Statistics 2: exam review Exam review Worked solutions

A rocket is launched from a platform. The table shows its height, h metres, t seconds after launch. A quadratic model is suggested. [5]

1. (a) Find a quadratic model \(h=at^2+bt+c\). Give each coefficient to 3 significant figures. [2]

   • M1 for using quadratic regression on the GDC
   • A1 a = -3.91, b = 21.9, c = 1.06

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

   • A1 1.00

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 \(R^2\) = 0.150
   • R1 quadratic fits better: 1.00 > 0.150

A disinfectant is added to a water sample. The table shows the number of bacteria, N hundred, in the sample t minutes later. An exponential model is suggested. [6]

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

   • M1 for using exponential regression on the GDC
   • A1 a = 95.4, b = -0.455

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

   • A1 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) Use the model to estimate the number of bacteria 3.5 minutes after the disinfectant was added. [1]

   • A1 19.4 hundred

4. (d) Use the model to find the time at which the number of bacteria falls to 2 hundred. Give your answer to 3 significant figures. [2]

   • M1 for setting the model equal to 2 and solving on the GDC (or graphing)
   • A1 t = 8.49 minutes

The metabolic rate, R watts, of six mammals of mass M kg is measured. A power model is suggested. [6]

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

   • M1 for using power regression on the GDC
   • A1 a = 67.9, b = 0.755

2. (b) By what factor does the model say the metabolic rate is multiplied when the mass 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 1.69 (using the stored value of b)

3. (c) Use the model to predict the metabolic rate of a mammal of mass 500 kg. [1]

   • A1 7410 W

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

   • R1 500 kg is far outside the data range (2 kg to 100 kg), so this is extrapolation and the prediction is not reliable

The mean monthly rainfall, y mm, at a weather station is recorded in eight months. Here t = 1 is January, t = 2 is February, and so on. 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 = 35.7, b = 0.506, c = -1.88, d = 58.3 (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 12.4 months: the rainfall pattern repeats every 12.4 months

3. (c) Use the model to predict the rainfall in month 6.5. [1]

   • A1 93.5 mm

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

   • A1 0.450 (data − model, using the stored model)

A catering firm records the cost, £C, of providing food for n guests at five events. Two models are suggested: Model 1 is \(C=2.2n+30\) and Model 2 is \(C=-0.004n^2+2.5n+28\). [6]

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

   • M1 for model values: 52.0, 74.0, 96.0, 118.0, 140.0
   • A1 residuals (data − model): 4.0, -1.0, 4.0, 0.0, 4.0
   • A1 \(SS_{res}\) = 49.0

2. (b) \(SS_{res}\) for Model 2 is 37.44. Which model fits the data better? Give a reason. [1]

   • R1 Model 2 has the smaller \(SS_{res}\)

3. (c) The student also finds that Pearson's product-moment correlation coefficient for the data is \(r=0.997\). Find \(R^2\) and say what it means in this context. [2]

   • A1 \(R^2=r^2\) = 0.994
   • R1 about 99.4% of the variation in the cost is explained by the model

The number of fish, N hundred, in a lake is estimated at the start of each of eight years (x = 0 to 7). A cubic model is suggested. [10]

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

   • M1 for using cubic regression on the GDC
   • A1 a = -0.0457, b = 0.832, c = -2.74, d = 19.9

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

   • A1 0.996

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

   • A1 quadratic \(R^2\) = 0.976
   • R1 cubic fits better: 0.996 > 0.976

4. (d) Use the cubic model to predict the number of fish at x = 20. [2]

   • M1 for substituting x = 20 into the stored model
   • A1 -67.8 hundred

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

   • R1 20 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 = 10.2, then falls
   • R1 the model gives a negative number of fish, which is impossible

The number of fibre broadband connections in a region, N thousand, is recorded at the end of each year. The time t is in years since the start of 2019, so t = 1 is the end of 2019. The number is modelled by \(N=\dfrac{2400}{1+Ce^{-kt}}\), where C and k are positive constants. The table gives the data. [16]

1. (a) In context, explain what the value 2400 means in the model. [1]

   • A1 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) Use the value of N at t = 2 to show that \(C=39e^{2k}\). [2]

