Date | May 2012 | Marks available | 6 | Reference code | 12M.1.sl.TZ2.4 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The random variable X has the following probability distribution, with \({\rm{P}}(X > 1) = 0.5\) .
Find the value of r .
Given that \({\rm{E}}(X) = 1.4\) , find the value of p and of q .
Markscheme
attempt to substitute \({\rm{P}}(X > 1) = 0.5\) (M1)
e.g. \(r + 0.2 = 0.5\)
\(r = 0.3\) A1 N2
[2 marks]
correct substitution into \({\rm{E}}(X)\) (seen anywhere) (A1)
e.g. \(0 \times p + 1 \times q + 2 \times r + 3 \times 0.2\)
correct equation A1
e.g. \(q + 2 \times 0.3 + 3 \times 0.2 = 1.4\) , \(q + 1.2 = 1.4\)
\(q = 0.2\) A1 N1
evidence of choosing \(\sum {{p_i} = 1} \) M1
e.g. \(p + 0.2 + 0.3 + 0.2 = 1\) , \(p + q = 0.5\)
correct working (A1)
\(p + 0.7 = 1\) , \(1 - 0.2 - 0.3 - 0.2\) , \(p + 0.2 = 0.5\)
\(p = 0.3\) A1 N2
Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award the final A mark for an inappropriate value of p.
[6 marks]
Examiners report
The majority of candidates were successful in earning full marks on this question.
In part (b), a small number of candidates did not use the correct formula for \({\rm{E}}(X)\) , even though this formula is given in the formula booklet. There were also a few candidates who incorrectly assumed that \(p = 0\) , forgetting that the sum of the probabilities must equal 1. There were a few candidates who left this question blank, which raises concerns about whether they had been exposed to probability distributions during the course.