2010 Applied Maths. Advanced Higher Statistics. Finalised Marking Instructions

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1 200 Applied Maths Advanced Higher Statistics Finalised Marking Instructions Scottish Qualifications Authority 200 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from the External Print Team, Centre Services, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s External Print Team, Centre Services, at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 General Marking Principles These principles describe the approach taken when marking Advanced Higher Applied Mathematics papers. For more detailed guidance please refer to the detailed Marking Instructions. The main principle is to give credit for the skills demonstrated and the criteria met. Failure to have a correct method may not preclude a candidate gaining credit for their solution. 2 The answer to one part of a question, even if incorrect, can be accepted as a basis for subsequent dependent parts of the question. 3 The following are not penalised: working subsequent to a correct answer (unless it provides firm evidence that the requirements of the question have not been met) legitimate variation in numerical values / algebraic expressions. 4 Full credit will only be given where the solution contains appropriate working. Where the correct answer might be obtained by inspection or mentally, credit may be given. 5 Sometimes the method to be used in a particular question is explicitly stated; no credit will be given where a candidate obtains the correct answer by an alternative method. 6 Where the method to be used in a particular question is not explicitly stated in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.) In the detailed Marking Instructions which follow, marks are shown alongside the line for which they are awarded. There are two codes used M and E. The code M indicates a method mark, so in question B(a), M means a method mark for the product rule. The code E refers to error, so in question B(b), up to 2 marks can be awarded but mark is lost for each error.

3 200 Statistics: Solutions A. (a) The reference range is µ ± 96σ. This yields (35 4, 37 0) as the required range. (b) Mean = Standard deviation = {NB excludes + 32} = A2. (a) Host opens B You open A You open C Car /2 Goat /2 Host opens C You open A You open B Car /2 Goat /2 You select A You select B Host opens C You open A You open B Car /6 Goat /6 You select C Host opens B You open A You open C Car /6 Goat /6 Completion of tree, Calculation of probabilities, (b) P (S Car) 2 / 6 P (Car S) = =, P (S) = 2 / 2 3 Contestants wishing to win a car should switch. A3. (a) Not every citizen will own a telephone so some citizens will not be represented thus potentially leading to bias. ˆ ˆ ˆ ˆ (b) y = p ( p) y = 2p = 0 when p = 0 5., y = 2 < 0 so the stationary point is a maximum with y max = With n = 000 the maximum margin of error will be i.e. just over 3% ,

4 A4. The observed and expected frequencies are: Left-handed Observed Expected H 0 : p = 0 H : p 0 X 2 = (O i E i ) 2 Right-handed x 2 (57 40)2 ( )2 = E i The critical value of chi-squared (df, % level) is < so the null hypothesis is rejected, at the % level, yielding strong evidence that the proportions of left and right-handed injured nurses are not in the ratio :9. It would be worth investigating whether or not the hypodermic equipment was designed for right-handed users. A5. P(X > Y) = P(X Y > 0) is required. Assuming that X and Y are independent X Y N( 3 0, ), P(X Y > 0) = P(Z > 2) 0 5 A6. SSR = S yy (S xy) 2 = S xx s 2 = SSR n x = 20 Ŷ A 95% confidence interval for E [Y x = 20] is Ŷ ± t 8, s n + (x x)2 S xx ± ( ) ± 3 4 = (30 93, 37 75) An assumption required is that ε i N(0,σ 2 ) in the linear model.

5 A7. (a) Assuming that waiting times are normally distributed: H 0 : µ = 4 and H : µ < 4 t = x µ s = = 2 3., / 7 94 n / 0 (b) Since 2 3 < 833, H 0 is rejected at the 5% level, furnishing evidence of a reduction in the mean waiting time. µ ± 3 σ n = 36 ± = 30 6, 4 4, (c) The sample size is large enough for the Central Limit Theorem to apply. A point plotting below the lower chart limit could signal a reduction in the mean waiting time. 2E A8. (a) Choice of display Labels/ Scales Appropriate comment on location (b) Strength Formulation A A A B A B B B A B B B B B B Rank H 0 : Median A = Median B H : Median A < Median B Rank sum for Formulation A is W = 20 E (W) = 2 n (n + m + ) = = 40 V (W) = 2 nm (n + m + ) = P (W 20 5) = P ( Z 8 6 ) = P (Z 2 39) = = Since < 0 0 we reject H 0 at the % level, indicating that the data provide evidence of greater median tensile strength for Formulation B.

6 A9 (a) F Bin (300, 0 002) (b) P (F 4) P (T 4) where T Poi (0 6) (c) = P (T 3) { e e e e } = e 0 6 [ ] = 86e 0 6 Compensation paid x Probability f (x) (d) E (X) = xf (x) In reality some passengers will travel in family groups so if one member was ill it is unlikely that other members would turn up. [END OF SECTION A]

7 Section B B. (a) f (x) = e 2x tan x f (x) = 2e 2x tan x + e 2x sec 2 x M,2E (b) g(x) = cos2x x 3 g (x) = ( 2 sin 2x) x3 (cos 2x) (3x 2 ) x 6 M,2E 2x sin 2x 3 cos 2x = x 4 The Product Rule correctly applied and executed would gain full marks. B2. The term in a 6 must come from a8. Thus the term is a 2 0 ( ) ( a) 2 (3a) 8, 2 = a 6 = a 6 B3. 3x (x + ) = A 2 (x + ) + B 2 (x + ) 3x = A + B(x + ) M x = 3 = A x = 0 0 = 3 + B B = 3 3x (x + ) = 3 2 (x + ) (x + ) 3x (x + ) dx = 3 2 (x + ) (x + ) dx 3 = + 3 ln (x + ) + c, x + B4. y.dy = 9te3t.dt M, y 2 2 = 3t 3e3t.dt 3e3t.dt M, = 3te 3t e 3t + c (0,2) 2 = 0 + c c = 3 y 2 = 2 (3te 3t e 3t + 3) y = 6te 3t 2e 3t + 6

8 B5. (a) (b) det A = m (m 0) (0 + 2) + (0 m) = m 2 m 2 = (m + ) (m 2) m =, M () () 0 0 (c) i.e. B = ( ( ) ( ) ( ( ( x ) y = z ) 2 x ) ( y = 2 2 z 7 ) 0 2 ) ( ) [END OF SECTION B SOLUTIONS]

2010 Maths. Advanced Higher. Finalised Marking Instructions

2010 Maths. Advanced Higher. Finalised Marking Instructions 00 Maths Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 00 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial