OCR S2 2007 June — Question 1 6 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2007
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeUnbiased estimator from summary statistics
DifficultyModerate -0.8 Part (i) is direct application of standard formulas for unbiased estimators (sample mean and variance with n-1 denominator) requiring only substitution of given values. Part (ii) tests conceptual understanding of the Central Limit Theorem—recognizing that with n=100, the CLT applies regardless of X's distribution—but requires no calculation. This is straightforward bookwork with minimal problem-solving.
Spec5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance

1 A random sample of observations of a random variable \(X\) is summarised by $$n = 100 , \quad \Sigma x = 4830.0 , \quad \Sigma x ^ { 2 } = 249 \text { 509.16. }$$
  1. Obtain unbiased estimates of the mean and variance of \(X\).
  2. The sample mean of 100 observations of \(X\) is denoted by \(\bar { X }\). Explain whether you would need any further information about the distribution of \(X\) in order to estimate \(\mathrm { P } ( \bar { X } > 60 )\). [You should not attempt to carry out the calculation.]

AnswerMarks Guidance
\(\mu = 48.3\)B1 48.3 seen
Biased estimate: 162.2016; can get B1M1M0M1
Multiply by \(n/(n-1)\)M1
\(= 163.84\)A1 Answer, 164 or 163.8 or 163.84
No, Central Limit theorem applies, so can assume distribution is normalB2 "No" with statement showing CLT is understood (though CLT does not need to be mentioned) [SR: No with reason that is not wrong: B1]
$\mu = 48.3$ | B1 | 48.3 seen
Biased estimate: 162.2016; can get B1M1M0 | M1 |
Multiply by $n/(n-1)$ | M1 |
$= 163.84$ | A1 | Answer, 164 or 163.8 or 163.84

No, Central Limit theorem applies, so can assume distribution is normal | B2 | "No" with statement showing CLT is understood (though CLT does not need to be mentioned) [SR: No with reason that is not wrong: B1]
1 A random sample of observations of a random variable $X$ is summarised by

$$n = 100 , \quad \Sigma x = 4830.0 , \quad \Sigma x ^ { 2 } = 249 \text { 509.16. }$$

(i) Obtain unbiased estimates of the mean and variance of $X$.\\
(ii) The sample mean of 100 observations of $X$ is denoted by $\bar { X }$. Explain whether you would need any further information about the distribution of $X$ in order to estimate $\mathrm { P } ( \bar { X } > 60 )$. [You should not attempt to carry out the calculation.]

\hfill \mbox{\textit{OCR S2 2007 Q1 [6]}}