S1 June 2018 Q7

EdexcelOld spec13 marksNormal Distribution Probability

7. Farmer Adam grows potatoes. The weights of potatoes, in grams, grown by Adam are normally distributed with a mean of 140 g and a standard deviation of 40 g.

Adam cannot sell potatoes with a weight of less than 92 g.

(a) Find the percentage of potatoes that Adam grows but cannot sell. (3)

The upper quartile of the weight of potatoes sold by Adam is \(q_3\)

(b) Find the probability that the weight of a randomly selected potato grown by Adam is more than \(q_3\) (2)

Betty selects a random sample of 3 potatoes sold by Adam.

(d) Find the probability that one weighs less than \(q_1\), one weighs more than \(q_3\) and one has a weight between \(q_1\) and \(q_3\) (3)

Mark scheme (a)

Scheme	Marks
\([W \sim \mathrm{N}(140, 40^2)]\) \(\mathrm{P}(W \lt 92) = \mathrm{P}\left(Z \lt \dfrac{92 - 140}{40}\right) = [\mathrm{P}(Z \lt -1.2)]\)	M1
\(= 1 - 0.8849\) = awrt 11.5 (%) or 0.115	dM1,A1
	(3)

Notes

Condone poor use of notation etc e.g. “P > \(q_1\)” for P(\(W\) > \(q_1\)) etc

1st M1 for standardising attempt with 92 or 188, 140 and 40 (o.e.) Accept \(\pm\) ignore inequality

2nd dM1 dependent on 1st M1, for attempting 1 – \(p\) where \(0.5 \lt p \lt 1\)

A1 for awrt 11.5 (%) or 0.115

Mark scheme (b)

Scheme	Marks
\([\mathrm{P}(W \gt q_3) = \mathrm{P}(W \gt 92) \times \mathrm{P}(W \gt q_3 \mid W \gt 92) =]\) \((1 - \text{(a)}) \times 0.25 = 0.8849 \times 0.25\)	M1
\(= 0.221225\) = awrt 0.221	A1
	(2)

Notes

M1 for (1 – their (a))\(\times\)0.25 or \(1 - [(1 - (a)) \times 0.75 + (a)] = 1 - [0.8849 \times 0.75 + 0.1151]\)

A1 for awrt 0.221

Sketch of normal curve for W: 0.115 shaded below 92, 0.221 between 92 and Q1 = 123, mean 140, 0.221 above Q3 = 171 — Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(W \lt q_1 \mid W \gt 92) = 0.25\) or \(\mathrm{P}(W \gt q_1 \mid W \gt 92) = 0.75\)	M1
\(\mathrm{P}(92 \lt W \lt q_1) = 0.25 \times 0.8849 = \text{"}0.221..\text{"}\) or \(\mathrm{P}(W \gt q_1) = 0.75 \times 0.8849 = 0.663675\)	M1
\(\mathrm{P}(W \lt q_1) = 0.221225 + 0.115\) = awrt 0.336 or \(\mathrm{P}(W \gt q_1) = 0.663675\) = awrt 0.664	A1
\(\dfrac{q_1 - 140}{40} = -0.42\) (calculator gives \(-0.422513 \sim -0.423404\) )	M1
so \(q_1 = 123.2\) = awrt 123 (g)	A1
	(5)

Mark scheme (d)

Scheme	Marks
\(\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{2} \times 3!\)	M1M1
\(= \dfrac{3}{16}\) or 0.1875	A1
	(3)
	(13 marks)

Notes

1st M1 for \(0.25 \times 0.25 \times 0.5\) (o.e.) e.g. \(\frac{1}{32}\) may be seen as decimals or fractions

2nd M1 for \(\times 3!\) or \(\times 6\) or adding all 6 cases. Must be multiplying probabilities.

A1 for \(\frac{3}{16}\) or any exact equivalent