only award if it follows from their numbers. Dependent on previous A mark being awarded SC using calculators M for sign of a suitable equation M A for awrt.66 provided equation is correct Edecel Internal Review

25 S Continuous random variables A correct comment [7] 5. (a) E(X) = (by symmetry) B Note B cao (b) <, gradient = = and equation is = y = so a B b passes through (, ) so b = B Note B for value of a. B for value of b (c) E( X ) ( ) d ( ) d + = MM = = + = or MA 6 Var(X) = ( X ) E( X ) [ ] =, E = so σ = =.86 (*) M Acso 7 ( ) A Note st M for attempt at a using their a. For attempt they need. Ignore limits. nd M for attempt at b a use their a and b. For attempt need to have either or. Ignore limits st A correct integration for both parts rd M for use of the correct limits on each part nd A for either getting and or awrt.67 somewhere or or awrt.67 th M for use of E(X ) [E(X)] must add Edecel Internal Review 5

26 S Continuous random variables both parts for E (X ) and only have subtracted the mean once. You must see this working rd Aσ = or or better with no incorrect working seen. (d) P( X q) = d =, q q = so q = =. Note M for attempting to find LQ, integral of either part of f() with their a and b =.5 awrt. MAA a Or their F() =.5 i.e. =.5 or a b +a b =.5 with their a and b If they add both parts of their F(), then they will get M. st A for a correct equation/epression using their a (e) σ =.8 so σ, + σ is wider than IQR, therefore greater than.5 M A Note nd A for or awrt. M for a reason based on their quartiles Possible reasons are P( σ <X< + σ) =.698 allow awrt.65.8 < LQ(.) A for correct answer >.5 NB you must check the reason and award the method mark. A correct answer without a correct reason gets M A [5] 6. (a) kt dt = or Area of triangle = M kt 5k = = or.5 k = or linear equation in k M Edecel Internal Review 6

27 S Continuous random variables k = 5 cso A kt (b) kt dt = M 6 6 = A 5 kt (c) E(T) = kt dt = M Var (T) = = 6 A kt kt dt 6 = ; 6 M; Mdep = 5 ( 6 ) 5 = 5 A 5 9 (d) B (e) B [] (a) F( ) = + d = [ ] + MA [ + ] [ + ] = = A (b) F() = < > BBft Edecel Internal Review 7

28 S Continuous random variables (c) F() =.75; or F(.5) = M; 9 9 = = = =.5 =.75 cso A and F() = = M = = quadratic terms = M dep 8 ± = M dep =. A 6 9 (d) Q Q > Q Q M Or mode = and mode < median Or mean = and median < mode Sketch of pdf here or be referred to if in a different part of the question Bo plot with Q, Q, Q values marked on Positive skew A [] 8. (a) d = = oe attempt to integrate both parts M k d k = k k oe both answer correct A + k k = adding two answers and putting = dmdep on previous M 5 k = k = * 5 A M for adding two integrals together =, ignore limits A for correct integration, ignore limits M using correct limits A cso Edecel Internal Review 8

29 S Continuous random variables (b) d = = 6 attempt to integrate f() for one part 6 M /6 A 5 d = = = or. 5 A E(X) = = = = A M attempting to use integral of f() A correct two integrals added with limits A correct integration ignore limits A awrt. (c) F() = tdt (for ) ignore limits for M M = must use limit of A F() = t dt; + tdt 5 (for < ) need limit of and variable upper limit; need limit M; M and = + 5 A < F( ) + < 5 > middle pair Bft ends B 7 Edecel Internal Review 9

