Date | November 2012 | Marks available | 3 | Reference code | 12N.1.sl.TZ0.2 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The following table shows the probability distribution of a discrete random variable X .
Find the value of k .
Find \({\text{E}}(X)\) .
Markscheme
evidence of summing to 1 (M1)
e.g. \(\sum\limits_{}^{} {p = 1{\text{, }}0.3 + k + 2k + 0.1 = 1} \)
correct working (A1)
e.g. \(0.4 + 3k{\text{, }}3k = 0.6\)
\(k = 0.2\) A1 N2
[3 marks]
correct substitution into formula \({\text{E}}(X)\) (A1)
e.g. \(0(0.3) + 2(k) + 5(2k) + 9(0.1){\text{, }}12k + 0.9\)
correct working
e.g. \(0(0.3) + 2(0.2) + 5(0.4) + 9(0.1){\text{, }}0.4 + 2.0 + 0.9\) (A1)
\({\text{E}}(X)\)= 3.3 A1 N2
[3 marks]
Examiners report
Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was \(5(0.4) = 0.2\) .
Overall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they were unfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, but then had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was \(5(0.4) = 0.2\) .