| Date | May 2018 | Marks available | 2 | Reference code | 18M.1.hl.TZ2.3 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The discrete random variable X has the following probability distribution, where p is a constant.
Find the value of p.
[2]
a.
Find μ, the expected value of X.
[2]
b.i.
Find P(X > μ).
[2]
b.ii.
Markscheme
equating sum of probabilities to 1 (p + 0.5 − p + 0.25 + 0.125 + p3 = 1) M1
p3 = 0.125 = \(\frac{1}{8}\)
p= 0.5 A1
[2 marks]
a.
μ = 0 × 0.5 + 1 × 0 + 2 × 0.25 + 3 × 0.125 + 4 × 0.125 M1
= 1.375 \(\left( { = \frac{{11}}{8}} \right)\) A1
[2 marks]
b.i.
P(X > μ) = P(X = 2) + P(X = 3) + P(X = 4) (M1)
= 0.5 A1
Note: Do not award follow through A marks in (b)(i) from an incorrect value of p.
Note: Award M marks in both (b)(i) and (b)(ii) provided no negative probabilities, and provided a numerical value for μ has been found.
[2 marks]
b.ii.
Examiners report
[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