30 S Continuous random variables M Att to integrate t (they need to increase the power by ). Ignore limits for method mark A allow use of t. must have used/implied use of limit of. This must be on its own without anything else added M att to integrate t dt and correct limits. 5 M tdt + Att to integrate using limits and. no need to see them put in. they must add this to their t dt. may be given if they add ¼ 5 Alternative method for these last two M marks Mfor att to t dt and putting + C 5 Muse of F() = to find C A + must be correct 5 B middle pair followed through from their answers. condone them using < or incorrectly they do not need to match up B end pairs. condone them using < or. They do not need to match up NB if they show no working and just write down the distribution. If it is correct they get full marks. If it is incorrect then they cannot get marks for any incorrect part. So if < < is correct they can get M A otherwise M A. if < < is correct they can get M AA otherwise M AA. you cannot award Bft if they show no working unless the middle parts are correct. (d) F(m) =.5 either eq M m + =.5 5 eq for their Aft m = 6 or.57 or awrt.57 A M either of their or + =. 5 5 A for their F(X) < < =.5 A cao If they add both their parts together and put =.5 they get M If they work out both parts separately and do not make the answer clear they can get M A A Edecel Internal Review

31 S Continuous random variables (e) negative skew B This depends on the previous B being awarded. One of the following statements which must be compatible with negative skew and their figures. If they use mode then they must have found a value for it Mean < Median Mean < mode Mean < median (< mode) Median < mode Sketch of the pdf db B negative skew only B Dependent on getting the previous B. their reason must follow through from their figures. [] 9. (a) K( + ) = M K = /8 A M putting F() = or F() F() = A cso. Must show substituting y = and the /8 (b) F(.5) = ( ) M 8 =.75 or A 88 M either attempting to find F(.5) may write and use F() F (.5) A awrt.75 Edecel Internal Review

32 S Continuous random variables (c) (y f ( y) = 9 + y) y otherwise M attempting to differentiate. Must see either a y n y n- at least once A for getting 9 (y + y) o.e and y allow < y < MA B B for the otherwise. Allow for y < and for y > Allow them to use any letter [7]. (a).5.5 f().5 5 Ma height of labelled and goes through (, ) shape must be between and and no other lines drawn (accept patios drawn) B B B B B correct shape B the graph must have a maimum of which must be labelled the line must be between and with not other line drawn ecept patios. They can get this mark even if the patio cannot be seen. the line must be straight and the right shape. (b) B B Only accept Edecel Internal Review

33 S Continuous random variables (c) ( )d = MA = A M attempt to find f()d for attempt we need to see n n+. ignore limits Al correct integration ignore limits Al accept or awrt.67 or. 6 m (d) ( )d =. 5 M m [ ] =. 5 m m + =.5 A m m +.5 = ± m = M m =.7 A M using f()d =.5 A m m + =.5 oe M attempting to solve quadratic. Al awrt.7 or + or + oe (e) Negative skew. B mean < median < mode Bdep First B for negative Second B for mean < median< mode. Need all or may eplain using diagram. [] Edecel Internal Review

34 S Continuous random variables. (a) f( ) ais.5 () ()*,,.5 * may be implied by start at y B both patio B must be straight B (b) Mode is = B (c) F() = tdt 6 (for ) ignore limits for M M = must use limit of A 5 F() = tdt; + tdt (for < ) 6 need limit of and variable upper M; M limit; need limit and = A < F( ) < > middle pair Bft ends B 7 (d) f(m) =.5 either eq M =.5 eq for their Aft = 6 =.5 6 or awrt.5 A [] Edecel Internal Review

35 S Continuous random variables,. (a) f() = β α, α < < β, otherwise function including inequality, otherwiseb,b (b) α + β α 5 =, = β α 8 or equivalent B, B α + β = α + 5β = ( β) + 5β = β = attempt to solve eqns M β = 6 α = both A (c) E(X) = 5 + = 75 cm 75 B (d) Standard deviation = (5 ) M =.7... cm 5 or awrt. A (e) P(X < ) + P(X > ) = st or at least one fraction, + or double M, M 6 = or or. or equivalent fraction 5 5 A []. (a) F(.) = (.. ) one minus required M =.87 A (b) F(.6) =.5 F(.59) =.98 both required awrt.5,.9 MA.5 lies between therefore median value lies between.59 and.6. B (c) f() = +,,, otherwise. attempt to differentiate, all correct MA Edecel Internal Review 5

