| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Rounded or discrete from continuous |
| Difficulty | Standard +0.3 This is a standard S2 normal distribution question covering routine techniques: z-score calculations for probabilities, inverse normal for percentiles, continuity correction, and a one-sample z-test. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Wing length} = 131) = P\left(\frac{130.5-130.5}{\sqrt{11.84}} \leq Z \leq \frac{131.5-130.5}{\sqrt{11.84}}\right)\) | B1 | For both limits correct, soi. e.g. use of 0.5 in probability calculation implies correct lower limit. |
| \(= P(0 < Z < 0.2906)\) \(= \Phi(0.2906) - \Phi(0)\) \(= 0.6143 - 0.5\) | M1 | For correct structure using their standardised values, i.e. finding the area between their z values found using \(\mu = 130.5\). Condone use of \(\sigma = 11.84\) if also used in part (i) or part (ii). |
| \(= 0.1143\) | A1 | CAO inc use of diff tables. Allow 0.1145. Allow 0.114 www. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 130.5\) \(H_1: \mu > 130.5\) | B1 | For both correct. Hypotheses in words must refer to population. Do not allow other symbols unless clearly defined as population mean. |
| Where \(\mu\) denotes the mean wing length (in the population) (of Scandinavian male blackbirds). | B1 | For definition of \(\mu\) in context. Do not allow "sample mean wing length" or "mean wing length of English blackbirds". |
| Test statistic \(= \frac{132.4 - 130.5}{\sqrt{11.84}/\sqrt{20}} = \frac{1.90}{0.7694}\) | M1* | Must include \(\sqrt{20}\). Condone use of \(\sigma = 11.84\) if also used in part (i), part (ii) or part (iii). Condone numerator reversed for max M1*A0B1M0depA0A0 (max 4/8). |
| \(= 2.469\) | A1 | Allow 2.47. |
| Upper 5% level 1 tailed critical value of \(z = 1.645\) | B1 | For 1.645. Must be positive. B0 if \(-1.645\) seen. No further A marks from here if wrong. |
| \(2.469 > 1.645\) | M1dep* | For sensible comparison leading to a conclusion. |
| The result is significant. There is sufficient evidence to reject \(H_0\) | A1 | For correct conclusion. e.g. for "significant" oe. FT only candidate's test statistic if cv \(= 1.645\). |
| There is sufficient evidence to suggest that the mean wing length (of this population of birds) is greater (than 130.5mm). | A1 | For non-assertive conclusion in context, consistent with their result. Condone use of "average" for "mean". FT only candidate's test statistic if cv \(= 1.645\). |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| With a 10% significance level rather than a 5% significance level: | ||
| Advantage: One is less likely to accept the null hypothesis when it is false. | E1 | Accept equivalent wording. |
| Disadvantage: One is more likely to reject the null hypothesis when it is true. | E1 | Note – Unless stated otherwise, assume the first comment relates to an advantage and the second comment relates to a disadvantage. |
| [2] |
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Wing length} = 131) = P\left(\frac{130.5-130.5}{\sqrt{11.84}} \leq Z \leq \frac{131.5-130.5}{\sqrt{11.84}}\right)$ | B1 | For both limits correct, soi. e.g. use of 0.5 in probability calculation implies correct lower limit. |
| $= P(0 < Z < 0.2906)$ $= \Phi(0.2906) - \Phi(0)$ $= 0.6143 - 0.5$ | M1 | For **correct structure** using their standardised values, i.e. finding the area between their z values found using $\mu = 130.5$. Condone use of $\sigma = 11.84$ if also used in part (i) or part (ii). |
| $= 0.1143$ | A1 | CAO inc use of diff tables. Allow 0.1145. Allow 0.114 www. |
| **[3]** | | |
---
## Question 3(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 130.5$ $H_1: \mu > 130.5$ | B1 | For both correct. Hypotheses in words must refer to population. Do not allow other symbols unless clearly defined as population mean. |
| Where $\mu$ denotes the **mean wing length** (in the population) (of Scandinavian male blackbirds). | B1 | For definition of $\mu$ in context. Do not allow "sample mean wing length" or "mean wing length of English blackbirds". |
| Test statistic $= \frac{132.4 - 130.5}{\sqrt{11.84}/\sqrt{20}} = \frac{1.90}{0.7694}$ | M1* | Must include $\sqrt{20}$. Condone use of $\sigma = 11.84$ if also used in part (i), part (ii) or part (iii). Condone numerator reversed for max M1*A0B1M0depA0A0 (max 4/8). |
| $= 2.469$ | A1 | Allow 2.47. |
| Upper 5% level 1 tailed critical value of $z = 1.645$ | B1 | For 1.645. Must be positive. B0 if $-1.645$ seen. No further A marks from here if wrong. |
| $2.469 > 1.645$ | M1dep* | For sensible comparison leading to a conclusion. |
| The result is significant. There is sufficient evidence to reject $H_0$ | A1 | For correct conclusion. e.g. for "significant" oe. FT only candidate's test statistic if cv $= 1.645$. |
| There is sufficient evidence to **suggest** that the mean wing length (of this population of birds) is greater (than 130.5mm). | A1 | For **non-assertive** conclusion in context, consistent with their result. Condone use of "average" for "mean". FT only candidate's test statistic if cv $= 1.645$. |
| **[8]** | | |
---
## Question 3(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| With a 10% significance level rather than a 5% significance level: | | |
| Advantage: One is less likely to accept the null hypothesis when it is false. | E1 | Accept equivalent wording. |
| Disadvantage: One is more likely to reject the null hypothesis when it is true. | E1 | Note – Unless stated otherwise, assume the first comment relates to an advantage and the second comment relates to a disadvantage. |
| **[2]** | | |
---3 The wing lengths of native English male blackbirds, measured in mm , are Normally distributed with mean 130.5 and variance 11.84.\\
(i) Find the probability that a randomly selected native English male blackbird has a wing length greater than 135 mm .\\
(ii) Given that $1 \%$ of native English male blackbirds have wing length more than $k \mathrm {~mm}$, find the value of $k$.\\
(iii) Find the probability that a randomly selected native English male blackbird has a wing length which is 131 mm correct to the nearest millimetre.
It is suspected that Scandinavian male blackbirds have, on average, longer wings than native English male blackbirds. A random sample of 20 Scandinavian male blackbirds has mean wing length 132.4 mm . You may assume that wing lengths in this population are Normally distributed with variance $11.84 \mathrm {~mm} ^ { 2 }$.\\
(iv) Carry out an appropriate hypothesis test, at the $5 \%$ significance level.\\
(v) Discuss briefly one advantage and one disadvantage of using a $10 \%$ significance level rather than a $5 \%$ significance level in hypothesis testing in general.
\hfill \mbox{\textit{OCR MEI S2 2014 Q3 [19]}}