Date | May 2015 | Marks available | 3 | Reference code | 15M.1.sl.TZ1.1 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
A discrete random variable X has the following probability distribution.
Find p.
Find E(X).
Markscheme
summing probabilities to 1 (M1)
eg,∑=1, 3+4+2+x=10
correct working (A1)
310+410+210+p=1, p=1−910
p=110 A1 N3
[3 marks]
correct substitution into formula for E(X) (A1)
eg0(310)+…+3(p)
correct working (A1)
eg410+410+310
E(X)=1110(1.1) A1 N2
[3 marks]
Total [6 marks]
Examiners report
Most candidates were able to find p, however expectation emerged as surprisingly more difficult. Quite often E(X)/4 was found or candidates wrote the formula with no further work.
Most candidates were able to find p, however expectation emerged as surprisingly more difficult. Quite often E(X)/4 was found or candidates wrote the formula with no further work.