OCR Further Statistics 2018 September — Question 3 7 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2018
SessionSeptember
Marks7
TopicCentral limit theorem
TypeDiscrete uniform distribution sample mean
DifficultyStandard +0.8 This is a straightforward application of the Central Limit Theorem to a discrete uniform distribution, requiring students to identify the appropriate normal approximation, calculate the mean and variance of the sample mean, and apply a continuity correction. While it tests understanding of CLT mechanics and justification, it follows a standard template with no novel problem-solving required, making it moderately above average difficulty for Further Maths students.
Spec2.04e Normal distribution: as model N(mu, sigma^2)5.01a Permutations and combinations: evaluate probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)
3 A discrete random variable \(X\) has the distribution \(\mathrm { U } ( 11 )\).
The mean of 50 observations of \(X\) is denoted by \(\bar { X }\).
Use an approximate method, which should be justified, to find \(\mathrm { P } ( \bar { X } \leqslant 6.10 )\).
AnswerMarks Guidance
\(E(X) = \frac{1}{2}(n+1) = 6\)B1 6 stated or implied
\(\text{Var}(X) = \frac{1}{12}(n^2-1)\)M1 using correct formula with \(n = 11\)
\(= \frac{1}{12}(11^2-1) = 10\)A1 Obtain 10, www
\(\bar{X} \sim N(6, \ldots 10 \div 50)\)M1 Normal stated or implied, their 6
A1FT1.1 Variance (their 10) \(\div\) 50
\(P(\bar{X} \leq 6.10) = \Phi\left(\frac{6.11-6}{\sqrt{0.2}}\right) = \Phi(0.2460)\)A1 2.2a
\(= 0.597146\)A1 [7] 
$E(X) = \frac{1}{2}(n+1) = 6$ | B1 | 6 stated or implied

$\text{Var}(X) = \frac{1}{12}(n^2-1)$ | M1 | using correct formula with $n = 11$

$= \frac{1}{12}(11^2-1) = 10$ | A1 | Obtain 10, www

$\bar{X} \sim N(6, \ldots 10 \div 50)$ | M1 | Normal stated or implied, their 6

| A1FT | 1.1 | Variance (their 10) $\div$ 50 | Allow SD/var muddles

$P(\bar{X} \leq 6.10) = \Phi\left(\frac{6.11-6}{\sqrt{0.2}}\right) = \Phi(0.2460)$ | A1 | 2.2a | Answer, awrt 0.597 | No continuity correction (0.588): A1

$= 0.597146$ | A1 [7] |

---
3 A discrete random variable $X$ has the distribution $\mathrm { U } ( 11 )$.\\
The mean of 50 observations of $X$ is denoted by $\bar { X }$.\\
Use an approximate method, which should be justified, to find $\mathrm { P } ( \bar { X } \leqslant 6.10 )$.

\hfill \mbox{\textit{OCR Further Statistics 2018 Q3 [7]}}
This paper (9 questions)
View full paper
Q1 4 Q2 7 Q3 7 Q4 9 Q5 8 Q6 10 Q7 11 Q8 8 Q9 11