Given that, on a randomly chosen day, the probability that he completes the crossword in less than \(a\) minutes is equal to 0.8, find the value of \(a\).

Find the probability that the total time taken for him to complete five randomly chosen crosswords exceeds 120 minutes.

Find the probability that, on a randomly chosen day, the time taken by Beatrice to complete the crossword is more than twice the time taken by Adam to complete the crossword. Assume that these two times are independent.

\(z = 0.841 \ldots \)    (A1)

\(a = \mu  + z\sigma \)    (M1)

\( = 26.2\)    A1

[3 marks]

let \(T\) denote the total time taken to complete 5 crosswords.

\(T\) is \({\text{N}}(110,{\text{ }}125)\)     (A1)(A1)

Note:     A1 for the mean and A1 for the variance.

\({\text{P}}(T > 120) = 0.186\)    A1

[3 marks]

consider the random variable \(U = Y - 2X\)     (M1)

\({\text{E}}(U) =  - 4\)    A1

\({\text{Var}}(U) = {\text{Var}}(Y) + 4{\text{Var}}(X)\)    (M1)

\( = 136\)    A1

\({\text{P}}(Y > 2X) = {\text{P}}(U > 0)\)    (M1)

\( = 0.366\)    A1

[6 marks]

Part (a) was very well answered with only a very few weak candidates using 0.8 instead of 0.841...

Part (b) was well answered with only a few candidates calculating the variance incorrectly.

Part (c) was again well answered. The most common errors, not often seen, were writing the variance of \(Y - 2X\) as either \({\text{Var}}(Y) + 2{\text{Var}}(X)\) or \({\text{Var}}(Y) - 2(or{\text{ }}4){\text{Var}}(X)\).