36 S Continuous random variables (d) f ( )d + = d attempt to integrate f() M = + sub in limits M = 7 or.58 or.58 or equivalent fraction A (e) df( ) = 6 + = attempt to differentiate f() and equate to M d = or. 6 or.667 A (f) mean < median < mode, therefore negative skew. Any pair, cao B,B []. (a) + d = k k f ( ) = M Area = + k k = correct integral / correct epression A k = cso A (b) P( X ) = ( + ) M f ( ) variable limit or +C + A correct integral + limit of = + (+ )( or May have k in ) A Edecel Internal Review 6

37 S Continuous random variables, < + F( ) = < Bft; B 5 middle; ends (c) = E( X ) ( + )d M valid attempt f ( ) and = + 6 correct integration A 9 = 7 = 5 = =.7... A awrt.7 (d) F( m ) =.5 + = M putting their F() = = or equiv ± 6. ( 7) = M attempt their term quadratic = ± i.e. = A awrt.8 (e) Mode = B Edecel Internal Review 7

38 S Continuous random variables (f) Mean < median < mode ( negative skew) B Or Mean < median allow numbers in place of words w diagram but line must not cross y ais [6] 5. (a) k ( ) d = f () = M k k = attempt need either or M correct A (9k 9k) ( 8k k) = k = =.75 cso A attempt () (b) E(X) = ( ) d f M correct A 6 =.6875 = =.69 (sf ) awrt.69 A 6 Edecel Internal Review 8

39 S Continuous random variables (c) F() = ( t t)dt f() with variable limit or +C M t t correct integral lower limit of or F() = or F() A = A = ( + ) A F() = ( + ) < < middle, ends Bft, B 6 F() = (d) ( + ) = + = =.75, + > their F() = / =.7, + < root between.7 and.75 M (or F(.7) =.5, F(.75) =.57 median between.7 and.75 M [5] 6. (a) k( ) d = M A f ( ) d =, all correct k = (*) A k(8 ) = cso A k = Edecel Internal Review 9

40 S Continuous random variables (b) E(X) =. ( )d M f ( ) d 5 = (*) 6 = 5 A 6.7 or or or (c) At mode,f () = M Implied = M Attempt to differentiate = or.5 or or A (d) At median, ( t t ) dt = F() = or f ( ) d = M = Attempt to integrate = = ± M Attempt to solve quadratic =.8 A Awrt.8 M [] Edecel Internal Review

41 S Continuous random variables (e) mean (.7) < median (.8) < mode (.5) M any pair negative skew cao A (f) f() lines < and >, labels, and B negative skew between and B [8] 7. (a) k ( + 5 ) d = M use of f ( ) d = 5 k + = A All correct integ. with limits (*) k = (*) 9 A c.s.o. Edecel Internal Review

42 S Continuous random variables (b) E(X) = ( + 5 )d M 9 use of f ( ) d = 9 = Correct integ n with limits cao A A d 5 (c) f ( ) = ( + 5) = ; Mode = d 9 Diff n of f() & = (d) F() = ( + 5 )d 9 use of f ( ) d M, A M = A Integ n with limits & symbol 5 = F() = < < ; > > A B B 5 (e) P(X.5) = F(.5) =.5 M A F(.5) or integral etc (f) Median =.5; Distribution is symmetrical Bcao, Bcao [7] Edecel Internal Review

43 S Continuous random variables 8 (a) E(X) = d + 8 d MM 5 f()d, terms added 5 8 = AA Epressions, limits = =.7 to sf or or 5 5 A 5 awrt.7 (b) F( ) = d = for < MA variable upper limit on f()d, F( ) = + 8 d for 5 their fraction + v.u.l on f()d & terms M = 8 + A 8 = ( + ) 5 A < F() = < B,B 7 ( + ) 5 > middle pair, ends Edecel Internal Review

44 S Continuous random variables (c) F(m) =.5 ( + ) = MAft 5 Their function =.5 m =.75 m =.8 to sf A awrt.8 (d) mean < median B Negative Skew dep B [7] 9. (a) k( 5 ) d = M Limits required 5 k = A 5 Sub in limits and solve to give ****k = 56 **** A Correct solution 5 (b) F() = = f( )d = (5 )d = M Variable upper limit required (5 ) A < F() = (5 ) Ends, middle. B,Bft > (5 )d = =.9 f ( )d,, (c) E() = 5 5 sf ( 7 ) MAA Edecel Internal Review

