A2 June 2023 Q4

4. The weights of eggs, \(E\) grams, follow a normal distribution, \(\mathrm{N}(60,\ 3^2)\)

The weights of empty small boxes, \(S\) grams, follow a normal distribution, \(\mathrm{N}(24,\ 1.8^2)\)

The weights of empty large boxes, \(L\) grams, follow a normal distribution, \(\mathrm{N}(40,\ 2.1^2)\)

Small boxes of eggs contain 6 randomly selected eggs.

Large boxes of eggs contain 12 randomly selected eggs.

(a) Find the probability that the total weight of a randomly selected small box of 6 eggs weighs less than 387 grams. (3)

(b) Find the probability that a randomly selected large box of 12 eggs weighs more than twice a randomly selected small box of 6 eggs. (5)

Mark scheme (a)

Scheme	Marks	AO
\(W = E_1 + E_2 + \ldots + E_6 + S \sim \mathrm{N}(6 \times 60 + 24,\ 6 \times 3^2 + 1.8^2)\) \([W \sim \mathrm{N}(384,\ 57.24)]\)	M1M1	3.3 1.1b
\(\mathrm{P}(W \lt 387) = 0.6541\ldots\) awrt 0.654	A1	3.4
	(3)

M1: setting up normal model with correct mean

M1: use of correct variance

A1: awrt 0.654

Mark scheme (b)

Scheme	Marks	AO
\(X = E_1 + E_2 + \ldots + E_{12} + L \sim \mathrm{N}(12 \times 60 + 40,\ 12 \times 3^2 + 2.1^2)\) \([X \sim \mathrm{N}(760,\ 112.41)]\)	M1	1.1b
\(2W \sim \mathrm{N}(\text{“}384\text{”} \times 2,\ 2^2 \times \text{“}57.24\text{”})\,[= \mathrm{N}(768,\ 228.96)]\)	M1	1.1b
\(X - 2W \sim \mathrm{N}(760 - 768,\ 112.41 + 228.96) \to \mathrm{N}(-8,\ 341.37)\)	M1A1	3.3 1.1b
\(\mathrm{P}(X - 2W \gt 0) = 0.3325\ldots\) awrt 0.333	A1	3.4
	(5)
	(8 marks)

M1: setting up normal model for large box of eggs, sight of \(12 \times 60 + 40\)

M1: sight or use of 2 times their mean from (a) and 4 times their variance from (a)

M1: setting up distribution for the difference in weights, with mean of their “760” – “768”

A1: correct mean and variance

A1: awrt 0.333