Continuous random variables — Paper 7 · A Level Mathematics 9709

7 The time in hours taken for clothes to dry can be modelled by the continuous random variable with probability density function given by k √t 1 ≤t ≤4, f(t) = 0 otherwise, where k is a constant. (i) Show that k = 14.3 [3] (ii) Find the mean time taken for clothes to dry. [4] (iii) Find the median time taken for clothes to dry. [3] (iv) Find the probability that the time taken for clothes to dry is between the mean time and the median time. [2]

12 marks

Mark scheme: Attempt to solve an equation in m, = 0.5 7 7 M1 m = 2.73 hours Correct answer (aef) A1 [3] .2726 5.0  .2 726 5.1 .2657 5.1  = − M1 Attempt to integrate using their mean and median(iv) ∫ kt dt 7 7 .2657   as limits = 0.0243 A1 [2] Correct answer accept between 0.0241 and 0.0257

This question in 9709/71 Oct/Nov 2008

5 The continuous random variable X has probability density function given by cos x 0 ≤x ≤14π, f(x) = (k0 otherwise, where k is a constant. (i) Show that k √2. [2] = [2] (ii) Find P(X > 0.4). (iii) Find the upper quartile of X. [3] (iv) Find the probability that exactly 3 out of 5 random observations of X have values greater than the upper quartile. [2]

9 marks

Mark scheme: 5 (i) = 1 M1 Equating to 1 and attempt to integrate with k∫ cos x dx 0 limits [k sin x ]π0 / 4 = 1 k sin (π/4) = 1 ⇒ k / 2 = 1 k = 2 AG A1 [2] Correct answer legit obtained (no decimals seen) π / 4 = [k sin x ]π4.0/ 4 M1 Attempt to integrate from 0.4 to π/4 o.e. (ii) ∫ k cos x dx 4.0 = 1 – k sin(0.4) = 0.449 A1 [2] Correct answer Q 3 M1 Equation with integral on one side and 0.75 (iii) ∫ k cos x dx = .075 0 on the other o.e. = 0.75 M1 Attempt to solve their integral for Q3 [k sin x ]Q0 3 k sinQ3 – 0 = 0.75 Q3 = 0.559 A1 [3] Correct answer (iv) 5C3 × (0.25)3 × (0.75)2 M1 Binomial expression involving 5C3, 0.25 and 0.75 = 0.0879 (45/512) A1 [2] Correct answer

This question in 9709/71 Oct/Nov 2009

6 The continuous random variable X has probability density function given by 13x(k −x) 1 ≤x ≤2, f(x) = 0 otherwise. (i) Show that the value of k is 32 . [3] 9 (ii) Find E(X). [2] (iii) Is the median less than or greater than 1.5? Justify your answer numerically. [3]

8 marks

Mark scheme: x 6 (i) ∫ 3 ( k − x ) dx = 1 M1 Equating to 1 and attempting to integrate, 1 ignore limits here. 2  kx 2 3  A1 Correct integration and correct limits.  −x  = 1 6 9   1  4 k 8   k 1  − − − = 1  6 9   6 9  k = 32/9 AG A1 32/9 correctly obtained. [3] 2 2 2 2 x x (ii) E(X) = ∫ 3 ( k − x ) dx M1 attempting to evaluate ∫ 3 ( k − x ) dx 1 1 2  kx 3 4  =  −x  9 12   1 = 1.52 (491/324) A1 correct answer. [2] 5.1 2 3 5.1 x  kx x  −  M1 attempt to integrate using 1 and 1.5 or 1.5 and  (iii) ∫ 3 ( k − x ) dx = 6 9 1   1 2 A1 = 0.477 (or 0.523) A1ft correct reasoning and conclusion, ft their this is < 0.5 so 1.5 is < median [3] integration -------------------------------------------------- --------- -------------------------------------------------------- Alternative method M1 ∫ f ( x ) dx limits 1 to m (or m to 2) and equate to 0.5 attempted. A1 Correct cubic in m in simplified form (no k’s). A1 m = 1.52(2), so median > 1.5, or, 2 trials showing that median > 1.5. 2

