Find the probability that a randomly chosen Supermug tea bag contains more than 3.9 g of tea.

Find the probability that, of two randomly chosen Megamug tea bags, one contains more than 5.4 g of tea and one contains less than 5.4 g of tea.

Find the probability that five randomly chosen Supermug tea bags contain a total of less than 20.5 g of tea.

Find the probability that the total weight of tea in seven randomly chosen Supermug tea bags is more than the total weight in five randomly chosen Megamug tea bags.

let S be the weight of tea in a random Supermug tea bag

\(S \sim {\text{N(4.2, 0.1}}{{\text{5}}^2})\)

\({\text{P}}(S > 3.9) = 0.977\)     (M1)A1

[2 marks]

let M be the weight of tea in a random Megamug tea bag

\(M \sim {\text{N(5.6, 0.1}}{{\text{7}}^2})\)

\({\text{P}}(M > 5.4) = 0.880 \ldots \)     (A1)

\({\text{P}}(M < 5.4) = 1 - 0.880 \ldots = 0.119 \ldots \)     (A1)

required probability \( = 2 \times 0.880 \ldots \times 0.119 \ldots = 0.211\)     M1A1

[4 marks]

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} < 20.5)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} = A\)     (M1)

\({\text{E}}(A) = 5{\text{E}}(S)\)

= 21     A1

\({\text{Var}}(A) = 5{\text{Var}}(S)\)

= 0.1125     A1

\(A \sim {\text{N(21, 0.1125}})\)

\({\text{P}}(A < 20.5) = 0.0680\)     A1

[4 marks]

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} - ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) > 0)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} - ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) = B\)     (M1)

\({\text{E}}(B) = 7{\text{E}}(S) - 5{\text{E}}(M)\)

= 1.4     A1

Note: Above A1 is independent of first M1.

\({\text{Var}}(B) = 7{\text{Var}}(S) + 5{\text{Var}}(M)\)     (M1)

= 0.302     A1

\({\text{P}}(B > 0) = 0.995\)     A1

[5 marks]

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).