Question
HLPaper 2
The random variable X has a binomial distribution with parameters n and p.
It is given that E(X) = 3.5.
1.[2]
Find the least possible value of n.
Verified
Solution
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
np= 3.5 (A1)
p ≤ 1 ⇒ least n= 4 A1
[2 marks]
2.[5]
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.
Verified
Solution
(1− p)n + np(1−p)n−1 = 0.09478 M1A1
attempt to solve above equation with np= 3.5 (M1)
n = 12, p = (=0.292) A1A1
Note: Do not accept n as a decimal.
[5 marks]