Date | November 2019 | Marks available | 6 | Reference code | 19N.2.SL.TZ0.S_7 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_7 | Adapted from | N/A |
Question
The following table shows the probability distribution of a discrete random variable X, where a⩾0 and b⩾0.
Show that b=0.3−a.
Find the difference between the greatest possible expected value and the least possible expected value.
Markscheme
correct approach A1
eg 0.2+0.5+b+a=1, 0.7+a+b=1
b=0.3−a AG N0
[1 mark]
correct substitution into E(X) (A1)
eg 0.2+4×0.5+a×b+(a+b−0.5)×a, 0.2+2+a×b−0.2a
valid attempt to express E(X) in one variable M1
eg 0.2+4×0.5+a×(0.3−a)+(−0.2)×a, 2.2+0.1a−a2,
0.2+4×0.5+(0.3−b)×b+(−0.2)×(0.3−b), 2.14+0.5b−b2
correct value of greatest E(X) (A1)
2.2025 (exact)
valid attempt to find least value (M1)
eg graph with minimum indicated, E(0) and E(0.3)
(0, 2.2) and (0.3, 2.14) if E(X) in terms of a
(0, 2.14) and (0.3, 2.2) if E(X) in terms of b
correct value of least E(X) (A1)
eg 2.14 (exact)
difference =0.0625 (exact) A1 N2
[6 marks]