45 S Continuous random variables (d) f () = 56 (5 ) = Mode =.5 MAA Attempt f (), (5 ) =,.5 (Or Sketch M, = & 5 A, Mode =.5 A) (e) F(.) =.9, F(.5) =.558 M,A Their F, awrt.9 &.558 or.98 & 6.5 F(m) =.5 m lies between. and.5 cso A (f) Mean (.9) < Median (..5) < Mode (.5) B Negative skew B dep [8]. (a) k ( + + )d = (limits needed and = ) M k + + = attempt at integration M A k = (*) A (b) E(X) =.f ( ) d M = ( )d limits needed A = + + integration and substituting limits dep M = A (c) ( )d = [ ] + + M = A < F() = > B B (d) P(. < X <.) = F(.) F(.) M =. A =.657 A [5] Edecel Internal Review 5

46 S Continuous random variables. (a) P(X >.7) = F(.7) =.67 M A d (b) f() = F() = M d = ( ) for A (c) E(X) = ( ) d = 5 5 M A 8 = =.6 5 A Var (X) = ( 5 8 ) d 5 M 6 = A 6 = 5 =.578 A 6 (d) f() = ( ) = M mode = skewness = =.8696 A 8 5 =.87 M A 6 5 [] Edecel Internal Review 6

47 S Continuous random variables f( ) B. (a) 7 7 B Labelled aes and <, > 6, 6, B 7 B (b) P(X 5) = P(X < 5) 5 and or area Δ M 5 = 5 (area of Δ) full method M 5 59 = 8 = 8 A 59 Probability it does not break down is ( ) M 8 59 probability it does break down is ( ) = (awrt).57 A 8 [9]. (a) F(.5) = k( (.5) (.5) ) = M 7 8 k = i.e. [ ] i.e. ( ) 6 k = k = 7 (*) A cso 6 (b) P(T > ) = F(), = 6 7 ( ) = 7 M, A Edecel Internal Review 7

48 S Continuous random variables (c) f(t) = F (t) =, 6 7 (6t t ) M, A i.e. f(t) = (t 7 t t ).5 otherwise Full definition B (d) E(T) =. 5.5 t f ( t) dt = ( t t ) d t t f ( t) M 7 = 7 t 5 t 5 = [( ) ()] A 8 = 5 9 =.9 (*) A cso (e) F(E(T)) = 6 7 (.9.9 ) =.75 evidence seen B P( T > ) part ( b) (f) P(T > T >.9) =, =, =.776 M, M P( T >.9) part ( e) accept awrt.776 A [] Edecel Internal Review 8

49 S Continuous random variables. Students seemed to fare better on the continuous distributions with the material spread over two questions rather than all together as previously. In part (a) a large majority of candidates were able to set F(m) =.5 and formulate a correct equation. However then many candidates were unable to manipulate the initial equation into a more suitable form. However candidates who showed their working were able to earn credit for using a correct method on an incorrect equation. Candidates who solved the quadratic on their calculator showed no such method and therefore lost two marks instead of one for the incorrect answer. A variety of methods were then used successfully to solve the equation namely the formula, completing the square and trial and improvement. Part (b) was well done with only a few candidates neglecting to put otherwise in the full definition. In part(c) the main problem was the result of confusion between discrete and continuous variables. It was not uncommon to see P( X ) = F(.) Many candidates were able to gain the method mark in part (d). Those who didn t put (.667) either repeated their answer to part (c) or multiplied it by, seemingly unconcerned by a probability greater than.. More candidates seemed to score full marks or nearly full marks than usual for this type of question. Some had problems with the concept of proof and some circular arguments were seen in part (b). There were also some problems in manipulating the algebraic fractions In part (a) many candidates used a proof by contradiction approach rather than starting from f(y) >. Some wrongly thought that a probability density function cannot be at any point and some thought that it can t be greater than. More attempted an eplanation in words than a symbolic proof. Part (b) saw many ecellent solutions. There was a lot of detail involved. Yes, the equation to be solved was only linear, but the coefficients were potentially forbidding to those of us who only use a calculator as a last resort. There were many admirable responses, where candidates displayed persistence and ecellent command of detail. The quantity of algebraic working seen varied substantially, from a few lines of genuine, succinct and accurate work, to a few pages of laboured inaccurate solutions. Part (c) was surprisingly badly done. A few candidates were confused by the variable being y rather than the more usual and so reflected their sketch in y =. The minority who took their time over the sketch got it correct while those who just saw the squared term assumed a parabola intersecting the -ais at and. The mode was usually identified from their sketch although, as ever, there were those who gave the y value rather than the value. Edecel Internal Review 9

