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Question
Sami is undertaking market research on packets of soap powder. He considers the brand “Gleam”. The weight of the contents of a randomly chosen packet of “Gleam” follows a normal distribution with mean 750 grams and standard deviation 20 grams.
The weight of the packaging follows a different normal distribution with mean 40 grams and standard deviation 5 grams.
Find:
(i) the probability that a randomly chosen packet of “Gleam” has a total weight exceeding 780 grams.
(ii) the probability that the total weight of the contents of five randomly chosen packets of “Gleam” exceeds 3800 grams.
Sami now considers the brand “Bright”. The weight of the contents of a randomly chosen packet of “Bright” follow a normal distribution with mean 650 grams and standard deviation 16 grams. Find the probability that the contents of six randomly chosen packets of “Bright” weigh more than the contents of five randomly chosen packets of “Gleam”.
Answer/Explanation
Markscheme
Note: In all parts accept answers which round to the correct 2sf answer.
(i) contents: \(X \sim N(750,{\text{ }}400)\)
packaging: \(Y \sim N(40,{\text{ }}25)\)
consider \(X + Y\) (M1)
\({\text{E}}(X + Y) = 790\) A1
\({\text{Var}}(X + Y) = 425\) A1
\({\text{P}}(X + Y > 780) = 0.686\) A1
(ii) Let \({X_1} + {X_2} + {X_3} + {X_4} + {X_5} = A\) M1
\({\text{E}}(A) = 5{\text{E}}(X) = 3750\) A1
\({\text{Var}}(A) = 5{\text{Var}}(X) = 2000\) A1
\({\text{P}}(A > 3800) = 0.132\) A1
Note: Condone the notation \(A = 5X\) if the variance is correct, M0 if not
contents of Bright: \(B \sim N(650,{\text{ }}256)\)
let \(G = {B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} – ({X_1} + {X_2} + {X_3} + {X_4} + {X_5})\) M1
\({\text{E}}(G) = 6 \times 650 – 5 \times 750 = 150\) A1
\({\text{Var}}(G) = 6 \times 256 + 5 \times 400 = 3536\) A1
\({\text{P}}(G > 0) = 0.994\) A1
Note: Condone the notation \(G = 6B – 5X\) if the variance is correct, M0 if not
Question
The lifetime, in years, of a randomly chosen basic vacuum cleaner is assumed to be modelled by the normal distribution \(B \sim {\text{N}}(14,{\text{ }}{3^2})\).
The lifetime, in years, of a randomly chosen robust vacuum cleaner is assumed to be modelled by the normal distribution \(R \sim {\text{N}}(20,{\text{ }}{4^2})\).
Find \({\text{P}}\left( {B > {\text{E}}(B) + \frac{1}{2}\sqrt {{\text{Var}}(B)} } \right)\).
Find the probability that the total lifetime of 7 randomly chosen basic vacuum cleaners is less than 100 years.
Find the probability that the total lifetime of 5 randomly chosen robust vacuum cleaners is greater than the total lifetime of 7 randomly chosen basic vacuum cleaners.
Answer/Explanation
Markscheme
\({\text{P}}(B > 15.5){\text{ }}\left( { = {\text{P}}(Z > 0.5)} \right)\) (M1)
\( = (1 – 0.69146) = 0.309\) A1
[2 marks]
consider \(V = {B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} + {B_7}\) (M1)
\({\text{E}}(V) = 98\) (A1)
\({\text{Var}}(V) = 63\) or equivalent (A1)
Note: No need to state \(V\) is normal.
\({\text{P}}(V < 100) = \left( {{\text{P}}\left( {Z < \frac{2}{{\sqrt {63} }} = 0.251976 \ldots } \right)} \right) = 0.599\) A1
[4 marks]
consider \(W = {R_1} + {R_2} + {R_3} + {R_4} + {R_5} – ({B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} + {B_7})\) (M1)
\({\text{E}}(W) = 2\) (A1)
\({\text{Var}}(W) = 80 + 63 = 143\) (A1)
\({\text{P}}(W > 0) = \left( {{\text{P}}\left( {Z < \frac{2}{{\sqrt {143} }}} \right)} \right)\) (M1)
\( = 0.566\) A1
[5 marks]