| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.hl.TZ2.8 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
The random variable X has a binomial distribution with parameters n and p.
It is given that E(X) = 3.5.
Find the least possible value of n.
[2]
a.
It is further given that P(X ≤ 1) = 0.09478 correct to 4 significant figures.
Determine the value of n and the value of p.
[5]
b.
Markscheme
np = 3.5 (A1)
p ≤ 1 ⇒ least n = 4 A1
[2 marks]
a.
(1 − p)n + np(1 − p)n−1 = 0.09478 M1A1
attempt to solve above equation with np = 3.5 (M1)
n = 12, p = \(\frac{7}{{24}}\) (=0.292) A1A1
Note: Do not accept n as a decimal.
[5 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.