50 S Continuous random variables. Most of candidates were able to gain the majority of the marks for this question. In part (a) most candidates either correctly substituted into the formula for F() or used F() F( ). A common error was to integrate F(X), which in many cases resulted in a correct answer but gained no marks as the method was incorrect. There were also a number of students who believed the distribution to be discrete and calculated F() accordingly. In part (b) there were a few candidates who integrated F() or used F() rather than differentiating to find f(). Some of those who differentiated correctly then failed to identify the regions in which the values of 6 and were valid. In part (c) a large number of candidates were unable to completely state the name of the distribution with common errors being to omit either the word continuous or the word uniform. In part (d) although most candidates were able to identify or calculate the mean, a few carried out complicated unnecessary calculations, which were usually incorrect. The candidates that used the formula usually achieved a correct solution for the variance. However, f mean often made errors or forgot to subtract the mean those that attempted to use ( ) squared. In part (d) many candidates failed to realise that the probability that X equals a single value in a continuous distribution is always.. Many candidates scored full marks in part (a). However, there was inevitably substantial variation in the style and presentation of their arguments. A small number of scripts were models of clarity and economy, while other candidates were not just long-winded and confusing, but their scripts contained incorrect and contradictory statements (e.g. 6k =, k =, 9k = ) on the way to a correct final solution. It is reassuring that many perfect solutions to part (b) were seen. The most common problem was the part of the distribution dealing with where many candidates simply worked out dt =. Of those who got the correct answer a miture of methods were used. Some use the approach dt + F( ) where ( ) integration: d = + C and ( ) F =. F =, while others chose indefinite Edecel Internal Review 5

51 S Continuous random variables In part (c) correct answers seemed relatively elusive. A significant number of candidates attempted to find E(X) by using f ( ) d for one part only, usually the f() for. There were a small number of candidates who averaged their answers to the two parts: Other candidates multiplied the F() by before integrating. Part (d) was well done by many candidates, even when there had been problems earlier in the question. Most understood what they were trying to do, and often had a correct version of the function to hand. However, some failed to provide an acceptable conclusion: so the median is between the two numbers or F(.6) < m < F(.7) are eamples which did not gain the mark. The most common error was to amalgamate their two functions from part (b). Others used the wrong function, i.e. their function from (b) for. Those candidates who used calculators to solve the cubic equation usually provided the required amount of supporting detail. Edecel Internal Review 5

