OCR S2 2015 June — Question 1 6 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeStandard two probabilities given
DifficultyStandard +0.3 This is a standard S2 question requiring students to set up two equations using inverse normal tables (finding z-scores for given probabilities) and solve simultaneously for μ and σ. While it involves multiple steps, the method is routine and commonly practiced in S2 courses, making it slightly easier than average overall but typical for this module.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

1 The random variable \(Y\) is normally distributed with mean \(\mu\) and variance \(\sigma ^ { 2 }\). It is found that \(\mathrm { P } ( Y > 150.0 ) = 0.0228\) and \(\mathrm { P } ( Y > 143.0 ) = 0.9332\). Find the values of \(\mu\) and \(\sigma\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{150-\mu}{\sigma} = 2.00\)M1 Standardise with \(\sigma\), \(\mu\) at least once, ignore cc, \(\sqrt{}\) errors, equate to \(z\)
\(\frac{143-\mu}{\sigma} = -1.5\)A1 Both LHS and signs of RHS correct
Both \(z\)-values correct to 3 SFB1
Correct method for solutionM1
\(\mu \in [145.95, 146.05]\) wwwA1
\(\mu = 146\), \(\sigma = 2\)A1 \(\sigma \in [1.995, 2.005]\) or \(\sigma^2 = 4\) www
Total: 6 \(z\) not used e.g. equated to 0.0228 and 0.9332 or 0.5092 and 0.8246: max M0M1. One \(z\), one not: M1A0B0. \(\sqrt{\sigma}\) or \(\sigma^2\): can get M1A0B1M1A1A0
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{150-\mu}{\sigma} = 2.00$ | M1 | Standardise with $\sigma$, $\mu$ at least once, ignore cc, $\sqrt{}$ errors, equate to $z$ |
| $\frac{143-\mu}{\sigma} = -1.5$ | A1 | Both LHS and signs of RHS correct |
| Both $z$-values correct to 3 SF | B1 | |
| Correct method for solution | M1 | |
| $\mu \in [145.95, 146.05]$ www | A1 | |
| $\mu = 146$, $\sigma = 2$ | A1 | $\sigma \in [1.995, 2.005]$ or $\sigma^2 = 4$ www |
| **Total: 6** | | $z$ not used e.g. equated to 0.0228 and 0.9332 or 0.5092 and 0.8246: max M0M1. One $z$, one not: M1A0B0. $\sqrt{\sigma}$ or $\sigma^2$: can get M1A0B1M1A1A0 |

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1 The random variable $Y$ is normally distributed with mean $\mu$ and variance $\sigma ^ { 2 }$. It is found that $\mathrm { P } ( Y > 150.0 ) = 0.0228$ and $\mathrm { P } ( Y > 143.0 ) = 0.9332$. Find the values of $\mu$ and $\sigma$.

\hfill \mbox{\textit{OCR S2 2015 Q1 [6]}}