Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Foundation c2 exam june 2013 resit 2 sols

402 views

Published on

Published in: Education
  • Be the first to comment

  • Be the first to like this

Foundation c2 exam june 2013 resit 2 sols

  1. 1. Subtract 1 2 mark for reducible fractions or incorrect rounding. Question marks: 4, 7, 6, 15, 5, 11, 16, 9, 9,18 1. (a) Simplify 6x3 + 8y − 7x3 − 14y + 2x3 (2 marks) x3 − 6y 1 mark per term (b) Evaluate y3 x when y = 2 and x = 4. (2 marks) 23 4 = 8 4 = 2 2 marks soln, 1 mark working but incorrect soln 2. (a) What is the equation of the line passing through the point (4, 3) with gradient 1 2 ? (3 marks) 3 = 1 2 × 4 + c (1 mark) c = 1 (1 mark) y = 1 2 x + 1 (1 mark) (b) Solve the simultaneous equations (4 marks) x −2y = −2 x −y = −4 y = −2 (2 marks) x = −6 (2 marks) 3. A farmer keeps track of how many eggs his hens lay. 2 2 4 3 5 4 4 2 4 (a) What is the mode? (1 marks) Mode is 4. (b) What is the median? (2 marks) Ordered data: 2 2 2 3 4 4 4 4 5 9 × 1 2 = 4.5 Median=4 1 mark ordered table 1 mark median (c) What is the interquartile range? (3 marks) 9 × 1 4 = 2.25 Q1 = 2 (1 mark) 9 × 3 4 = 6.75 Q3 = 4 (1 mark) IQR= 4 − 2 = 2 (1 mark) 4. Sloths should have a mean of 16.5 hours sleep a day. A sloth owner wants to see if his sloths are sleeping too much and collects the following data. 17 18 20 19 16 (a) State the null hypothesis and the alternative hypothesis. (2 marks) H0 : µ = 16.5 H1 : µ > 16.5 1 mark for 16.5, 1 mark for >, -1 mark if no µ 1
  2. 2. (b) What is the mean of the test data? (2 marks) µ(X) = 90 5 = 18 2 marks for correct soln, 1 mark if correct working incorrect sol (c) What is the sample standard deviation of the test data? (4 marks) σ2 (X) = 1630−5×182 4 = 2.5 σ(X) = 1.5811 1 mark sum of squares, 1 mark working, 1 mark sample var, 1 mark soln (d) What is the T test statistic? (3 marks) T = 18−16.5 1.5811/ √ 5 = 2.121 1 mark identifying vars, 1 mark working, 1 mark soln (e) The study is taken with a 1% level of significance. What is the critical T value? (2 marks) df = 5 − 1 = 4 (1 mark) 1%, 1 tail, df = 4 C = 3.747 (1 mark) (f) What can we deduce? (2 marks) 2.121 < 3.747 (1 mark) Accept H0 (1 mark) 5. Consider the following triangle. 34 60 h (a) What is the length h to 2 d.p.? (2 mark) sin(60) = h 3 (1 mark) h = 2.60 (1 mark) (b) What is the area of the triangle 1 d.p.? (3 marks) Right triangle: Area = 1 2 × 2.60 × 3 × sin(30) = 1.95 (1 mark) Left triangle:√ 42 − 2.602 = 3.04 Area = 1 2 × 3.04 × .2.60 = 3.95 (1 mark) Area= 3.04 + 3.95 = 7. (1 mark) 6. Solve the following quadratic equations using the method stated. No points will be awarded if another method is used. Leave answers in surd form. (a) 3x2 + 12x + 5 = 0. Solve by using the quadratic formula. (4 marks) a = 3, b = 12, c = 5 x = −12± √ 122−4×3×5 2×3 = −6± √ 21 3 1 mark identifying a,b,c, 1 mark working, 2 marks soln (b) x2 − x − 6 = 0. Solve by factorising. (2 marks) (x − 3)(x + 2) = 0 (1 mark) 2
  3. 3. x = −2, 3 (1 mark) (c) x2 + 10x + 6 = 0. Solve by completing the square. (5 marks) (x + 5)2 − 25 + 6 = 0 (2 marks) (x + 5)2 = 19 (1 mark) x + 5 = √ 19 (1 mark) x = −5 ± √ 19 (1 mark) 7. A researcher wants to show that height and baldness are independent. He collects the following data; Short Average Tall Luxuriant 6 6 8 Thinning 7 8 5 Bald 2 16 2 (a) State the null hypothesis and the alternative hypothesis. (2 marks) H0 : There is no correlation. (1 mark) H1 : there is a correlation. (1 mark) 1 mark if reversed (b) What is the χ2 test statistic? (10 marks) Totals: Short Average Tall Row total Luxuriant 6 6 8 20 Thinning 7 8 5 20 Bald 2 16 2 20 Column total 15 30 15 60 Fit: 5 10 5 5 10 5 5 10 5 Residual: 1 -4 3 2 -2 0 -3 6 -3 χ2 table: 0.2 1.6 1.8 0.8 0.4 0 1.8 3.6 1.8 χ2 = 12 2 marks per table, −1 2 per error, −1 2 per table with rounding errors, 2 marks for soln (c) If we test at a 5% level of significance what is the critical χ2 value? (2 marks) df = (3 − 1)(3 − 1) = 4 (1 mark) C = 9.49 (1 mark) (d) What can we deduce? (2 marks) 9.49 < 12 (1 mark) Reject H0. (1 mark) 3