   • M1 for \(60=\dfrac{2400}{1+Ce^{-2k}}\), so \(1+Ce^{-2k}=2400\div60=40\)
   • A1 \(Ce^{-2k}=39\), so \(C=39e^{2k}\) (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 \(C=7e^{7k}\) from \(N=\dfrac{2400}{1+Ce^{-7k}}\) with N = 300
   • A1 k = 0.344 (equating the two expressions and solving on the GDC)
   • A1 C = 77.5 (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 = 160.9 (using the stored C and k)

5. (e) The regulator will use the model if E < 0.8, 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: 0.4, -0.6, 0.8, -0.5
   • A1 \(SS_{res}\) = 1.41
   • R1 \(E=\sqrt{1.41\div4}\) = 0.594 < 0.8, so the model will be used

6. (f) It is estimated that 6% of connections need a technical visit each year, and that one technician can make 80 visits in a year. Use the model to find an expression for the number of technicians, R, needed at time t. [2]

   • M1 for finding the number of visits: 0.06 × 1000N = 60N, then dividing by 80 visits per technician
   • A1 \(R=\dfrac{1800}{1+77.5e^{-0.344t}}\)

7. (g) At the end of 2023 there were 150 technicians, and at the end of 2025 there were 210. Find the average number of technicians added per year between the end of 2023 and the end of 2025. [1]

   • A1 \((210-150)\div2\) = 30

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

   • M1 for the supply \(S=150+30(t-5)\) and the demand R
   • A1 t = 6.56 (first intersection with t ≥ 5 on the GDC)
   • A1 t = 6.56 is 6 complete years after the start of 2019, so the year is 2025

A baker records the mass, X grams, of a loaf and the mass, Y grams, of a bag of rolls. The variable X has mean 820 and variance 144. The variable Y has mean 250 and variance 64. The masses of different items are independent. [9]

1. (a) The cost, C pence, of the ingredients in a loaf is \(C=0.5X+30\). Find E(C) and Var(C). [2]

   • A1 E = 0.5×820+30 = 440
   • A1 Var = 0.5²×144 = 36 (the constant adds nothing)

2. (b) Find the mean and variance of \(D=X-Y\), the amount by which a loaf is heavier than a bag of rolls. [2]

   • A1 E = 820−250 = 570
   • A1 Var = 144+64 = 208 (variances add even for a difference)

3. (c) A tray holds three loaves with masses \(X_1, X_2, X_3\). 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) = 3×820 = 2460
   • M1 for Var(T) = 3Var(X) = 432
   • R1 Var(3X) = 9Var(X) = 1296: T adds three independent observations, but 3X triples one observation

4. (d) A shop sells two loaves and three bags of rolls. 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 = 890, Var = 480

In a game, the score, X, on one turn takes the values in the table, 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(1/1+1/2+1/3+1/4) = 1
   • A1 k = 12/25

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

   • A1 E(X) = Σ x×P(x) = 4k = 1.92 (follow through the value of k)

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

   • M1 for E(X²) − (E(X))² with E(X²) = 4.800
   • A1 Var(X) = 1.11

4. (d) A random sample of 36 turns is taken and the sample mean score, \(\bar X\), is found. Find \(E(\bar X)\) and the standard error of \(\bar X\). [2]

   • A1 E(X̄) = E(X) = 1.92
   • A1 standard error = √(1.114÷36) = 0.176

5. (e) In another game, the number of wins, B, in 20 plays has distribution \(B\sim\mathrm{B}(20,0.15)\). The mean, \(\bar B\), of 40 independent observations of B is found. Find \(E(\bar B)\) and \(\mathrm{Var}(\bar B)\). [2]

   • M1 for E(B) = 20×0.15 = 3.00 and Var(B) = 20×0.15×0.85 = 2.55
   • A1 E(B̄) = 3.00, Var(B̄) = 2.55÷40 = 0.0638

The mass of one egg, X grams, has mean 58 and standard deviation 3. The masses of different eggs are independent. A box holds 12 eggs, and the empty box has a fixed mass of 45 grams. [10]

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

   • M1 for mean \(12\times58+45\) and variance \(12\times3^2=108\) (the fixed mass adds nothing to the variance)
   • A1 mean 741 g, standard deviation \(\sqrt{108}\) = 10.4 g

