AQA Paper 3 2021 June — Question 11 1 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2021
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeFinding binomial parameters from properties
DifficultyEasy -1.2 This is a straightforward recall question requiring only the binomial distribution formulas (mean = np, variance = np(1-p)) and simple algebra to find p = 1 - 144/225 = 0.64. It's a single-step calculation with multiple choice answers provided, making it easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities

The random variable \(X\) is such that \(X \sim B(n, p)\) The mean value of \(X\) is 225 The variance of \(X\) is 144 Find \(p\). Circle your answer. [1 mark] 0.36 \quad 0.6 \quad 0.64 \quad 0.8

Question 11:
AnswerMarks Guidance
11Circles correct answer 1.1b
Total1
QMarking Instructions AO
Question 11:
11 | Circles correct answer | 1.1b | B1 | 0.36
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
The random variable $X$ is such that $X \sim B(n, p)$

The mean value of $X$ is 225

The variance of $X$ is 144

Find $p$.

Circle your answer.
[1 mark]

0.36 \quad 0.6 \quad 0.64 \quad 0.8

\hfill \mbox{\textit{AQA Paper 3 2021 Q11 [1]}}