A-Level Mathematics October/November 2024 Q2: The masses, in kilograms, of small and large bags of wheat have the independent distrib…

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  Let $S$ be the mass of a small bag and $L$ be the mass of a large bag. We are given: $S \sim N(16.0, 0.4)$ $L \sim N(51.0, 0.9)$

  We need to find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag. This can be written as $P(S_1 + S_2 + S_3 > L)$.

  Let's define a new random variable $T = S_1 + S_2 + S_3 - L$. We want to find $P(T > 0)$. Since $S_1, S_2, S_3$ and $L$ are independent normal distributions, their linear combination $T$ is also a normal distribution.

  First, we find the mean (expectation) of $T$: $E(T) = E(S_1 + S_2 + S_3 - L)$ $E(T) = E(S_1) + E(S_2) + E(S_3) - E(L)$ $E(T) = 3 \times E(S) - E(L)$ $E(T) = 3 \times 16.0 - 51.0 = 48 - 51 = -3$

  Next, we find the variance of $T$: $Var(T) = Var(S_1 + S_2 + S_3 - L)$ Since the variables are independent, we add the variances: $Var(T) = Var(S_1) + Var(S_2) + Var(S_3) + Var(L)$ $Var(T) = 3 \times Var(S) + Var(L)$ $Var(T) = 3 \times 0.4 + 0.9 = 1.2 + 0.9 = 2.1$

  So, the distribution of $T$ is $T \sim N(-3, 2.1)$.

  Now we need to find $P(T > 0)$. We standardise this using the Z-formula, $Z = \frac{X - \mu}{\sigma}$. $P(T > 0) = P\left(Z > \frac{0 - (-3)}{\sqrt{2.1}}\right)$ $P(T > 0) = P\left(Z > \frac{3}{\sqrt{2.1}}\right)$ $P(T > 0) = P(Z > 2.07019...)$

  We look for the probability corresponding to $Z > 2.070$. This is the upper tail of the distribution. $P(Z > 2.070) = 1 - P(Z < 2.070)$ $P(Z > 2.070) = 1 - \Phi(2.070)$ Using the normal distribution tables, $\Phi(2.070) = 0.9808$.

  $P(T > 0) = 1 - 0.9808 = 0.0192$

  The probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag is 0.0192 (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly calculating the expectation of the combined distribution: E(T) = 3 × 16.0 - 51.0 = -3.
  • M1 — Correctly calculating the variance of the combined distribution by adding all individual variances: Var(T) = 3 × 0.4 + 0.9 = 2.1.
  • M1 — Correctly standardising the value 0 using their calculated mean and variance: (0 - (-3)) / √2.1.
  • M1 — Finding the correct area of the normal distribution consistent with their standardised value. For a positive z-value and a 'greater than' probability, this involves calculating 1 - Φ(z).
  • A1 — Obtaining the final correct answer of 0.0192, or an answer that rounds to it.

  Common mistakes

  • Subtracting the variance of the large bag instead of adding it, i.e., calculating Var(T) = 3 × 0.4 - 0.9 = 0.3. Variances are always added for combinations of independent variables.
  • Incorrectly calculating the variance of the three small bags as 3² × 0.4 instead of 3 × 0.4. This confuses Var(S₁+S₂+S₃) with Var(3S).
  • Using the variance (2.1) instead of the standard deviation (√2.1) in the denominator of the standardisation formula.
  • Finding the wrong area from the normal distribution tables, for example calculating Φ(2.070) directly instead of 1 - Φ(2.070) for a 'greater than' probability.

  Examiner tip: When combining independent random variables, always add their variances, regardless of whether you are adding or subtracting the variables themselves.

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