S1 January 2012 Q7

EdexcelOld spec10 marksNormal Distribution

7. A manufacturer fills jars with coffee. The weight of coffee, \(W\) grams, in a jar can be modelled by a normal distribution with mean 232 grams and standard deviation 5 grams.

(a) Find \(\mathrm{P}(W \lt 224)\). (3)

(b) Find the value of \(w\) such that \(\mathrm{P}(232 \lt W \lt w) = 0.20\) (4)

Two jars of coffee are selected at random.

Mark scheme (a)

Scheme	Marks
\(\mathrm{P}(W \lt 224) = \mathrm{P}\left(z \lt \dfrac{224 - 232}{5}\right)\) \(= \mathrm{P}(z \lt -1.6)\)	M1
\(= 1 - 0.9452\)	M1
\(= 0.0548\) awrt 0.0548	A1
	(3)

Notes

M1 for standardising with 232 and 5. (i.e. not \(5^2\) or \(\sqrt{5}\)). Accept \(\pm\dfrac{w - 232}{5}\).
M1 for finding (1- a probability > 0.5)
A1 awrt 0.0548

Mark scheme (b)

Scheme	Marks
0.5 – 0.2 = 0.3 0.3 or 0.7 seen	M1
\(\dfrac{w - 232}{5} = 0.5244\) 0.5244 seen	B1; M1
\(w = 234.622\) awrt 235	A1
	(4)

Notes

M1 Can be implied by use of \(\pm\) 0.5244 or \(\pm\) (0.52 to 0.53)
B1 for \(\pm\) 0.5244 only.
Second M1 standardise with 232 and 5 and equate to \(z\) value of (0.52 to 0.53) or (0.84 to 0.85)
1 – z used award second M0.
Require consistent signs i.e. \(\dfrac{232 - w}{5} = -0.5244\) or negative z value for M1.
A1 dependent upon second M mark for awrt 235 but see note below.
Common errors involving probabilities and not z values:
P(Z<0.2) = 0.5793 used instead of \(z\) value gives awrt 235 but award M0B0M0A0
P(Z<0.8) = 0.7881 used instead of \(z\) value award M0B0M0A0.
M1B0M0A0 for 0.6179, M1B0M0A0 for 0.7580

Scheme	Marks
\(0.2 \times (1 - 0.2)\)	M1
\(2 \times 0.8 \times (1 - 0.8) = 0.32\)	M1 A1
	(3)
	(10 marks)