This question in 9709/72 Oct/Nov 2009

1 f()t k 0 12 time waiting ()t Fred arrives at random times on a station platform. The times in minutes he has to wait for the next train are modelled by the continuous random variable for which the probability density function f is shown above. (i) State the value of k. [1] (ii) Explain briefly what this graph tells you about the arrival times of trains. [1]

2 marks

Mark scheme: 1 (i) 1/12 B1 Accept 0.0833 [1] (ii) trains arrive every 12 minutes B1 must have ‘every 12 minutes’ [1]

This question in 9709/71 May/June 2010

1 f()t k 0 12 time waiting ()t Fred arrives at random times on a station platform. The times in minutes he has to wait for the next train are modelled by the continuous random variable for which the probability density function f is shown above. (i) State the value of k. [1] (ii) Explain briefly what this graph tells you about the arrival times of trains. [1]

2 marks

Mark scheme: 1 (i) 1/12 B1 Accept 0.0833 [1] (ii) trains arrive every 12 minutes B1 must have ‘every 12 minutes’ [1]

This question in 9709/72 May/June 2010

5 The random variable T denotes the time in seconds for which a firework burns before exploding. The probability density function of T is given by ke0.2t 0 ≤t ≤5, f(t) = 0 otherwise, where k is a constant. 1 (i) Show that k = [3] 5(e −1). (ii) Sketch the probability density function. [2] (iii) 80% of fireworks burn for longer than a certain time before they explode. Find this time. [3]

8 marks

Mark scheme: M1 Equating to 1 and attempting to integrate5 (i) ∫ ke 2.0 t dt = 1 0  k 0.1   k 0  e − e = 1 A1 Correct integrand and limits  2.0   2.0  k (e − 1) = 1 2.0 1 k = AG A1 Correct answer legitimately obtained 5e( − )1 [3] (ii) B1 Correct curve shape 0 5 B1 Correct horizontal lines (need to see a 5) [2] T M1 Equation relating T and 0.2 or 0.8 (iii) ∫ ke 2.0 tdt = 2.0 0 2.0T [5 ke ] − [5 k ] = 2.0 A1 Correct equation (can be in ‘k’) 2.0 T 2.0 e = + 1 = .1344 5 k T = 1.48 (seconds) A1 Correct answer [3] GCE AS/A LEVEL – May/June 2010 9709 72

This question in 9709/72 May/June 2010

4 f()x 1 0.5 0 x 0 1 2 The diagram shows the graph of the probability density function, f, of a random variable X which takes values between 0 and 2 only. (i) Find P(1 < X < 1.5). [2] (ii) Find the median of X. [3] (iii) Find E(X). [2]

7 marks

Mark scheme: x dx M1 Attempt find correct area eg 1 squ + 24 (i) 0.5(0.5 + 0.75)×0.5 or ∫1 1/4 squ = 5/16 or 0.3125 or 0.313 A1 [2] or integral with correct limits any f(x) m x d x M1 Attempt area from 0 to m (or m to 2) 2 (ii) 1/2 m × m/2 or ∫ 0 their f(x) = 1/2 M1 Expression for area = 1/2. Ignore limits m = √2 or 1.41 A1 [3] 2 x (iii) 2 2 d x M1 Attempt ∫ xf( x )dx . Ignore limits ∫ 0 = 4/3 oe A1 [2]

This question in 9709/71 Oct/Nov 2010

4 The random variable X has probability density function given by ke−x 0 ≤x ≤1, f(x) = 0 otherwise. e (i) Show that k = [3] e −1. (ii) Find E(X) in terms of e. [4]