52 S Continuous random variables 5. A minority of candidates achieved a high rate of success on this question. In part (a) most candidates were able to write E(X) = without difficulty. A variety of methods were seen in part (b). The method of the mark scheme was seen, perhaps f d = only from a minority of candidates. Many candidates preferred to use calculus: ( ). However, the use of calculus requires more subtlety and sensitivity than was available to many of the candidates. Answers of a = ½ and b = seemed to be not uncommon, resulting from the a = b a = incorrect methods: d and ( ) d. There were candidates who obtained the correct answers using calculus, but it often took considerable working, in contrast to the epected method. There were some candidates who obtained full marks to part (c) with solutions that were confident, fluent and accurate. Furthermore, many of these responses were also efficient: four or five lines of working provided a solution that was not just correct but contained all the required details. However, a wide variety of alternative responses were also seen. Some were indeed correct, but inefficient. Other candidates used an incorrect strategy. Some candidates only worked with the domain. Others worked with both domains, but wanted to keep the domains separate, resulting in two separate versions ofvar(x ) = d = : Var( X ) d d and Var( X ) d Some candidates then calculated the average of these two versions of the variance. Many candidates also found E(X) from scratch in this part rather than using the answer they had in part (a). Not only did this waste time, but whilst they often had it correct in part (a) they gained an incorrect value by integration in this part which they then went on to use. It must be noted that where the answer to a question is given, marks cannot be gained by restating this without sufficient working. Some attempts were made to describe the answer as proven even though no real working had been done. A reasonable number of correct solutions to part (d) were seen. Some candidates went so far as to specify fully the cumulative distribution function before using the correct part to find the lower quartile. Even though this etra work was not required, strictly speaking, it did provide these candidates with a ready method for part (e). It would appear that whilst most candidates attempted part (e), their responses consisted of a simple statement, usually greater than.5 together with an irrelevant reason. A tiny minority of candidates responded in the manner intended. A few provided a clear diagram to illustrate this same argument. However, the majority of successful candidates preferred to evaluate the probability. This was not straightforward, ecept for those who had already obtained a full and correct version of the cumulative distribution function. Part (e) seemed to challenge all but the most able candidates. Edecel Internal Review 5

53 S Continuous random variables 6. Part (a), with its answer given, produced fewer problems than similar questions in previous papers. Most candidates were able to obtain the required value of k. Part (b) was generally well done. There were a wide variety of methods used such as finding the area of a trapezium, others found the area of the triangle and subtracted from. Others obtained F(). The most common fault was the use of incorrect limits, 7 was often seen as the lower limit. Part (c) was a good source of marks for a majority of candidates. A few lost marks as a result of not writing their answers as an eact number. However, many provided answers as both eact fractions and as approimated decimals. The most common error was to finde(t ), call it Var(T) and then stop. Part (d) was not popular. Of those who attempted it, there were some long-winded methods involving calculus and ultimately incorrect answers. The most successful candidates did a quick sketch of the p.d.f. to find the mode. There were a few good sketches in part (e) however; there were all sorts of alternatives. Many just gave a sketch of the original p.d.f. 7. This question proved challenging in parts to some candidates but was attempted in full by many, with a high degree of success. In part (a) most candidates were aware that they needed to integrate the given function and did so successfully, including the fractions. Problems generally arose in the use of the correct limits. It was common to see candidates use limits of or and rather than using a variable upper limit. Several candidates chose to use a constant c rather than limits but often did not proceed to use F() = or F() = to find the value of c. A large number of candidates who got the correct answer went on to multiply their epression by 9. In (b) F() was defined well candidates seem to be more aware of the need for the and and there were a limited number who had the wrong ranges for these. The majority of correct answers in (c) were found by solving the quadratic rather than by the easier method of substituting.5 into the equation. Many of those who used the quadratic formula used complicated coefficients. Most went on to correctly find Q There is still a great deal of confusion in the minds of some candidates over skewness with a number writing reasons such as Q <Q <Q. There was a tendency to write wordy eplanations rather than the succinct Q Q >Q Q. This gained the marks but many candidates were unable to epress themselves clearly. There is still confusion between positive and negative skewness with a few candidates doing correct calculations but concluding it was negative. A few candidates calculated the mean or mode and used mean > median > mode. These gained full marks if correctly found but used precious time doing unnecessary calculations. Edecel Internal Review 5

