OCR S2 2011 June — Question 2 7 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind k given probability statement
DifficultyStandard +0.3 This is a slightly above-average S2 question requiring understanding of normal distribution symmetry to find μ and σ from equal tail probabilities, then using tables to find a specific percentile. The symmetry insight is straightforward, and the calculations are routine once parameters are identified.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
2 The random variable \(Y\) has the distribution \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). It is given that $$\mathrm { P } ( Y < 48.0 ) = \mathrm { P } ( Y > 57.0 ) = 0.0668 .$$ Find the value \(y _ { 0 }\) such that \(\mathrm { P } \left( Y > y _ { 0 } \right) = 0.05\).
Question 2:
Part \(\alpha\):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mu = \frac{48+57}{2} = 52.5\)M1, A1 Use symmetry to find \(\mu\); obtain \(\mu = 52.5\)
\(\Phi^{-1}(0.9332) = 1.5\)B1 1.5 seen, e.g. in \(4.5 \div 1.5\)
\(4.5 \div 1.5 \quad [\sigma = 3]\)M1 \(4.5 \div\) their \(\Phi^{-1}\), or \(1.645 \div\) their \(\Phi^{-1}\), must be \(+\)ve, allow cc
Part \(\beta\):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{57-\mu}{\sigma} = 1.5,\ \frac{48-\mu}{\sigma} = -1.5\)M1, A1 M1 for one, ignoring cc, \(\sigma^2\), sign or "\(1-\)" errors, RHS must be \(\Phi^{-1}\) (*not* \(\Phi\)). [e.g. 0.8246 or 0.5267] or 0.0668 or 0.9332; A1 for both completely correct except for value of \(z\)
Solve simultaneously: \(\mu = 52.5 \quad [\sigma = 3]\)B1, A1 \(z = 1.5\) or \(-1.5\) in at least one equation. Solve without obvious errors, get \(\mu = 52.5\), OK from wrong \(z\)
\(\mu + \frac{4.5}{1.5} \times 1.645 = \mathbf{57.4}(35)\)M1, B1, A1 7 
# Question 2:

## Part $\alpha$:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = \frac{48+57}{2} = 52.5$ | M1, A1 | Use symmetry to find $\mu$; obtain $\mu = 52.5$ |
| $\Phi^{-1}(0.9332) = 1.5$ | B1 | 1.5 seen, e.g. in $4.5 \div 1.5$ |
| $4.5 \div 1.5 \quad [\sigma = 3]$ | M1 | $4.5 \div$ their $\Phi^{-1}$, or $1.645 \div$ their $\Phi^{-1}$, must be $+$ve, allow cc |

## Part $\beta$:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{57-\mu}{\sigma} = 1.5,\ \frac{48-\mu}{\sigma} = -1.5$ | M1, A1 | M1 for one, ignoring cc, $\sigma^2$, sign or "$1-$" errors, RHS must be $\Phi^{-1}$ (*not* $\Phi$). [e.g. 0.8246 or 0.5267] or 0.0668 or 0.9332; A1 for both completely correct except for value of $z$ |
| Solve simultaneously: $\mu = 52.5 \quad [\sigma = 3]$ | B1, A1 | $z = 1.5$ or $-1.5$ in at least one equation. Solve without obvious errors, get $\mu = 52.5$, OK from wrong $z$ |
| $\mu + \frac{4.5}{1.5} \times 1.645 = \mathbf{57.4}(35)$ | M1, B1, A1 | **7** | $\mu + z\sigma$ [their $\mu$ and $\sigma$, anything recognisable as $z$]; $z = 1.645$ seen; answer in range [57.4, 57.45], cwo |

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2 The random variable $Y$ has the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. It is given that

$$\mathrm { P } ( Y < 48.0 ) = \mathrm { P } ( Y > 57.0 ) = 0.0668 .$$

Find the value $y _ { 0 }$ such that $\mathrm { P } \left( Y > y _ { 0 } \right) = 0.05$.

\hfill \mbox{\textit{OCR S2 2011 Q2 [7]}}
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