7 marks

Mark scheme: − M1 Int 1, ignore limits 4 (i) ∫0 ke xd x = 1 = 1 [− ke − x ]01 A1 Correct integral & limits, & = 1 (= –ke–1 – (–ke0) ) e − 1 = k × = 1 or k(e – 1) = e e e A1 Correctly obtained, no errors seen k = AG e − 1 [3] GCE AS/A LEVEL – October/November 2011 9709 73 e 1 − M1 Attempt ∫xf(x)dx, ignore limits (ii) e − 1 ∫0 xe xd x e − x 1 1 − x M1* Attempt integration by parts the correct = – ∫0 ( − e )dx ) way round, ignore limits ([x ( − e ) ]0 e − 1 e x 1 − x 1 M1dep* Attempt second integral of the form = ([− xe− – [e ) ±∫e–xdx, ignore limits e − 1 ]0 ]0 e (= (–e–1 – 0 – (e–1 – 1) ) ) e − 1 e 2 e − 2 A1 e = (1 – ) or oe Accept k instead of throughout e − 1 e e − 1 e − 1 [4] except ans

This question in 9709/73 Oct/Nov 2011

4 A random variable X has probability density function given by D k 3 −x 1 ≤x ≤2, f x = 0 otherwise, where k is a constant. (i) Show that k = 2 [3] 3. (ii) Find the median of X. [4]

7 marks

Mark scheme: Attempt ∫f(x) = 1, ignore limits or4 (i) k ∫ (3 − x ) d x = 1 M1 1 k (h1 + h2) = 1 2 2  2  A1 Correct integration & limits or k 3 x −x  = 1 2 k   1 (2 + 1) = 1 2 (k(6 – 2 – (3 – 0.5)) = 1) 3 1 k × 1.5 = 1 or k × = 1 or k = oe 2 5.1 k = 2 AG A1 [3] No errors seen 3 m 2 (ii) 3 ∫ ( 3 − x ) d x = 5.0 oe ∫ from m to 2 M1* Attempt Int f(x) = 0.5, ignore limits oe 1 Or use of area of trapezium  2 m   2  x   3 x −  = 5.0  3 2     1  2  2  dep M1* Sub of correct limits into their integral. 3m −m − 5.2 = 5.0 3 2 Or trapezium using 1 and m/m and 2   Any correct 3-term QE = 0 or (m–3)² =2.5 2 A1 m −m6 + 5.6 = 0 oe  6 ± 36 − 4 × 5.6   m = = .1 42 or .4 58  2   6 − 10 A1 [4] or oe; single correct ans m = 1.42 (3 sf) 2 Total: 7

This question in 9709/72 Oct/Nov 2015

4 The time, X hours, taken by a large number of runners to complete a race is modelled by the probability density function given by t k 0 ≤x ≤a, f x = x + 1 2 0 otherwise, where k and a are constants. a + 1 (i) Show that k = . [3] a … … … … … … … … … … … … … … (ii) State what the constant a represents in this context. [1] … … … … … Three quarters of the runners take half an hour or less to complete the race. (iii) Find the value of a. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) 2 ( 1) 0 d + ∫ a k x x = 1 M1 – ( 1) 0 +     k x a = 1 –k( 1 1) 1 − + a = 1 M1 Attempt subst correct limits into correct integral k × 1 + a a = 1 and k = 1 + a a AG A1 No errors seen 3 Question Answer Marks Guidance 4(ii) Max time allowed by model (for runners to finish) B1 Allow: All runners finish in time a or less or Longest time (taken by any runner) oe 1 4(iii) 1 + a a 2 0.5 1 ( 1) 0 + ∫ x dx = 3 4 M1 Attempt integ f(x) and = 3 4 ; ignore limits oe. Condone missing / incorrect k – 1 + a a 1 ( 1) 0.5 0 +     x = 3 4 – 1 + a a ( 2 3 1 −) = 3 4 M1 Attempt subst correct limits into correct integral. Condone missing / incorrect k a = 0.8 oe A1 3

This question in 9709/72 Oct/Nov 2018