OCR S2 2012 January — Question 1 4 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2012
SessionJanuary
Marks4
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TopicMeasures of Location and Spread
TypeStandard unbiased estimates calculation
DifficultyEasy -1.2 This is a straightforward application of standard formulas for unbiased estimates of mean and variance from summary statistics. It requires only direct substitution into well-rehearsed formulas (x̄ = Σx/n and s² = [Σx² - (Σx)²/n]/(n-1)) with minimal calculation, making it easier than average and purely procedural with no problem-solving element.
Spec5.05b Unbiased estimates: of population mean and variance
A random sample of 50 observations of the random variable \(X\) is summarised by $$n = 50, \Sigma x = 182.5, \Sigma x^2 = 739.625.$$ Calculate unbiased estimates of the expectation and variance of \(X\). [4]
AnswerMarks Guidance
\(\bar{u} = \bar{x} = 3.65\)M1 Correct formula for biased estimate used, award if 1.47 seen
\(s^2 = \frac{739.625}{50} - 3.65^2 = [1.47]\)M1 \(n/(n-1)\) factor used, or if wrong single formula, M1 if \(n-1\) divisor anywhere. Correct single formula: M2
\(\sigma^2 = \frac{50}{49}s^2\)
\(= 1.5\)A1 Answer 1.5 or exact equivalent only
[4] 
$\bar{u} = \bar{x} = 3.65$ | M1 | Correct formula for biased estimate used, award if 1.47 seen
$s^2 = \frac{739.625}{50} - 3.65^2 = [1.47]$ | M1 | $n/(n-1)$ factor used, or if wrong single formula, M1 if $n-1$ divisor anywhere. Correct single formula: M2
$\sigma^2 = \frac{50}{49}s^2$ | | 
$= 1.5$ | A1 | Answer 1.5 or exact equivalent only
| [4] |
A random sample of 50 observations of the random variable $X$ is summarised by

$$n = 50, \Sigma x = 182.5, \Sigma x^2 = 739.625.$$

Calculate unbiased estimates of the expectation and variance of $X$. [4]

\hfill \mbox{\textit{OCR S2 2012 Q1 [4]}}
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