A-Level Mathematics October/November 2024 Q5(a): The weights of the green apples sold by a shop are normally distributed with mean 90 gr…

1 answer

• ▲0
  accepted ✓

  Let $X$ be the weight of a randomly chosen green apple in grams. The distribution is given as $X \sim N(90, 8^2)$.

  We need to find the probability $P(83 < X < 95)$.

  First, we standardise the values $x_1 = 83$ and $x_2 = 95$ to find the corresponding $z$-scores.

  $z_1 = \frac{x_1 - \mu}{\sigma} = \frac{83 - 90}{8} = -\frac{7}{8} = -0.875$

  $z_2 = \frac{x_2 - \mu}{\sigma} = \frac{95 - 90}{8} = \frac{5}{8} = 0.625$

  So, we are looking for $P(-0.875 < Z < 0.625)$, where $Z \sim N(0, 1)$.

  This can be calculated as $\Phi(0.625) - \Phi(-0.875)$. Using the symmetry property $\Phi(-z) = 1 - \Phi(z)$, we have:

  $P(-0.875 < Z < 0.625) = \Phi(0.625) - (1 - \Phi(0.875))$ $= \Phi(0.625) + \Phi(0.875) - 1$

  From the standard normal distribution tables: $\Phi(0.625) = 0.7340$ $\Phi(0.875) = 0.8092$

  Therefore, the probability is: $0.7340 + 0.8092 - 1 = 1.5432 - 1 = 0.5432$

  The probability that a randomly chosen green apple weighs between 83 grams and 95 grams is $0.543$ (to 3 s.f.).

  ---

  How the marks are awarded

  • M1 — Correctly applying the standardisation formula $\frac{x - \mu}{\sigma}$ to either 83 or 95, as shown in the steps $\frac{83 - 90}{8}$ or $\frac{95 - 90}{8}$. No continuity correction is used.
  • A1 — Obtaining both correct z-scores, $-0.875$ and $0.625$. These must be seen for the mark to be awarded.
  • M1 — Correctly formulating the area calculation for the probability $P(-0.875 < Z < 0.625)$. This is shown by the expression $\Phi(0.625) + \Phi(0.875) - 1$ or an equivalent correct method.
  • A1 — Obtaining the final correct answer of $0.543$ or $0.5432$. The answer must be the result of a correct method.

  Common mistakes

  • Using the variance (64) instead of the standard deviation (8) in the standardisation formula, e.g., calculating $\frac{95-90}{64}$.
  • Incorrectly calculating the area from the z-scores, for example, calculating $\Phi(0.625) - \Phi(0.875)$ or $1 - \Phi(0.625) - \Phi(0.875)$.
  • Applying an unnecessary and incorrect continuity correction by using values like 82.5 or 95.5.
  • Making a sign error when standardising, for instance calculating $\frac{90-83}{8}$ which gives a positive z-score instead of a negative one.

  Examiner tip: Mastering the standardisation formula and correctly manipulating areas under the normal curve, especially for $P(a < Z < b)$, is fundamental to solving normal distribution problems.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.