54 S Continuous random variables 8. A minority of candidates achieved a high rate of success on this question. Part (a) was often badly done with lots of fiddling of figures involved in getting to 5.The work was often very poorly organised with lots of crossing out so that candidates did not really know what they had got. Part (b) was often at least half correct. Quite a few candidates stopped at. but many managed to gain the correct answer. Part (c) was completed correctly by a minority of candidates. Many candidates did not put in limits although they subsequently used them. An incorrect answer of + was common and cropped up in a several cases 5 showing an inability to deal with fractions correctly. Part (d) was answered well by the more able but many candidates equated to a half rather than +. 5 In part (e) if a candidate attempted this part they generally got mark and often marks. 9. The majority of candidates were able to attempt this question with a high degree of success (a) Many candidates had a number of attempts at this part before getting a solution. In some cases, responses showed a lack of understanding between the p.d.f. and the c.d.f. This occurred when the candidate differentiated the given function then proceeded to integrate it. The most common error was to interpret the given function as the p.d.f., integrate it and put the answer equal to. A small number of candidates took the value of k = /8 and used it to work backwards. (b) (c) The most common errors were to find F(.5) or integrate the given function. There were many correct solutions with a minority of candidates being unsuccessful. Marks were mainly lost through, having differentiated correctly to find the function, not specifying the p.d.f. fully. A few candidates tried integrating to find the p.d.f.. The majority of candidates attempted this question. (a) Most sketches were clearly labelled with a few omitting the value on the y-ais. Candidates should draw their sketch in the space in the question book, not on graph paper. (b) A few gave the mode as or. (c) Most were able to find E(X) with only occasional errors in using f() = or in substituting the limits. (d) Finding the median proved challenging for a sizeable minority. Although most wrote that F(m) =.5, finding F(m) proved difficult. There were many eemplary solutions but those candidates who struggled got, but then failed to use the limits correctly or made arithmetic mistakes. It was common for those who had no real understanding to put =.5 and solve to get =.5. Edecel Internal Review 5

55 S Continuous random variables (e) In many cases the answers to this part reflected confusion in understanding the concept of skewness. In many cases where responses were incorrect there was little or no evidence of using the results found, or positive skewness was stated but the reason related to negative skewness.. This question has been a good discriminator. The majority of candidates attempted this question with varying degrees of success. In part (a) there were many good sketches with clear labelling but many lost a mark through not marking the patios clearly. In part (b) the most common mistake is to give the value of f() i.e. ½. Part (c) was a problem for a majority of candidates. It was evident that many candidates were not competent in finding the CDF of a function given in two parts. Finding F() for was reasonably well answered, but quite often candidates did not use the limits or simply wrote down the answers without showing any working. Candidates were less successful in finding F() for, with few using limits correctly and many not taking into consideration the answer to the first part. Candidates who used the alternative method, using +c and F() = and F() = were generally more successful in getting the correct F(). Responses to part (d) would seem to reflect a lack of understanding of what the median is. Candidates quoted F() =.5 and the proceeded to put F() for =.5 and solve. It was rare to find evidence of candidates checking which part to use before setting up an equation. Many candidates solved F() =.5 for both parts and then not said which answer was the median. Another common error was adding F() for and F() for and then solving.. The first two parts of this question caused more difficulties for candidates than the later parts. Standard calculations for a uniform distribution are well understood but applying them to a problem caused difficulties for weaker candidates. Stronger candidates had little problem with part (a) but others failed to give a full statement, missing either the values outside the range of the interval or the ranges for the different parts of the density function. In part (b) only better candidates were able to state correctly, and then solve, the two simultaneous equations. In the final part many candidates found P(X < ) =. but then failed to double this. A common alternative solution was to use the interval < < 75. Edecel Internal Review 55

