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

Statistics 2: exam review Exam review Answers

Section A: short questions

    1. (a) \(h = -3.91t^2 + 21.9t + 1.06\)
    2. (b) \(R^2\) = 1.00
    3. (c) \(R^2\) = 0.150: the quadratic model fits better because its \(R^2\) is larger
    1. (a) \(N = 95.4e^{-0.455t}\)
    2. (b) a is the number of bacteria, in hundreds, in the sample when the disinfectant was added (the value of N at t = 0)
    3. (c) 19.4 hundred
    4. (d) t = 8.49 minutes
    1. (a) \(R = 67.9M^{0.755}\)
    2. (b) 1.69
    3. (c) 7410 W
    4. (d) 500 kg is far outside the data range (2 kg to 100 kg), so this is extrapolation and the prediction is not reliable
    1. (a) \(y = 35.7\sin(0.506t - 1.88) + 58.3\)
    2. (b) 12.4 months: the rainfall pattern repeats every 12.4 months
    3. (c) 93.5 mm
    4. (d) 0.450
    1. (a) 49.0
    2. (b) Model 2: its \(SS_{res}\) (37.44) is smaller
    3. (c) \(R^2\) = 0.994: about 99.4% of the variation in the cost is explained by the linear model

Section B: long questions

    1. (a) \(N = -0.0457x^3 + 0.832x^2 - 2.74x + 19.9\)
    2. (b) \(R^2\) = 0.996
    3. (c) Quadratic \(R^2\) = 0.976; the cubic fits better because its \(R^2\) is larger
    4. (d) -67.8 hundred
    5. (e) 20 is outside the data range (0 to 7), so this is extrapolation; the cubic reaches a maximum at x = 10.2 and then falls, and the model gives a negative number of fish, which is impossible
    1. (a) 2400 thousand connections is the maximum number the model allows (the carrying capacity); N gets closer to 2400 as t increases but never reaches it
    2. (b) \(C=39e^{2k}\)
    3. (c) \(C=7e^{7k}\); k = 0.344, C = 77.5
    4. (d) a = 160.9
    5. (e) \(SS_{res}\) = 1.41; E = 0.594 < 0.8, so the model will be used
    6. (f) \(R=\dfrac{1800}{1+77.5e^{-0.344t}}\)
    7. (g) 30 technicians per year
    8. (h) t = 6.56, in the year 2025

Section C: HL questions (HL students only)

    1. (a) E = 440, Var = 36
    2. (b) E = 570, Var = 208
    3. (c) E = 2460, Var = 432; Var(3X) = 1296. 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 = 890, Var = 480
    1. (a) k = 12/25
    2. (b) E(X) = 1.92
    3. (c) Var(X) = 1.11
    4. (d) E(sample mean) = 1.92; standard error = 0.176
    5. (e) E = 3, Var = 0.0638
    1. (a) mean = 741 g, standard deviation = 10.4 g
    2. (b) mean = 0 g, standard deviation = 4.24 g
    3. (c) \(2X\): mean 116, standard deviation 6; \(X_1+X_2\): mean 116, standard deviation 4.24; \(2X\) varies more
    4. (d) n = 37
    1. (a) \(T=X+Y\) ~ N(67, 25)
    2. (b) 0.0548
    3. (c) D = X − Y ~ N(-3, 25); 0.274
    4. (d) 0.228
    5. (e) \(\bar W\) is approximately N(12, 0.5) because n = 50 > 30 (central limit theorem)
    1. (a) 0.996
    2. (b) \(\bar X\sim\mathrm{N}(1.2,0.00375)\); probability = 0.102
    3. (c) 0.103
    4. (d) n = 20 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 = 4.88; unbiased variance = 0.536
    2. (b) 6.17
    3. (c) n = 10
    4. (d) Remove it: it is extreme, and it was caused by a known error so it is not a genuine call length
    5. (e) The question is leading (it calls the service excellent), so it does not measure what customers really think: it has poor validity
    6. (f) Test-retest
    1. (a) 6.011
    2. (b) \(s_x\) = 2.452: the sample is closer to its own mean than to the true mean, so dividing by n underestimates the variance
    3. (c) (23.9, 26.7)
    4. (d) (24.0, 26.6)
    1. (a) (39.3, 41.6)
    2. (b) The interval (39.3, 41.6) contains 40, so the data are consistent with the claim
    3. (c) (44.8, 46.8)
    4. (d) A t-interval is used because the population standard deviation is unknown (estimated from a small sample); the lifetimes 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 (38.7, 42.2), width 3.45 against 2.40)
    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) (-4.52, -0.732): the interval does not contain 0, so there is evidence of a change in the mean
    2. (b) mean = 65.9, margin of error = 2.7, \(s_{n-1}\) = 4.25
    3. (c) width = 2.47 (against 5.4): the larger sample gives a narrower interval, about 0.46 of the width

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