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IB Mathematics: Applications and Interpretation HL · AHL 8: Statistics 2 · AHL 8.13

Statistics 2: exam review (second paper) Exam review: second paper Answers

Section A: short questions

    1. (a) \(y = -2.55x^2 + 30.5x - 38.5\)
    2. (b) \(R^2\) = 0.997
    3. (c) \(R^2\) = 0.289: the quadratic model fits better because its \(R^2\) is larger
    1. (a) \(V = 1170e^{-0.276t}\)
    2. (b) a is the value of the drone, in pounds, when it was bought (the value of V at t = 0)
    3. (c) 447
    4. (d) t = 7.46 years
    1. (a) \(P = 0.755v^{3.03}\)
    2. (b) 8.14
    3. (c) 22300 kW
    4. (d) 30 km/h is far outside the data range (4 km/h to 14 km/h), so this is extrapolation and the prediction is not reliable
    1. (a) \(y = 0.614\sin(0.636t + 0.376) + 0.115\)
    2. (b) 9.88 seconds: the wave pattern repeats every 9.88 seconds
    3. (c) 0.0562 m
    4. (d) 0.0504
    1. (a) 0.230
    2. (b) Model 1: its \(SS_{res}\) (18.31) is smaller
    3. (c) \(R^2\) = 0.998: about 99.8% of the variation in the fare is explained by the linear model

Section B: long questions

    1. (a) \(y = -0.0742x^3 + 1.13x^2 - 3.05x + 11.9\)
    2. (b) \(R^2\) = 0.997
    3. (c) Quadratic \(R^2\) = 0.968; the cubic fits better because its \(R^2\) is larger
    4. (d) -437 million
    5. (e) 24 is outside the data range (0 to 7), so this is extrapolation; the cubic reaches a maximum at x = 8.54 and then falls, and the model gives a negative revenue, which is impossible
    1. (a) 480 thousand users is the maximum number the model allows (the carrying capacity); N gets closer to 480 as t increases but never reaches it
    2. (b) \(C=29e^{2k}\)
    3. (c) \(C=4e^{7k}\); k = 0.396, C = 64.1
    4. (d) a = 48.8
    5. (e) \(SS_{res}\) = 0.63; E = 0.397 < 0.5, so the model will be used
    6. (f) \(R=\dfrac{19200}{1+64.1e^{-0.396t}}\)
    7. (g) 300 scooters per year
    8. (h) t = 5.54, in the year 2025

Section C: HL questions (HL students only)

    1. (a) E = 79.5, Var = 81
    2. (b) E = 27, Var = 61
    3. (c) E = 135, Var = 108; Var(3X) = 324. For T three separate independent observations are added, so the variances add (3Var(X)); for 3X one observation is tripled, so Var(3X) = 9Var(X)
    4. (d) E = 36, Var = 147
    1. (a) k = 60/77
    2. (b) E(X) = 3.12
    3. (c) Var(X) = 1.19
    4. (d) E(sample mean) = 3.12; standard error = 0.155
    5. (e) E = 5, Var = 0.133
    1. (a) mean = 480 g, standard deviation = 9.68 g
    2. (b) mean = 0 g, standard deviation = 3.54 g
    3. (c) \(2X\): mean 48, standard deviation 5; \(X_1+X_2\): mean 48, standard deviation 3.54; \(2X\) varies more
    4. (d) n = 40
    1. (a) \(T=X+Y\) ~ N(150, 41)
    2. (b) 0.217
    3. (c) D = X − Y ~ N(6, 41); 0.826
    4. (d) 0.110
    5. (e) \(\bar W\) is approximately N(9, 0.267) because n = 60 > 30 (central limit theorem)
    1. (a) 0.436
    2. (b) \(\bar X\sim\mathrm{N}(45,7.2)\); probability = 0.136
    3. (c) 0.149
    4. (d) n = 15 is less than 30 and the distribution of the C items is not known to be normal, so the total is not approximately normal
    1. (a) unbiased mean = 6.97; unbiased variance = 0.605
    2. (b) 10.5
    3. (c) n = 8
    4. (d) Remove it: it is extreme, and it was caused by a known error so it is not a genuine waiting time
    5. (e) The question is leading (it suggests the answer), so it does not measure what pupils really think: it has poor validity
    6. (f) Test-retest
    1. (a) 3.179
    2. (b) \(s_x\) = 1.783: the sample is closer to its own mean than to the true mean, so dividing by n underestimates the variance
    3. (c) (11.6, 13.5)
    4. (d) (11.9, 13.7)
    1. (a) (207, 216)
    2. (b) The interval (207, 216) does not contain 220, so there is evidence that the mean is not 220
    3. (c) (197, 200)
    4. (d) A t-interval is used because the population standard deviation is unknown (estimated from a small sample); the volumes are assumed to be normally distributed
    5. (e) Wider: a higher confidence level needs a wider interval to be more sure of containing the mean (the 99% interval is (205, 218), width 12.6 against 8.77)
    6. (f) If many samples were taken and an interval found from each, about 95% of the intervals would contain the true population mean
    1. (a) (1.45, 6.55): the interval does not contain 0, so there is evidence of a change in the mean
    2. (b) mean = 252, margin of error = 9, \(s_{n-1}\) = 12.6
    3. (c) width = 8.05 (against 18): the larger sample gives a narrower interval, about 0.45 of the width

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