Edexcel S3 — Question 5 11 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
DifficultyStandard +0.8 This S3 question requires understanding of distributions of differences and sums of independent normal variables. Part (a) needs recognizing that the difference of two normal variables follows N(0, 2σ²), then finding P(|X₁-X₂| > 4). Part (b) requires summing six egg weights plus box weight and applying normal distribution properties. These are non-trivial applications requiring conceptual understanding beyond routine calculations, placing it moderately above average difficulty.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions

An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams.
  1. Find the probability that two of the farm's eggs chosen at random differ in weight by more than 4 grams. [5]
The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams.
  1. Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams. [6]

AnswerMarks Guidance
Part (a)M1 A1, M1, M1 A1 let \(W =\) weight of egg; let \(A = W_1 - W_2\) \(\therefore A \sim N(0, 2 \times 3.9^2) = \sim N(0, 30.42)\); require \(2 \times P(A > 4) = 2 \times P\left(Z > \frac{4-0}{\sqrt{30.42}}\right)\); \(= 2 \times P(Z > 0.73) = 2 \times (1 - 0.7673) = 0.465\)
Part (b)M1 A2, M1, M1 A1, (11) let \(T =\) total weight of box and eggs \(\therefore T \sim N(28 + 6 \times 55, 1.2^2 + 6 \times 3.9^2) = \sim N(358, 92.7)\); \(P(T < 350) = P\left(Z < \frac{350-358}{\sqrt{92.7}}\right)\); \(= P(Z < -0.83) = 1 - 0.7967 = 0.2033\)
**Part (a)** | M1 A1, M1, M1 A1 | let $W =$ weight of egg; let $A = W_1 - W_2$ $\therefore A \sim N(0, 2 \times 3.9^2) = \sim N(0, 30.42)$; require $2 \times P(A > 4) = 2 \times P\left(Z > \frac{4-0}{\sqrt{30.42}}\right)$; $= 2 \times P(Z > 0.73) = 2 \times (1 - 0.7673) = 0.465$

**Part (b)** | M1 A2, M1, M1 A1, (11) | let $T =$ total weight of box and eggs $\therefore T \sim N(28 + 6 \times 55, 1.2^2 + 6 \times 3.9^2) = \sim N(358, 92.7)$; $P(T < 350) = P\left(Z < \frac{350-358}{\sqrt{92.7}}\right)$; $= P(Z < -0.83) = 1 - 0.7967 = 0.2033$
An organic farm produces eggs which it sells through a local shop. The weight of the eggs produced on the farm are normally distributed with a mean of 55 grams and a standard deviation of 3.9 grams.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that two of the farm's eggs chosen at random differ in weight by more than 4 grams. [5]
\end{enumerate}

The farm sells boxes of six eggs selected at random. The weight of the boxes used are normally distributed with a mean of 28 grams and a standard deviation of 1.2 grams.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the probability that a randomly chosen box with six eggs in weighs less than 350 grams. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S3  Q5 [11]}}