Statistics 2: exam review (second paper) Exam review: second paper One question at a time

A stall owner records the daily profit, £y, when a phone case is sold at £x each. A quadratic model is suggested. [5]

1. (a) Find a quadratic model \(y=ax^2+bx+c\). Give each coefficient to 3 significant figures. [2]

   • M1 for using quadratic regression on the GDC
   • A1 a = -2.55, b = 30.5, c = -38.5

2. (b) Write down the value of \(R^2\) for the quadratic model. [1]

   • A1 0.997

3. (c) Find \(R^2\) for a linear model for the same data. Use the two values of \(R^2\) to say which model fits better. [2]

   • A1 linear \(R^2\) = 0.289
   • R1 quadratic fits better: 0.997 > 0.289

The value, £V, of a drone is recorded t years after it was bought new. An exponential model is suggested. [6]

1. (a) Find the exponential model \(V=ae^{bt}\). Give a and b to 3 significant figures. [2]

   • M1 for using exponential regression on the GDC
   • A1 a = 1170, b = -0.276

2. (b) Write down the meaning of the value of a in this context. [1]

   • A1 a is the value of the drone, in pounds, when it was bought (the value of V at t = 0)

3. (c) Use the model to estimate the value of the drone, in pounds, 3.5 years after it was bought. [1]

   • A1 447

4. (d) Use the model to find the time, in years, at which the value of the drone falls to £150. Give your answer to 3 significant figures. [2]

   • M1 for setting the model equal to 150 and solving on the GDC (or graphing)
   • A1 t = 7.46 years

The engine power, P kW, needed to drive a boat at a steady speed of v km/h is measured for six speeds. A power model is suggested. [6]

1. (a) Find the power model \(P=av^b\). Give a and b to 3 significant figures. [2]

   • M1 for using power regression on the GDC
   • A1 a = 0.755, b = 3.03

2. (b) By what factor does the model say the power is multiplied when the speed is doubled? Give your answer to 3 significant figures. [2]

   • M1 for \(a(2x)^b = 2^b\times ax^b\), so the factor is \(2^b\)
   • A1 8.14 (using the stored value of b)

3. (c) Use the model to predict the power needed at a speed of 30 km/h. [1]

   • A1 22300 kW

4. (d) Comment on the reliability of your prediction. [1]

   • R1 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

The vertical displacement, y metres, of a marker floating on water is measured from the still-water level at eight times, t seconds after the first reading, as waves pass. A sine model is suggested. [6]

1. (a) Find the sine model \(y=a\sin(bt+c)+d\). Use radians, with a and b positive. Give each value to 3 significant figures. [2]

   • M1 for using sine regression on the GDC in radians
   • A1 a = 0.614, b = 0.636, c = 0.376, d = 0.115 (c may differ by a multiple of \(2\pi\))

2. (b) Find the period of the model and say what it means in this context. [2]

   • M1 for \(2\pi\div b\)
   • A1 9.88 seconds: the wave pattern repeats every 9.88 seconds

3. (c) Use the model to predict the displacement at t = 4.5. [1]

   • A1 0.0562 m

4. (d) Find the residual at t = 3. [1]

   • A1 0.0504 (data − model, using the stored model)

A taxi firm records the fare, £F, for five journeys of d km. Two models are suggested: Model 1 is \(F=1.5d+4.2\) and Model 2 is \(F=1.1d+5\). [6]

1. (a) Find \(SS_{res}\) for Model 1. [3]

   • M1 for model values: 7.2, 10.2, 13.2, 16.2, 19.2
   • A1 residuals (data − model): 0.2, -0.1, 0.1, -0.4, 0.1
   • A1 \(SS_{res}\) = 0.230

2. (b) \(SS_{res}\) for Model 2 is 18.31. Which model fits the data better? Give a reason. [1]

   • R1 Model 1 has the smaller \(SS_{res}\)

3. (c) The student also finds that Pearson's product-moment correlation coefficient for the data is \(r=0.999\). Find \(R^2\) and say what it means in this context. [2]

   • A1 \(R^2=r^2\) = 0.998
   • R1 about 99.8% of the variation in the fare is explained by the model

A streaming service's revenue, £y million, is recorded in each of eight quarters (x = 0 to 7). A cubic model is suggested. [10]

1. (a) Find a cubic model \(y=ax^3+bx^2+cx+d\). Give each coefficient to 3 significant figures. [2]

   • M1 for using cubic regression on the GDC
   • A1 a = -0.0742, b = 1.13, c = -3.05, d = 11.9

