S1 June 2011 Q2

EdexcelOld spec5 marksNormal Distribution

2. The random variable \(X \sim \mathrm{N}(\mu, 5^2)\) and \(\mathrm{P}(X \lt 23) = 0.9192\)

(a) Find the value of \(\mu\). (4)

(b) Write down the value of \(\mathrm{P}(\mu \lt X \lt 23)\). (1)

Mark scheme (a)

Scheme	Marks
awrt \(\pm\) 1.40	B1
\(\dfrac{23 - \mu}{5} = \text{"}1.40\text{"}\) (o.e)	M1A1ft
\(\mu = 16\) (or awrt 16.0)	A1
	(4)

Notes

B1 for awrt \(\pm\) 1.40 or better seen anywhere. Condone 1.4 instead of 1.40
M1 for attempting to standardise with 23 and 5 and \(\mu\), accept \(\pm\)
e.g. \(\dfrac{23 - \mu}{25} = 1.40\) can score B1M0 (since using 25 not 5 for standardising)
\(\dfrac{23 - \mu}{5} = 0.9192\) can score B0M1 (since have correct standardisation)
Can accept equivalent equations e.g. \(23 - \mu = 5 \times \text{"}1.40\text{"}\)

1st A1ft for standardised expression = to a \(z\) value (\(|z| \gt 1\)). Signs must be compatible.
Follow through their \(z\)
e.g. \(\dfrac{23 - \mu}{5}\) = their \(z\) where \(z \gt 1\) or \(\dfrac{\mu - 23}{5}\) = their \(z\) where \(z \lt -1\)
2nd A1 for 16 or awrt 16.0 if they are using a more accurate \(z\)

Correct answer only scores 4/4 but if any working is seen apply scheme

Scheme	Marks
0.4192	B1
	(1)
	(5 marks)