2. (b) Two eggs 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 \(58-58=0\)
   • A1 variance \(3^2+3^2=18\) (variances add for a difference), standard deviation 4.24 g

3. (c) Alain takes the mass of one egg and doubles it, \(2X\). Beatrice takes the masses of two eggs 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 116, variance \(2^2\times3^2=36\), standard deviation 6
   • A1 \(X_1+X_2\): mean 116, variance \(3^2+3^2=18\), standard deviation 4.24
   • R1 \(2X\) varies more (6 > 4.24): doubling one value is not the same as adding two independent values

4. (d) The mean mass of a random sample of n eggs is \(\bar X\). Find the smallest n for which the standard deviation of \(\bar X\) is less than 0.5 grams. [2]

   • M1 for \(3\div\sqrt{n}<0.5\), so \(n>36\)
   • A1 n = 37

A cyclist's morning ride takes X minutes, where \(X\sim\mathrm{N}(32,3^2)\), and the evening ride takes Y minutes, where \(Y\sim\mathrm{N}(35,4^2)\). The two times are independent. Let T be the total time for the two rides. [9]

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

   • M1 for E = 32+35 = 67 and Var = 3²+4² = 25 (variances add; standard deviations are squared first)
   • A1 \(T=X+Y\) ~ N(67, 25)

2. (b) Find \(P(T>75)\). [1]

   • A1 0.0548 (normal cdf on the GDC with mean 67 and standard deviation √25 = 5.000)

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

   • A1 D = X − Y ~ N(-3, 25) (variances add for a difference)
   • A1 P(D > 0) = 0.274

4. (d) A random sample of 5 values of X is taken, with sample mean \(\bar X\). Find \(P(\bar X<31)\). [2]

   • M1 for \(\bar X\) ~ N(32, 3²/5) = N(32, 1.800)
   • A1 0.228

5. (e) The time, W minutes, the cyclist waits at crossings in a week has an unknown distribution, with mean 12 and standard deviation 5. Giving a reason, state the approximate distribution of the mean, \(\bar W\), of 50 weeks. [2]

   • A1 N(12, 0.5) = N(12, 5²/50)
   • R1 the population is not normal but n = 50 > 30, so the central limit theorem applies

A fruit packer buys pineapples from three sources. The mass of an item from source A is \(X\sim\mathrm{N}(1.2,0.15^2)\) kg and the mass of an item from source B is \(Y\sim\mathrm{N}(1.45,0.2^2)\) kg. All masses are independent. [10]

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

   • M1 for \(W=2X-Y\) and the event \(W>0\): mean \(2\times1.2-1.45\) = 0.95
   • A1 variance \(2^2\times0.15^2+0.2^2\) = 0.13
   • A1 \(P(W>0)\) = 0.996

2. (b) A batch of 6 A items is taken, with mean mass \(\bar X\). State the distribution of \(\bar X\). The batch is rejected if \(\bar X\) differs from 1.2 by more than 0.1. Find the probability that a batch is rejected. [3]

   • A1 \(\bar X\sim\mathrm{N}(1.2,0.15^2/6)=\mathrm{N}(1.2,0.00375)\)
   • M1 for \(P(\bar X>1.3)+P(\bar X<1.1)\), or twice \(P(\bar X>1.3)\) by symmetry
   • A1 0.102

3. (c) Items from source C have mean mass 1.3 and standard deviation 0.25, with an unknown distribution. A crate holds 40 items from C. Find the approximate probability that the total mass of the crate is greater than 54. [3]

   • M1 for the CLT: the total is approximately normal with mean \(40\times1.3=52\) and variance \(40\times0.25^2=2.5\)
   • M1 for \(P(T>54)\) on the GDC with standard deviation \(\sqrt{2.5}\) = 1.581
   • A1 0.103

4. (d) A smaller crate holds only 20 items from C. Explain why the central limit theorem cannot be used to answer part (c) for this crate. [1]

   • R1 n = 20 < 30 and the distribution of the C items is not known to be normal, so the central limit theorem does not apply

A call centre records the length, in minutes, of eight randomly chosen calls: 4.2, 5.1, 3.8, 6.0, 4.7, 5.5, 4.4, 5.3. [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 = 4.88 and variance = \((0.7324)^2\) = 0.536