2. (b) Write down the value of \(R^2\) for the cubic model. [1]

   • A1 0.997

3. (c) Find \(R^2\) for a quadratic model. Which of the two models fits the data better? Give a reason. [2]

   • A1 quadratic \(R^2\) = 0.968
   • R1 cubic fits better: 0.997 > 0.968

4. (d) Use the cubic model to predict the revenue at x = 24. [2]

   • M1 for substituting x = 24 into the stored model
   • A1 -437 million

5. (e) Comment on the reliability of your prediction. [3]

   • R1 24 is outside the data range (0 to 7), so this is extrapolation
   • R1 the cubic has a negative leading coefficient and reaches a maximum at x = 8.54, then falls
   • R1 the model gives a negative revenue, which is impossible

The number of registered users of an e-scooter scheme, N thousand, is recorded at the end of each year. The time t is in years since the start of 2020, so t = 1 is the end of 2020. The number is modelled by \(N=\dfrac{480}{1+Ce^{-kt}}\), where C and k are positive constants. The table gives the data. [16]

1. (a) In context, explain what the value 480 means in the model. [1]

   • A1 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) Use the value of N at t = 2 to show that \(C=29e^{2k}\). [2]

   • M1 for \(16=\dfrac{480}{1+Ce^{-2k}}\), so \(1+Ce^{-2k}=480\div16=30\)
   • A1 \(Ce^{-2k}=29\), so \(C=29e^{2k}\) (AG)

3. (c) Find a second expression for C using the value of N at t = 7. Hence find k and C. [3]

   • M1 for \(C=4e^{7k}\) from \(N=\dfrac{480}{1+Ce^{-7k}}\) with N = 96
   • A1 k = 0.396 (equating the two expressions and solving on the GDC)
   • A1 C = 64.1 (using the stored k)

4. (d) The model's predictions for t = 3 to 6 are shown in the table, with one value a missing. Find a, the value the model gives at t = 5, correct to 1 decimal place. [1]

   • A1 a = 48.8 (using the stored C and k)

5. (e) The city will use the model if E < 0.5, where \(E=\sqrt{SS_{res}/n}\) and n is the number of predictions. Find \(SS_{res}\) for t = 3 to 6, and by finding E show that the model will be used. [3]

   • M1 for residuals (data − model) at t = 3 to 6: -0.3, 0.3, 0.6, -0.3
   • A1 \(SS_{res}\) = 0.63
   • R1 \(E=\sqrt{0.63\div4}\) = 0.397 < 0.5, so the model will be used

6. (f) It is estimated that 20% of registered users ride on any one day, and that one scooter is needed for every 5 daily riders. Use the model to find an expression for the number of scooters, R, needed at time t. [2]

   • M1 for finding the number of daily riders: 0.2 × 1000N = 200N, then dividing by 5 riders per scooter
   • A1 \(R=\dfrac{19200}{1+64.1e^{-0.396t}}\)

7. (g) At the end of 2024 there were 2200 scooters, and at the end of 2026 there were 2800. Find the average number of scooters added per year between the end of 2024 and the end of 2026. [1]

   • A1 \((2800-2200)\div2\) = 300

8. (h) Assume the number of scooters keeps increasing at the rate you found in part (g). Find the value of t, where t ≥ 5, at which the number of scooters first becomes insufficient, and the year in which this happens. [3]

   • M1 for the supply \(S=2200+300(t-5)\) and the demand R
   • A1 t = 5.54 (first intersection with t ≥ 5 on the GDC)
   • A1 t = 5.54 is 5 complete years after the start of 2020, so the year is 2025

A courier company records the time, X minutes, of a delivery run and the time, Y minutes, to load a van. The variable X has mean 45 and variance 36. The variable Y has mean 18 and variance 25. The times for different jobs are independent. [9]

1. (a) The cost, \(\pounds C\), of a delivery run is \(C=1.5X+12\). Find E(C) and Var(C). [2]

   • A1 E = 1.5×45+12 = 79.5
   • A1 Var = 1.5²×36 = 81 (the constant adds nothing)

2. (b) Find the mean and variance of \(D=X-Y\), the amount by which a delivery run takes longer than loading. [2]

   • A1 E = 45−18 = 27
   • A1 Var = 36+25 = 61 (variances add even for a difference)

3. (c) A driver makes three runs with times \(X_1, X_2, X_3\). Find the mean and variance of \(T=X_1+X_2+X_3\). Say why \(\mathrm{Var}(3X)\) is not equal to \(\mathrm{Var}(T)\). [3]

