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”.

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

Part (a)(i) was well answered in general. In (a)(ii) and (b), however, many candidates made the fairly common error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) with \(nX\) which gives an incorrect variance. This is an important distinction which needs to be emphasized.

Part (a)(i) was well answered in general. In (a)(ii) and (b), however, many candidates made the fairly common error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) with \(nX\) which gives an incorrect variance. This is an important distinction which needs to be emphasized.