S3 June 2008 Q4

EdexcelOld spec11 marksContinuous Random Variables

4. The weights of adult men are normally distributed with a mean of 84 kg and a standard deviation of 11 kg.

(a) Find the probability that the total weight of 4 randomly chosen adult men is less than 350 kg. (5)

The weights of adult women are normally distributed with a mean of 62 kg and a standard deviation of 10 kg.

(b) Find the probability that the weight of a randomly chosen adult man is less than one and a half times the weight of a randomly chosen adult woman. (6)

Mark scheme (a)

Scheme	Marks
\(X = M_1 + M_2 + M_3 + M_4 \sim \mathrm{N}(336, 22^2)\) \(\mu = \) 336	B1
\(\sigma^2 = 22^2\) or 484	B1
\(\mathrm{P}(X \lt 350) = \mathrm{P}\left(Z \lt \dfrac{350 - 336}{22}\right)\)	M1
\(= \mathrm{P}(Z \lt 0.64)\) awrt 0.64	A1
\(=\) awrt 0.738 or 0.739	A1
	(5)

Notes

2^nd B1 for \(\sigma = 22\) or \(\sigma^2 = 22^2\) or 484
M1 for standardising with their mean and standard deviation (ignore direction of inequality)

Mark scheme (b)

Scheme	Marks
\(M \sim \mathrm{N}(84, 121)\) and \(W \sim \mathrm{N}(62, 100)\) Let \(Y = M - 1.5W\)	M1
\(\mathrm{E}(Y) = 84 - 1.5 \times 62 = -9\)	A1
\(\mathrm{Var}(Y) = \mathrm{Var}(M) + 1.5^2\,\mathrm{Var}(W)\)	M1
\(= 11^2 + 1.5^2 \times 10^2 = 346\)	A1
\(\mathrm{P}(Y \lt 0),\ = \mathrm{P}(Z \lt 0.48\ldots) =\) awrt 0.684 ~ 0.686	M1, A1
	(6)
	(11 marks)

Notes

1^st M1 for attempting to find \(Y\). Need to see \(\pm(M - 1.5W)\) or equiv. May be implied by \(\mathrm{Var}(Y)\).
1^st A1 for a correct value for their \(\mathrm{E}(Y)\) i.e. usually \(\pm 9\). Do not give M1A1 for a “lucky” \(\pm 9\).

2^nd M1 for attempting \(\mathrm{Var}(Y)\) e.g. \(\ldots + 1.5^2 \times 10^2\) or \(11^2 + 1.5^2 \times \ldots\)

3^rd M1 for attempt to calculate the correct probability. Must be attempting a probability > 0.5.
Must attempt to standardise with a relevant mean and standard deviation

Using \(\sigma^2_M = 11\) or \(\sigma^2_W = 10\) is not a misread.