| Date | November 2007 | Marks available | 5 | Reference code | 07N.2.sl.TZ0.1 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Copy | Question number | 1 | Adapted from | N/A |
Question
When Geraldine travels to work she can travel either by car (C), bus (B) or train (T). She travels by car on one day in five. She uses the bus 50 % of the time. The probabilities of her being late (L) when travelling by car, bus or train are 0.05, 0.12 and 0.08 respectively.
It is not necessary to use graph paper for this question.
Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.
Find the probability that Geraldine travels by bus and is late.
Find the probability that Geraldine is late.
Find the probability that Geraldine travelled by train, given that she is late.
Sketch the curve of the function \(f (x) = x^3 − 2x^2 + x − 3\) for values of \(x\) from −2 to 4, giving the intercepts with both axes.
On the same diagram, sketch the line \(y = 7 − 2x\) and find the coordinates of the point of intersection of the line with the curve.
Find the value of the gradient of the curve where \(x = 1.7\) .
Markscheme
Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages. (A5)
[5 marks]
0.06 (accept \(0.5 \times 0.12\) or 6%) (A1)(ft)
[1 mark]
for a relevant two-factor product, either \(C \times L\) or \(T \times L\) (M1)
for summing three two-factor products (M1)
\((0.2 \times 0.05 + 0.06 + 0.3 \times 0.08)\)
0.094 (A1)(ft)(G2)
[3 marks]
\(\frac{{0.3 \times 0.08}}{{0.094}}\) (M1)(A1)(ft)
award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution
= 0.255 (A1)(ft)(G2)
[3 marks]
(G3)
[3 marks]
line drawn with –ve gradient and +ve y-intercept (G1)
(2.45, 2.11) (G1)(G1)
[3 marks]
\(f ' (1.7) = 3(1.7)^2 - 4(1.7) + 1\) (M1)
award (M1) for substituting in their \(f' (x)\)
2.87 (A1)(G2)
[2 marks]
Examiners report
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.
This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.
The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.
