A large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.

A large can and a small can are selected at random. Find the probability that the large can contains at least \(30\) milliliters more than five times the amount contained in the small can.

A large can and five small cans are selected at random. Find the probability that the large can contains at least \(30\) milliliters less than the total amount contained in the small cans.

\({\text{P}}(L \ge 4995) = 0.785\)     (M1)A1

Note:     Accept any answer that rounds correctly to \(0.79\).

Award M1A0 for \(0.78\).

Note:     Award M1A0 for any answer that rounds to \(0.55\) obtained by taking \({\text{SD}} = 40\).

[2 marks]

we are given that \(L \sim {\text{N}}(5000,{\text{ }}40)\) and \(S \sim {\text{N}}(1000,{\text{ }}25)\)

consider \(X = L - 5S\) (ignore \( \pm 30\))     (M1)

\({\text{E}}(X) = 0\) (\( \pm 30\) consistent with line above)     A1

\({\text{Var}}(X) = {\text{Var}}(L) + 25{\text{Var}}(S) = 40 + 625 = 665\)     (M1)A1

require \({\text{P}}(X \ge 30)\;\;\;({\text{or P}}(X \ge 0){\text{ if }} - 30{\text{ above}})\)     (M1)

obtain \(0.122\)     A1

Note:     Accept any answer that rounds correctly to \(2\) significant figures.

[6 marks]

consider \(Y = L - ({S_1} + {S_2} + {S_3} + {S_4} + {S_5})\) (ignore \( \pm 30\))     (M1)

\({\text{E}}(Y) = 0\) (\( \pm 30\) consistent with line above)     A1

\({\text{Var}}(Y) = 40 + 5 \times 25 = 165\)     A1

require \({\text{P}}(Y \le  - 30){\text{ (or P}}(Y \le 0){\text{ if }} + 30{\text{ above)}}\)     (M1)

obtain \(0.00976\)     A1

Note:     Accept any answer that rounds correctly to \(2\) significant figures.

Note:     Condone the notation \(Y = L - 5S\) if the variance is correct.

[5 marks]

Total [13 marks]

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).