OCR S2 2011 January — Question 2 6 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2011
SessionJanuary
Marks6
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Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeFinding n from sample mean distribution
DifficultyStandard +0.3 This is a standard S2 CLT application requiring students to set up two equations using the sampling distribution N(μ, 25/n), convert probabilities to z-scores, and solve simultaneous equations. While it involves multiple steps and algebraic manipulation, it follows a well-practiced procedure with no conceptual surprises—slightly easier than average for A-level.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem
2 The random variable \(H\) has the distribution \(\mathrm { N } \left( \mu , 5 ^ { 2 } \right)\). The mean of a sample of \(n\) observations of \(H\) is denoted by \(\bar { H }\). It is given that \(\mathrm { P } ( \bar { H } > 53.28 ) = 0.0250\) and \(\mathrm { P } ( \bar { H } < 51.65 ) = 0.0968\), both correct to 4 decimal places. Find the values of \(\mu\) and \(n\).
AnswerMarks Guidance
\(\frac{53.28}{5/\sqrt{n}} = 1.96\)M1dep Standardise with \(\sqrt{n}\) once & equate to \(z\), allow sign, square/√ errors
\(\mu - 51.65 = 1.3\) / \(5 / \sqrt{n}\)A1 twice, signs correct, \(z\) s may be wrong
\(\sqrt{n} = 10\), \(n = 100\)B1 Both correct; \(z\) values seen
\(\mu = 52.3\)A1 \(n = 100\), not from wrong signs
B1a.r.t. 52.3, right arithmetic needed but \(\sqrt{n}\) can be omitted
depM1Solve to get \(\sqrt{n}\) or \(\mu\), needs first M1
6 
$\frac{53.28}{5/\sqrt{n}} = 1.96$ | M1dep | Standardise with $\sqrt{n}$ once & equate to $z$, allow sign, square/√ errors
$\mu - 51.65 = 1.3$ / $5 / \sqrt{n}$ | A1 | twice, signs correct, $z$ s may be wrong
$\sqrt{n} = 10$, $n = 100$ | B1 | Both correct; $z$ values seen
$\mu = 52.3$ | A1 | $n = 100$, not from wrong signs
| B1 | a.r.t. 52.3, right arithmetic needed but $\sqrt{n}$ can be omitted
| depM1 | Solve to get $\sqrt{n}$ or $\mu$, needs first M1
| | 6
2 The random variable $H$ has the distribution $\mathrm { N } \left( \mu , 5 ^ { 2 } \right)$. The mean of a sample of $n$ observations of $H$ is denoted by $\bar { H }$. It is given that $\mathrm { P } ( \bar { H } > 53.28 ) = 0.0250$ and $\mathrm { P } ( \bar { H } < 51.65 ) = 0.0968$, both correct to 4 decimal places. Find the values of $\mu$ and $n$.

\hfill \mbox{\textit{OCR S2 2011 Q2 [6]}}
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