S3 June 2013 Q5
5. Blumen is a perfume sold in bottles. The amount of perfume in each bottle is normally distributed. The amount of perfume in a large bottle has mean 50 ml and standard deviation 5 ml. The amount of perfume in a small bottle has mean 15 ml and standard deviation 3 ml.
One large and 3 small bottles of Blumen are chosen at random.
A large bottle and a small bottle of Blumen are chosen at random.
| Scheme | Marks |
|---|---|
| Let \(L \sim \mathrm{N}(50, 25)\) and \(S \sim \mathrm{N}(15, 9)\) Let \(X = L - (S_1 + S_2 + S_3)\) | B1 |
| \(\mathrm{E}(X) = 50 - 3 \times 15 = 5\) | B1 |
| \(\mathrm{Var}(X) = 25 + 3 \times 9 = 52\) | M1A1 |
| \(\mathrm{P}(X \lt 0) \quad = \mathrm{P}\left(Z \lt \dfrac{-5}{\sqrt{52}}\right)\) | dM1 |
| \(= \mathrm{P}(Z \lt -0.693..)\) \(= 0.244\) or 0.2451 (tables) (awrt 0.244 ~ 0.245) | A1 |
| (6) |
Notes
1st B1 for forming a suitable variable \(X\) explicitly seen. Do not give for \(L - 3S\) but allow \(L - (S + S + S)\)
2nd B1 for \(\mathrm{E}(X) = 5\) (or – 5 if their \(X\) is defined the other way around)
1st M1 for an attempt at \(\mathrm{Var}(X) = \mathrm{Var}(L) + 3\mathrm{Var}(S)\). Do not condone 5 for “25” or 3 for “9”
1st A1 for 52
2nd dM1 for attempting the correct probability and standardising with their mean and sd.
This mark is dependent on 1st M1 so if \(X\) is not being used or wrong variance score M0
If their method is not crystal clear then they must be attempting P(\(Z\) < -ve value) or P(\(Z\) > +ve value) i.e. their probability after standardisation should lead to a prob. < 0.5
2nd A1 for awrt 0.244 ~ 0.245
Correct ans. only scores 5/6 (or 6/6 if 1st B1) but must be clearly labelled as (a) or the first answer.
| Scheme | Marks |
|---|---|
| Let \(Y = L - 3S\) | B1 |
| \(\mathrm{E}(Y) = 50 - 3 \times 15 = 5\) | B1 |
| \(\mathrm{Var}(Y) = 25 + 3^2 \times 9 = 106\) | M1A1 |
| \(\mathrm{P}(Y \gt 0) \quad = \mathrm{P}\left(Z \gt \dfrac{-5}{\sqrt{106}}\right)\) | dM1 |
| \(= \mathrm{P}(Z \gt -0.4856..)\) \(= 0.686\) or 0.6879 (tables) (awrt 0.686 ~ 0.688) | A1 |
| (6) | |
| (12 marks) |
Notes
1st B1 for defining a new variable [\(Y =\) ]\(\pm\) (\(L - 3S\)). May be implied by a correct variance.
2nd B1 for \(\mathrm{E}(Y) = 5\) (or – 5 if their \(Y\) is defined as \(Y = 3S - L\) )
1st M1 for an attempt at \(\mathrm{Var}(Y) = \mathrm{Var}(L) + 3^2\,\mathrm{Var}(S)\). Do not condone 5 for “25” or 3 for “9”
1st A1 for 106 only
2nd dM1 for attempting the correct probability and standardising with their mean and sd.
This mark is dependent on 1st M1 so if \(Y\) is not being used or wrong variance score M0
If their method is not crystal clear then they must be attempting P(\(Z\) > -ve value) or P(\(Z\) < +ve value) i.e. their probability after standardisation should lead to a prob. > 0.5
2nd A1 for an awrt 0.686 ~ 0.688
Correct answer only scores 6/6 but must be clearly labelled as (b) or the second answer.