   • A1 E(T) = 3×45 = 135
   • M1 for Var(T) = 3Var(X) = 108
   • R1 Var(3X) = 9Var(X) = 324: T adds three independent observations, but 3X triples one observation

4. (d) A shift has two delivery runs and three loadings. Find the mean and variance of \(K=(X_1+X_2)-(Y_1+Y_2+Y_3)\). [2]

   • M1 for Var(K) = 2Var(X)+3Var(Y) with the coefficients not squared, as the observations are separate
   • A1 E = 36, Var = 147

A machine records the number of faults, X, in a metre of cloth. The table shows its distribution, where \(P(X=x)=\dfrac{k}{x}\) and k is a constant. [9]

1. (a) Find the value of k. [2]

   • M1 for setting the probabilities to sum to 1: k(1/2+1/3+1/4+1/5) = 1
   • A1 k = 60/77

2. (b) Find E(X). [1]

   • A1 E(X) = Σ x×P(x) = 4k = 3.12 (follow through the value of k)

3. (c) Find Var(X). [2]

   • M1 for E(X²) − (E(X))² with E(X²) = 10.91
   • A1 Var(X) = 1.19

4. (d) A random sample of 50 metres is taken and the sample mean number of faults, \(\bar X\), is found. Find \(E(\bar X)\) and the standard error of \(\bar X\). [2]

   • A1 E(X̄) = E(X) = 3.12
   • A1 standard error = √(1.194÷50) = 0.155

5. (e) In another cloth, the number of faulty rolls, B, in a batch of 25 has distribution \(B\sim\mathrm{B}(25,0.2)\). The mean, \(\bar B\), of 30 independent batches is found. Find \(E(\bar B)\) and \(\mathrm{Var}(\bar B)\). [2]

   • M1 for E(B) = 25×0.2 = 5.00 and Var(B) = 25×0.2×0.8 = 4.00
   • A1 E(B̄) = 5.00, Var(B̄) = 4.00÷30 = 0.133

The mass of one tile, X grams, has mean 24 and standard deviation 2.5. The masses of different tiles are independent. A pack holds 15 tiles, and the empty pack has a fixed mass of 120 grams. [10]

1. (a) Find the mean and the standard deviation of the total mass of a full pack. [3]

   • M1 for mean \(15\times24+120\) and variance \(15\times2.5^2=93.75\) (the fixed mass adds nothing to the variance)
   • A1 mean 480 g, standard deviation \(\sqrt{93.75}\) = 9.68 g

2. (b) Two tiles are chosen at random, with masses \(X_1\) and \(X_2\). Find the mean and the standard deviation of \(X_1-X_2\). [2]

   • A1 mean \(24-24=0\)
   • A1 variance \(2.5^2+2.5^2=12.5\) (variances add for a difference), standard deviation 3.54 g

3. (c) Carla takes the mass of one tile and doubles it, \(2X\). Dev takes the masses of two tiles and adds them, \(X_1+X_2\). Find the mean and the standard deviation of each, and say which varies more. [3]

   • A1 \(2X\): mean 48, variance \(2^2\times2.5^2=25\), standard deviation 5
   • A1 \(X_1+X_2\): mean 48, variance \(2.5^2+2.5^2=12.5\), standard deviation 3.54
   • R1 \(2X\) varies more (5 > 3.54): doubling one value is not the same as adding two independent values

4. (d) The mean mass of a random sample of n tiles is \(\bar X\). Find the smallest n for which the standard deviation of \(\bar X\) is less than 0.4 grams. [2]

   • M1 for \(2.5\div\sqrt{n}<0.4\), so \(n>39.06\)
   • A1 n = 40

A swimmer's 100 m breaststroke takes X seconds, where \(X\sim\mathrm{N}(78,4^2)\), and the 100 m backstroke takes Y seconds, where \(Y\sim\mathrm{N}(72,5^2)\). The two times are independent. Let T be the total time for the two swims. [9]

1. (a) State the distribution of \(T=X+Y\). [2]

   • M1 for E = 78+72 = 150 and Var = 4²+5² = 41 (variances add; standard deviations are squared first)
   • A1 \(T=X+Y\) ~ N(150, 41)

2. (b) Find \(P(T<145)\). [1]

   • A1 0.217 (normal cdf on the GDC with mean 150 and standard deviation √41 = 6.403)

3. (c) Find \(P(X>Y)\). [2]

   • A1 D = X − Y ~ N(6, 41) (variances add for a difference)
   • A1 P(D > 0) = 0.826