2. (b) A second sample of 15 values has standard deviation \(s_n=2.4\). Find the unbiased estimate of the population variance. [2]

   • M1 for \(s_{n-1}^2 = \dfrac{n}{n-1}s_n^2 = \dfrac{15}{14}\times2.4^2\)
   • A1 6.17

3. (c) A third sample has variance \(s_n^2=5.76\) and unbiased estimate \(s_{n-1}^2=6.4\). Find the sample size n. [2]

   • M1 for \(\dfrac{n}{n-1}=6.4\div5.76\)
   • A1 n = 10

4. (d) A ninth call is recorded as 61 minutes. The manager finds that the line was left connected by mistake after the customer hung up. 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 call length

5. (e) A customer survey asks: “How much did you enjoy our excellent service?” Comment on the validity of this question. [1]

   • R1 The question is leading (it calls the service excellent), so it does not measure what customers really think: it has poor validity

6. (f) The centre asks the same customers to complete the same survey again a week later and compares the answers. Name this test of reliability. [1]

   • R1 Test-retest

A gardener measures the heights, in cm, of 10 seedlings. The sum of the heights is 253.0 and the sum of the squares of the heights is 6455. [9]

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

   • M1 for \(\dfrac{\sum x^2-(\sum x)^2/n}{n-1}=\dfrac{6455.0-253.0^2/10}{9}\)
   • A1 6.011

2. (b) The calculator gives two standard deviations: \(\sigma_x\) = 2.326 and \(s_x\) = 2.452. 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 \(s_x\) = 2.452 (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 height of a seedling in cm. 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 \(\bar x\) = 25.3, \(s_{n-1}\) = 2.452, n = 10, C = 0.90
   • A1 (23.9, 26.7)

4. (d) A second sample of 16 seedlings has mean 25.3 and \(s_{n-1}\) = 2.4. Without using the interval function of your calculator, find a 95% confidence interval for its population mean. The critical value of t with 15 degrees of freedom is 2.131. Give the limits to 3 significant figures. [3]

   • M1 for \(25.3\pm2.131\times\dfrac{2.4}{\sqrt{16}}\)
   • A1 margin of error = 1.28
   • A1 (24.0, 26.6)

A random sample of 10 batteries of one brand is tested. Their lifetimes, in hours, are 38.2, 41.5, 39.8, 42.1, 40.6, 37.9, 43.0, 39.4, 41.8, 40.2. The lifetimes are assumed to be normally distributed. The maker claims that the mean lifetime is 40 hours. [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 (39.3, 41.6)

2. (b) Say whether your interval supports the maker’s claim, giving a reason. [1]

   • R1 The interval (39.3, 41.6) contains 40, so the data are consistent with the claim

3. (c) A second brand has lifetimes with known population standard deviation 4.2 hours. A random sample of 50 batteries has mean lifetime 45.8 hours. Find a 90% confidence interval for the mean lifetime of this brand. [2]

   • M1 for a z-interval on the GDC with σ = 4.2, ̅x = 45.8, n = 50, C = 0.90
   • A1 (44.8, 46.8)

4. (d) For the first brand, 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 lifetimes 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 (38.7, 42.2))

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 coach measures the resting heart rate, in beats per minute, of 8 students before and after a six-week training programme. 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 programme. Give the limits to 3 significant figures. [3]

   • M1 for finding the differences -3, -2, -1, -5, 1, -5, -5, -1 and using the t-interval on the GDC
   • A1 (-4.52, -0.732)
   • 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 resting heart rate, in beats per minute, of a second group of 12 adults, based on a t-distribution, is (63.2, 68.6). 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 = 65.9 and margin of error = (upper − lower) ÷ 2 = 2.7
   • M1 for margin of error = \(t\times s_{n-1}/\sqrt{n}\) with \(t\) = 2.201 (df = 11)
   • A1 \(s_{n-1}\) = 4.25

3. (c) Another sample of 48 adults 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 \(=2.012\times4.249/\sqrt{48}\) = 1.23, so width = 2.47
   • A1 narrower than 5.4: a larger sample size makes the interval narrower