Mathematics

1. 1.

   A random sample of 36 independent observations from a normal population has mean 52. The population standard deviation is known to be 6. Test H0: mu = 50 against H1: mu > 50 at the 5% level using a z-test.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The sampling distribution of the mean is normal with standard deviation sigma/sqrt(n), not sigma.
   2. The alternative specifies an upper-tail test. A low upper-tail probability under H0 supports rejection, not proof that H1 is certainly true.

   Marking points

   • Standard error = 6/sqrt(36) = 1.
   • z = (52 - 50)/1 = 2.
   • One-sided p-value = 1 - Phi(2) = 0.0228 approximately, below 0.05.
   • Reject H0; there is evidence at the 5% level that the population mean exceeds 50.

   Examiner tip: Compare the one-sided p-value with 0.05; do not double it for this alternative.

2. 2.

   A particle moves on a straight line with velocity v(t) = 3t^2 m s^-1 for 0 <= t <= 2, where t is in seconds. Find its displacement over this interval and acceleration at t = 2.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Velocity integrates to a change in position and differentiates to acceleration, so the two requested quantities require opposite calculus operations.
   2. Velocity is nonnegative throughout this interval, so the distance travelled also equals the displacement of 8 m.

   Marking points

   • Displacement = integral from 0 to 2 of 3t^2 dt = [t^3] from 0 to 2.
   • Displacement = 8 m.
   • a(t) = dv/dt = 6t, hence a(2) = 12 m s^-2.

   Examiner tip: Evaluating v(2) gives a velocity, not the displacement over two seconds.

3. 3.

   In ten independent trials, nine successes are observed. Test H0: p = 0.5 against H1: p > 0.5 at 5% significance using the exact binomial upper-tail probability.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. More extreme outcomes for the stated alternative are those with at least as many successes as observed, so include ten successes as well.
   2. This is a test of an unknown population probability; the observed fraction 0.9 is not the assumed probability in the null distribution.

   Marking points

   • Under H0, X follows Binomial(10,0.5).
   • P(X >= 9) includes X = 9 and X = 10.
   • p-value = (C(10,9) + C(10,10))/2^10 = 11/1024 = 0.010742... .
   • Since this is below 0.05, reject H0; the data provide evidence that p > 0.5.

   Examiner tip: Do not calculate only P(X = 9), or substitute 0.9 for the null probability.

4. 4.

   A fill amount X is modelled as normal with mean 16 and standard deviation 2. Find P(X < 14), then find c such that P(X < c) = 0.05. Give both answers to three significant figures.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The first task evaluates a cumulative probability at a known cutoff; the second reverses that process using an inverse normal calculation.
   2. A lower-tail cutoff is below the mean. Before rounding its value is about 12.7103, which checks the direction and scale.

   Marking points

   • At 14, z = (14 - 16)/2 = -1.
   • P(X < 14) = Phi(-1) = 0.159 to three significant figures.
   • The lower 5% standard-normal quantile is approximately -1.64485.
   • c = 16 + 2(-1.64485) = 12.7 to three significant figures.

   Examiner tip: The 5% lower quantile is negative; using +1.64485 gives the upper 95% cutoff.

5. 5.

   A population model is P(t) = 1000/(1 + 9e^(-0.4t)), t >= 0 in years. Find P(0), the initial growth rate, and when it reaches 500. Using P' = 0.4P(1 - P/1000), find the maximum growth rate on t >= 0.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The growth-rate expression is a downward-opening quadratic in population size, so completing the square identifies its maximum.
   2. The trajectory starts at 100 and increases toward 1000, so 500 is actually reached in the allowed time domain. The carrying capacity is a model limit, not a guaranteed real population.

   Marking points

   • P(0) = 1000/10 = 100.
   • Initial growth rate = 0.4(100)(1 - 100/1000) = 36 per year.
   • P = 500 gives 9e^(-0.4t) = 1.
   • t = ln(9)/0.4 = 5.49 years to three significant figures.
   • P' = 100 - 0.0004(P - 500)^2, so the maximum is 100 per year at P = 500, which the model reaches.

   Examiner tip: Maximum population size and maximum growth rate are different quantities.

6. 6.

   For x > 0 a model cost is C(x) = x^2 + 100/x. Find the exact minimising x and minimum cost, and justify that the minimum is global.

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Multiplying the stationary equation by x^2 is valid because x is positive. Use the resulting x^3 = 50 relation to simplify the cost without rounding.
   2. The cost tends to infinity both as x approaches zero from above and as x tends to infinity. This also confirms that neither boundary can improve the unique minimum.

   Marking points

   • C'(x) = 2x - 100/x^2.
   • C' = 0 gives x^3 = 50, so x = 50^(1/3).
   • C''(x) = 2 + 200/x^3 > 0 for all x > 0.
   • Strict convexity on the whole domain makes this unique stationary point the global minimum.
   • Since 100/x = 2x^2 there, minimum cost = 3(50^(2/3)).

   Examiner tip: A local minimum test at one point alone does not establish a global optimum.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.