4. (d) A random sample of 6 values of X is taken, with sample mean \(\bar X\). Find \(P(\bar X>80)\). [2]

   • M1 for \(\bar X\) ~ N(78, 4²/6) = N(78, 2.667)
   • A1 0.110

5. (e) The time, W minutes, the swimmer spends stretching in a week has an unknown distribution, with mean 9 and standard deviation 4. Giving a reason, state the approximate distribution of the mean, \(\bar W\), of 60 weeks. [2]

   • A1 N(9, 0.267) = N(9, 4²/60)
   • R1 the population is not normal but n = 60 > 30, so the central limit theorem applies

A courier company delivers parcels from three depots. The mass of an item from source A is \(X\sim\mathrm{N}(45,6^2)\) minutes and the mass of an item from source B is \(Y\sim\mathrm{N}(92,9^2)\) minutes. All masses are independent. [10]

1. (a) Find the probability that the total mass of two randomly chosen A items is greater than the mass of one randomly chosen B item. [3]

   • M1 for \(W=X_1+X_2-Y\) and the event \(W>0\): mean \(2\times45-92\) = -2
   • A1 variance \(6^2+6^2+9^2\) = 153
   • A1 \(P(W>0)\) = 0.436

2. (b) A batch of 5 A items is taken, with mean mass \(\bar X\). State the distribution of \(\bar X\). The batch is rejected if \(\bar X\) differs from 45 by more than 4. Find the probability that a batch is rejected. [3]

   • A1 \(\bar X\sim\mathrm{N}(45,6^2/5)=\mathrm{N}(45,7.2)\)
   • M1 for \(P(\bar X>49)+P(\bar X<41)\), or twice \(P(\bar X>49)\) by symmetry
   • A1 0.136

3. (c) Items from source C have mean mass 30 and standard deviation 8, with an unknown distribution. A crate holds 36 items from C. Find the approximate probability that the total mass of the crate is greater than 1130. [3]

   • M1 for the CLT: the total is approximately normal with mean \(36\times30=1080\) and variance \(36\times8^2=2304\)
   • M1 for \(P(T>1130)\) on the GDC with standard deviation \(\sqrt{2304}\) = 48.00
   • A1 0.149

4. (d) A smaller crate holds only 15 items from C. Explain why the central limit theorem cannot be used to answer part (c) for this crate. [1]

   • R1 n = 15 < 30 and the distribution of the C items is not known to be normal, so the central limit theorem does not apply

A school canteen records the waiting time, in minutes, of eight randomly chosen pupils: 6.5, 7.2, 5.8, 8.1, 6.9, 7.7, 6.2, 7.4. [9]

1. (a) Find unbiased estimates of the population mean and variance. [2]

   • M1 for using the GDC one-variable statistics, with \(s_{n-1}\) rather than \(s_n\)
   • A1 mean = 6.97 and variance = \((0.7778)^2\) = 0.605

2. (b) A second sample of 12 values has standard deviation \(s_n=3.1\). Find the unbiased estimate of the population variance. [2]

   • M1 for \(s_{n-1}^2 = \dfrac{n}{n-1}s_n^2 = \dfrac{12}{11}\times3.1^2\)
   • A1 10.5

3. (c) A third sample has variance \(s_n^2=4.9\) and unbiased estimate \(s_{n-1}^2=5.6\). Find the sample size n. [2]

   • M1 for \(\dfrac{n}{n-1}=5.6\div4.9\)
   • A1 n = 8

4. (d) A ninth waiting time is recorded as 48 minutes. The recorder finds that the stopwatch was left running when the pupil was called away. Say whether this value should be removed before estimating the mean, giving a reason. [1]

   • R1 Remove it: it is extreme, and it was caused by a known error so it is not a genuine waiting time

5. (e) A pupil survey asks: “Don’t you agree that the canteen food is great value?” Comment on the validity of this question. [1]

   • R1 The question is leading (it suggests the answer), so it does not measure what pupils really think: it has poor validity

6. (f) The school gives the same pupils the same survey twice, two weeks apart, and compares the answers. Name this test of reliability. [1]

   • R1 Test-retest

A technician measures the resistance, in ohms, of 12 components. The sum of the resistances is 150.6 and the sum of the squares of the resistances is 1925. [9]

1. (a) Find an unbiased estimate of the population variance. [2]

   • M1 for \(\dfrac{\sum x^2-(\sum x)^2/n}{n-1}=\dfrac{1925.0-150.6^2/12}{11}\)
   • A1 3.179

