S2 January 2007 Q3

EdexcelOld spec15 marksBinomial Distribution Normal Distribution

3. For a particular type of plant 45% have white flowers and the remainder have coloured flowers. Gardenmania sells plants in batches of 12. A batch is selected at random.

Calculate the probability that this batch contains

(a) exactly 5 plants with white flowers, (3)

(b) more plants with white flowers than coloured ones. (2)

Gardenmania takes a random sample of 10 batches of plants.

Due to an increasing demand for these plants by large companies, Gardenmania decides to sell them in batches of 50.

(d) Use a suitable approximation to calculate the probability that a batch of 50 plants contains more than 25 plants with white flowers. (7)

Mark scheme (a)

Scheme	Marks
Let \(W\) represent the number of white plants. \(W \sim \mathrm{B}(12, 0.45)\)	B1
\(\mathrm{P}(W = 5) = \mathrm{P}(W \leqslant 5) - \mathrm{P}(W \leqslant 4)\)	M1
\(= 0.5269 - 0.3044\) \(= 0.2225\)	A1
	(3)

Notes

M1 use of \({}^{12}\mathrm{C}_5\,0.45^5\,0.55^7\) or equivalent award B1M1

values from correct table implies B

A1 awrt 0.222(5)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(W \geqslant 7) = 1 - \mathrm{P}(W \leqslant 6) \quad\) or \(= 1 - \mathrm{P}(W \lt 7)\)	M1
\(= 1 - 0.7393\) \(= 0.2607\)	A1
	(2)

Notes

M1 \(1 - 0.7393\) implies method

A1 awrt 0.261

Scheme	Marks
P(3 contain more white than coloured) \(= \dfrac{10!}{3!7!}(0.2607)^3(1 - 0.2607)^7\)	M1A1ft
\(= 0.256654\ldots\)	A1
	(3)

Mark scheme (d)

Scheme	Marks
mean \(= np = 22.5\); var \(= npq = 12.375\)	B1B1
\(\mathrm{P}(W \gt 25) \approx \mathrm{P}\left(Z \gt \dfrac{25.5 - 22.5}{\sqrt{12.375}}\right)\)	M1;M1
\(\approx \mathrm{P}(Z \gt 0.8528..)\)	A1
\(\approx 1 - 0.8023\)	M1
\(\approx 0.1977\)	A1
	(7)
	(15 marks)

Notes

M1;M1 \(\pm\) standardise with \(\sigma\) and \(\mu\); \(\pm 0.5\) c.c.

1st A1 awrt 0.85

3rd M1 ‘one minus’

2nd A1 awrt 0.197 or 0.198