56 S Continuous random variables. There has been a steady improvement over the years in candidates approach to the questions using given distributions. Weaker candidates are still prone to confusion, but many are able to identify and use the formulas for mean, median and mode correctly. In part (a) most candidates attempted to substitute. into the given cumulative distribution function but some did not take their answer from to achieve the correct solution. Many candidates substituted the given values in part (b) correctly but not all eplained fully why this demonstrated that the median lay between them. When a solution is suggested, then care should be taken that an adequate clear eplanation is given. The correct derivative was stated by many candidates in part (c) but some failed to give a full statement of the distribution, missing either the limits or the regions outside the given interval. Integrating f() correctly and using the limits caused problems for weaker candidates in part (d). A small but appreciable number also made the statement that 6/ 9/ = 5/. Not all those who differentiated the probability density function placed the differential equal to to obtain the mode. Those who did so usually attained the correct solution. Nearly all candidates who attempted the final part of the question compared at least of the mean, median and mode. The majority who had calculated these correctly were able to identify the skew as negative.. This question was well answered by a high percentage of students who gained full marks or only dropped up to four marks. The vast majority of candidates attempted all parts of this question. Evidence of its challenging nature to a number of candidates was the large amount of crossings out and untidy working. This said, however, it was clear candidates had been well prepared for a question of this type. Part (a)was done well with very few fudged solutions seen. Most candidates scored full marks. Part (b) was problematic for a number of candidates who simply wrote an incorrect answer without any working, hence losing up to four marks. Others integrated f() but without a variable upper limit or lower limit of. A minority of candidates had difficulty with integration. A significant number of candidates lost many marks on this answer through using k instead of /k in their working for parts (b), (c), and (d). Candidates who lost marks on part (b) often gained marks later for parts (c) and (d) through working from the original function rather than using their answer to part (b). In part (c) most candidates knew how to find the mean, although a few tried to integrate F() rather than f(). In part (d) many candidates knew what to do to find the median with the majority of marks lost because the wrong epression for F(X) was used. A few poor solutions of quadratic solutions were seen but it was good to see many candidates correctly discard the unwanted solution. In part (e) many candidates differentiated to find the mode which was inappropriate in this case. However quite a number drew a good sketch and used this to correctly identify the mode. In part (f) the inequality mean<median<mode was generally known and quoted, often in spite of conflict with their answers to the previous parts! 5. There is an increasing number of candidates who are able to attempt questions of this type successfully. However, many still have difficulty with the calculus required and/or calculating and using the cumulative distribution function. Part (a) was completed successfully by many candidates who showed clearly how the given solution was achieved. Edecel Internal Review 56

57 S Continuous random variables In part (b) only the weaker candidates had problems in calculating the integral required. In part (c) the question was completed entirely successfully by a minority of candidates. The most common error was the failure to use an upper variable limit and a lower limit of for the integration. Those using the equivalence method to find the constant of integration were usually successful. A few candidates failed to show the full distribution function F() omitting and. Some candidates did some quite remarkable lengthy, but incorrect, algebraic work on the cumulative distribution function in part (d). Generally however candidates did manage to substitute but a number forgot to say that F () = ½. 6. Most candidates responded well to this question. However, a smallish number were rather confused; integrating instead of differentiating, and vice versa, confusing mode and median and substituting =.5 in F() instead of solving F() =.5 in part (d). The majority of candidates provided satisfactory solutions to part (a) and earned all four marks, although a very small number attempted to fake the proof. The most common error was the omission of =. There were a large number of perfect solutions to part (b) with concise and accurate working. The overall response to part (c) was again satisfactory, although a sizeable minority used f() or integrated. In part (d) a majority of candidates established the quartic equation. However, this was as far as most of them went. A small proportion went on to solve the disguised quadratic, usually using the formula, although completing the square was seen on a few occasions. A minority attempted numerical techniques to obtain an answer correct to three significant figures. All too many candidates produced this incorrect solution; (8 ) = 8 = 8 or 8 = 8 = 8 or = In part(e) the appropriate comparisons were generally made although this did not always lead to the correct conclusions being made. There were many correct answers, although if the evidence of the responses from parts (e) and part (f) were considered together, it could be construed that some candidates were guessing. The first mark in part (f) was mostly for administration; labels and the horizontal sections. The lines < and > were often omitted and labels were often incorrect. For the second mark, too many candidates drew small, inaccurate sketches that were symmetrical with the mode at = rather than slight negative skew. 7. Many correct solutions to this question were seen, but there were also some poor solutions resulting from untidy working and poor arithmetic when substituting limits. The candidates seemed to know what methods to use but they could not always apply them accurately. Their integration and differentiation techniques were fine but using them in the various parts was at times disappointing. More care and attention to detail was needed. Edecel Internal Review 57