2. (b) The calculator gives two standard deviations: \(\sigma_x\) = 1.707 and \(s_x\) = 1.783. Say which of these is the better estimate of the population standard deviation, and explain why dividing by n gives an estimate that is too small. [2]

   • A1 \(s_x\) = 1.783 (this is \(s_{n-1}\))
   • R1 the deviations are measured from the sample mean, which is closer to the data than the true mean, so \(\sum(x-\bar x)^2\) is too small and dividing by n underestimates; dividing by n − 1 corrects this

3. (c) Find a 90% confidence interval for the population mean resistance of a component in ohms. Use the sample mean and the value of \(s_x\) from part (b), and give the limits to 3 significant figures. [2]

   • M1 for the t-interval on the GDC with \(\bar x\) = 12.55, \(s_{n-1}\) = 1.783, n = 12, C = 0.90
   • A1 (11.6, 13.5)

4. (d) A second sample of 21 components has mean 12.8 and \(s_{n-1}\) = 1.9. Without using the interval function of your calculator, find a 95% confidence interval for its population mean. The critical value of t with 20 degrees of freedom is 2.086. Give the limits to 3 significant figures. [3]

   • M1 for \(12.8\pm2.086\times\dfrac{1.9}{\sqrt{21}}\)
   • A1 margin of error = 0.865
   • A1 (11.9, 13.7)

A random sample of 10 bottles from one filling machine is measured. The volumes, in ml, are 212, 205, 219, 208, 215, 203, 221, 210, 207, 217. The volumes are assumed to be normally distributed. The label says that the mean volume is 220 ml. [9]

1. (a) Find a 95% confidence interval for the population mean. Give the limits to 3 significant figures. [2]

   • M1 for using the t-interval on the GDC (raw data, 95%)
   • A1 (207, 216)

2. (b) Say whether your interval supports the label, giving a reason. [1]

   • R1 The interval (207, 216) does not contain 220, so there is evidence that the mean is not 220

3. (c) A second machine fills bottles with known population standard deviation 6.5 ml. A random sample of 40 bottles has mean volume 198.4 ml. Find a 90% confidence interval for the mean volume of this machine. [2]

   • M1 for a z-interval on the GDC with σ = 6.5, ̅x = 198.4, n = 40, C = 0.90
   • A1 (197, 200)

4. (d) For the first machine, explain why a t-interval is used rather than a z-interval, and state an assumption needed. [2]

   • R1 the population standard deviation is unknown, so it is estimated by \(s_{n-1}\) and the t-interval is used (the sample is small)
   • R1 assume the volumes are normally distributed

5. (e) A 99% confidence interval is found from the same data. State, with a reason, whether it is wider or narrower than the 95% interval. [1]

   • R1 wider: higher confidence needs a wider interval (the 99% interval is (205, 218))

6. (f) Explain what is meant by saying that an interval is a 95% confidence interval. [1]

   • R1 about 95% of intervals found in this way from repeated samples would contain the true mean

A teacher records the scores of 9 students on a vocabulary test before and after a revision course. The differences are assumed to be normally distributed. [8]

1. (a) Find a 95% confidence interval for the mean of the paired differences (after − before), and say what it shows about the effect of the course. Give the limits to 3 significant figures. [3]

   • M1 for finding the differences 4, 2, -1, 8, 6, -1, 7, 5, 6 and using the t-interval on the GDC
   • A1 (1.45, 6.55)
   • R1 the interval does not contain 0, so there is evidence of a change in the mean

2. (b) A 95% confidence interval for the mean reaction time, in milliseconds, of a second group of 10 students, based on a t-distribution, is (243.0, 261.0). Find the sample mean, the margin of error and the unbiased estimate \(s_{n-1}\) of the standard deviation. [3]

   • M1 for mean = (lower + upper) ÷ 2 = 252 and margin of error = (upper − lower) ÷ 2 = 9
   • M1 for margin of error = \(t\times s_{n-1}/\sqrt{n}\) with \(t\) = 2.262 (df = 9)
   • A1 \(s_{n-1}\) = 12.6

3. (c) Another sample of 40 students has the same mean and the same value of \(s_{n-1}\) as in part (b). Find the width of the 95% confidence interval from this sample, and compare it with the width of the interval in part (b). [2]

   • M1 for margin of error \(=2.023\times12.58/\sqrt{40}\) = 4.02, so width = 8.05
   • A1 narrower than 18: a larger sample size makes the interval narrower