A2 October 2020 Q7

EdexcelCurrent spec17 marksContinuous Random Variables

7. Fence panels come in two sizes, large and small. The lengths of the large panels are normally distributed with mean 198 cm and standard deviation 5 cm. The lengths of the small panels are normally distributed with mean 74 cm and standard deviation 3 cm.

(a) Find the probability that the total length of a random sample of 3 large panels is greater than the total length of a random sample of 8 small panels. (6)

One large panel and one small panel are selected at random.

(b) Find the probability that the length of the large panel is more than \(\dfrac{8}{3}\) times the length of the small panel. (5)

Rosa needs 1000 cm of fencing. The large panels cost £80 each and the small panels cost £30 each. Rosa’s plan is to buy 5 large panels and measure the total length. If the total length is less than 1000 cm she will then buy one small panel as well.

Mark scheme (a)

Scheme	Marks	AO
\(X = L_1 + L_2 + L_3 \sim \mathrm{N}\left(594,\ \sqrt{75}^{\,2}\right)\) \(Y = S_1 + \ldots + S_8 \sim \mathrm{N}\left(592,\ \sqrt{72}^{\,2}\right)\)	M1 A1 A1	3.3 1.1b 1.1b
\(\mathrm{P}(X \gt Y) = \mathrm{P}(D \gt 0)\) where \(D \sim \mathrm{N}\left(2,\ \sqrt{147}^{\,2}\right)\)	M1 A1ft	2.1 1.1b
\(= 0.56551\ldots\) awrt 0.566	A1	3.4
	(6)

Notes

1^st M1 for an attempt at \(X\) or \(Y\) – expression or implied by one correct distribution

1^st A1 for a correct distribution for \(X\) or implied by \(\mathrm{E}(D) = 2\)

2^nd A1 for a correct distribution for \(Y\) or implied by \(\mathrm{Var}(D) = 147\)

2^nd M1 for a correct strategy – attempt \(X - Y\) and \(\mathrm{P}(D \gt 0)\) statement

3^rd A1ft for a correct distribution for \(D\) ft their \(X\) and \(Y\)

4^th A1 for awrt 0.566

Mark scheme (b)

Scheme	Marks	AO
\(W = L - \dfrac{8}{3}S \quad \Rightarrow \quad W \sim \mathrm{N}\left(\dfrac{2}{3},\ 25 + \dfrac{64}{9} \times 9\right) = \mathrm{N}\left(\dfrac{2}{3},\ \sqrt{89}^{\,2}\right)\)	M1 M1,A1	3.3 2.1,1.1b
\(\mathrm{P}(W \gt 0) = 0.528168\ldots\) awrt 0.528	M1A1	3.4,1.1b
	(5)

Notes

1^st M1 for attempt at a correct model (normal and mean)

2^nd M1 for correct expression for variance of their model provided of the form \(L - kS\) or \(kL - S\)

1^st A1 for a fully correct distribution

3^rd M1 for a correct probability statement using their distribution

2^nd A1 for awrt 0.528

Mark scheme (c)

Scheme	Marks	AO
\(F = L_1 + \ldots + L_5 \sim \mathrm{N}\left(990,\ \sqrt{125}^{\,2}\right)\)	M1 A1	3.1b 1.1b
\(\mathrm{P}(F \lt 1000) = 0.814455\ldots\) (o.e.)	A1	3.4
E(cost of Rosa’s plan) = \(430 \times \text{“}0.814\ldots\text{”} + 400 \times \left(1 - \text{“}0.814\ldots\text{”}\right)\)	M1	2.1
= £ 424.43	A1	1.1b
Buying 14 small panels cost \(14 \times 30 =\) £420 So Rosa’s plan is likely to be more expensive	A1	3.2a
	(6)
	(17 marks)

Notes

1^st M1 for a correct start to solve the problem attempt at \(F\) and correct mean

1^st A1 for a correct distribution

2^nd A1 for using this model to find \(\mathrm{P}(F \lt 1000)\) = awrt 0.814 or \(\mathrm{P}(F \gt 1000)\) = awrt 0.186

2^nd M1 for a correct strategy to solve the problem i.e. attempt at expected cost ft their prob

3^rd A1 for awrt £424

4^th A1 for a correct conclusion must have comparison with £420 and reject Rosa’s plan

[(c) is an extended problem and a 3.1, 3.2 question]