  4. 4. 8. (a) Show that there is a solution to x6 − 5x + 2 = 0 for some x between 0 and 1. (3 marks) Set f(x) = x6 − 5x + 2 f(0) = 2 (1 mark) f(1) = −2 (1 mark) f(0) > 0, f(1) < 0 so there is a soln to f(x) = 0 for some x between 0 and 1. (1 mark) (b) Use trial and improvement to find a solution to 1 d.p. (6 marks) x f(x) Soln between 0.5 -0.48 0 and 0.5 0.3 0.5 0.3 and 0.5 0.4 0.004 0.4 and 0.5 0.45 -0.24 0.4 and 0.45 Soln is 0.4 to 1 d.p. 3 marks working, 1 mark midpoint, 2 marks soln 9. An airship has speed v = 4t − 100t−2 . (a) What is the speed of the airship when t = 10? (1 mark) v(10) = 4 × 10 − 100 × 10−2 = 39 (b) What is the airships acceleration as a function of time? (3 marks) a = dv dt = 4 + 200t−3 1 mark differentiating, 1 mark per term (c) When t = 10 the distance x = 262. Write the distance the rocket travels as a function of time. (5 marks) x = vdt = 4t − 100t−2 dt = 2t2 + 100t−1 + C 262 = 2 × 102 + 100 × 10−1 + C C = 52 x = 2t2 + 100t−1 + 52 2 marks for integrating, 1 mark for substituting, 1 mark for c, 1 mark for soln 10. (a) Differentiate y = 8x3 − 2x. (2 marks) dy dx = 24x2 − 2 1 mark per term (b) What are the x and y co-ordinates of the stationary points of the graph y = 8x3 − 2x? (4 marks) dy dx = 0 24x2 − 2 = 0 x2 = 1 12 x = ± 1√ 12 = ±0.29 (2 marks) y( 1√ 12 ) = − 4 3 √ 12 = −0.38 (1 mark) y(− 1√ 12 ) = 4 3 √ 12 = 0.38 (1 mark) (c) What are the natures of these stationary points? (3 marks) d2 y dx2 = 48x (1 mark) d2 y dx2 ( 1√ 12 ) = 4 √ 12 > 0 Minimum (1 mark) d2 y dx2 (− 1√ 12 ) = −4 √ 12 < 0 Maximum (1 mark) 4
  5. 5. (d) Sketch the graph y = 8x3 − 2x making sure to label your sketch clearly. (5 marks) −0.5 −0.29 0.29 0.5 −0.38 0.38 x y 1 mark per stat point, 2 marks for intercepts, 1 mark for shape (e) By integrating, find the area under the graph y = 8x3 − 2x between the values x = −0.5 and x = 0.5. (4 marks) 0.5 −0.5 8x3 − 2xdx = [2x4 − x2 ]0.5 −0.5 = (2 × 0.54 − 0.52 ) − (2 × (−0.5)4 − (−0.5)2 ) = 0 2 marks integral, 1 mark subs, 1 mark soln 5
  6. 6. Formulae Let X be a list of data of size n. Mean: µ(X) = n i=1 X[i] n Variance σ2 (X) = n i=1(X[i])2 n − µ2 (X) Z-statistic Z = µ(X) − µ σ/ √ n Sample Variance σ2 (X) = n i=1(X[i])2 − nµ2 (X) n − 1 T-statistic T = µ(X) − µ σ(X)/ √ n Alternative notation Mean ¯x = x n Variance V ar = x2 n − ¯x2 Z-statistic Z = ¯x − A σ/ √ n Sample Variance s2 = x2 − n¯x2 n − 1 T-statistic T = ¯x − A s/ √ n 6
  7. 7. Pythagoras’ Theorem a2 + b2 = c2 tan(A) = opp adj , cos(A) = adj hyp , sin(A) = opp hyp Sine rule a sin(A) = b sin(B) = c sin(C) Cosine rule a2 = b2 + c2 − 2bc cos(A) Area Area = 1/2ab sin(C) Quadratic formula x = −b ± √ b2 − 4ac 2a Equation of a straight line y = mx + c Gradient of a straight line m = y2 − y1 x2 − x1 χ2 Process 1. We refer to the entry in the ith column and the jth row as M(i, j). 2. Calculate the row totals Ri, the column totals Ci and the overall total T. 3. Construct the fit table. The entry in the ith column and jth row is given by: F(i, j) = Ci × Rj T 4. Construct the residual table. The entry in the ith column and jth row is given by: R(i, j) = M(i, j) − F(i, j) 5. Construct the χ2 -table. The entry in the ith column and jth row is given by: χ2 (i, j) = R(i, j)2 F(i, j) 7
  8. 8. Tables Critical Z-values Sig. Lev. 5% Sig. Lev. 1% One-tail Two-tail One-tail Two-tail Probability 0.05 0.025 0.01 0.005 Critical value 1.65 1.96 2.33 2.58 Critical T-values Sig. Lev. 5% Sig. Lev. 1% One-tail Two-tail One-tail Two-tail d.f. 0.05 0.025 0.01 0.005 1 6.314 12.706 31.821 63.656 2 2.920 4.303 6.965 9.925 3 2.353 3.182 4.541 5.841 4 2.132 2.776 3.747 4.604 5 2.015 2.571 3.365 4.032 6 1.943 2.447 3.143 3.707 7 1.895 2.365 2.998 3.499 8 1.860 2.306 2.896 3.355 9 1.833 2.262 2.821 3.250 10 1.812 2.228 2.764 3.169 Critical χ2 value 5% significance 1% Significance df Probability 0.05 Probability 0.01 1 3.84 6.63 2 5.99 9.21 3 7.81 11.3 4 9.49 13.3 5 11.1 15.1 6 12.6 16.8 7 14.1 18.5 8 15.5 20.1 9 16.9 21.7 10 18.3 23.2 11 19.7 24.7 12 21.0 26.2 8

×