| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two or more different variables |
| Difficulty | Standard +0.3 This is a standard S3 question testing linear combinations of normal variables with straightforward applications: basic normal probability calculations, sum of independent normals, and a routine one-sample t-test. All parts follow textbook procedures with no novel insight required, though part (iii) requires careful setup of the mass formula. Slightly above average difficulty due to multiple parts and the need to correctly handle scaling/combining distributions. |
| Spec | 5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(V_G < 60) = P(Z < \frac{60-56.5}{2.9} = 1.2069)\) | M1, A1 | For standardising. Award once, here or elsewhere. |
| \(= 0.8862\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V_T \sim N(56.5 + 38.4 = 94.9,\) | B1 | Mean. |
| \(2.9^2 + 1.1^2 = 9.62)\) | B1 | Variance. Accept sd \((= 3.1016)\). |
| \(P(\text{this} > 100) = P(Z > \frac{100-94.9}{3.1016} = 1.6443)\) | ||
| \(= 1 - 0.9499 = 0.0501\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(W_T \sim N(3.1 \times 56.5 + 0.8 \times 38.4 = 205.87,\) | M1, A1 | Use of "mass = density × volume". Mean. |
| \(3.1^2 \times 2.9^2 + 0.8^2 \times 1.1^2 = 81.5945)\) | M1, A1 | Variance. Accept sd \((= 9.0330)\). |
| \(P(200 < \text{this} < 220)\) | M1 | Formulation of requirement. |
| \(= P(\frac{200-205.87}{9.0330} < Z < \frac{220-205.87}{9.0330})\) | ||
| \(= P(-0.6498 < Z < 1.5643)\) | ||
| \(= 0.9411 - (1 - 0.7422) = 0.6833\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 200,\ H_1: \mu > 200\) | ||
| Test statistic \(= \frac{205.6-200}{\frac{8.51}{\sqrt{10}}} = 2.081\) | M1, A1 | Allow alternative: \(200 + (\text{c's } 1.833) \times \frac{8.51}{\sqrt{10}} (= 204.933)\). Use of \(200 - \bar{x}\) scores M1A0, but ft. |
| Refer to \(t_9\) | M1 | No ft from here if wrong. \(P(t > 2.081) = 0.0336\). |
| Single-tailed 5% point is 1.833. | A1 | No ft from here if wrong. |
| Significant. | A1 | ft only c's test statistic. |
| Seems that the required reduction of the mean weight has not been achieved. | A1 | ft only c's test statistic. |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(V_G < 60) = P(Z < \frac{60-56.5}{2.9} = 1.2069)$ | M1, A1 | For standardising. Award once, here or elsewhere. |
| $= 0.8862$ | A1 | |
**Total: 3 marks**
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $V_T \sim N(56.5 + 38.4 = 94.9,$ | B1 | Mean. |
| $2.9^2 + 1.1^2 = 9.62)$ | B1 | Variance. Accept sd $(= 3.1016)$. |
| $P(\text{this} > 100) = P(Z > \frac{100-94.9}{3.1016} = 1.6443)$ | | |
| $= 1 - 0.9499 = 0.0501$ | A1 | c.a.o. |
**Total: 3 marks**
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $W_T \sim N(3.1 \times 56.5 + 0.8 \times 38.4 = 205.87,$ | M1, A1 | Use of "mass = density × volume". Mean. |
| $3.1^2 \times 2.9^2 + 0.8^2 \times 1.1^2 = 81.5945)$ | M1, A1 | Variance. Accept sd $(= 9.0330)$. |
| $P(200 < \text{this} < 220)$ | M1 | Formulation of requirement. |
| $= P(\frac{200-205.87}{9.0330} < Z < \frac{220-205.87}{9.0330})$ | | |
| $= P(-0.6498 < Z < 1.5643)$ | | |
| $= 0.9411 - (1 - 0.7422) = 0.6833$ | A1 | c.a.o. |
**Total: 6 marks**
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 200,\ H_1: \mu > 200$ | | |
| Test statistic $= \frac{205.6-200}{\frac{8.51}{\sqrt{10}}} = 2.081$ | M1, A1 | Allow alternative: $200 + (\text{c's } 1.833) \times \frac{8.51}{\sqrt{10}} (= 204.933)$. Use of $200 - \bar{x}$ scores M1A0, but ft. |
| Refer to $t_9$ | M1 | No ft from here if wrong. $P(t > 2.081) = 0.0336$. |
| Single-tailed 5% point is 1.833. | A1 | No ft from here if wrong. |
| Significant. | A1 | ft only c's test statistic. |
| Seems that the required reduction of the mean weight has not been achieved. | A1 | ft only c's test statistic. |
**Total: 6 marks**
---2 A factory manufactures paperweights consisting of glass mounted on a wooden base. The volume of glass, in $\mathrm { cm } ^ { 3 }$, in a paperweight has a Normal distribution with mean 56.5 and standard deviation 2.9. The volume of wood, in $\mathrm { cm } ^ { 3 }$, also has a Normal distribution with mean 38.4 and standard deviation 1.1. These volumes are independent of each other. For the purpose of quality control, paperweights for testing are chosen at random from the factory's output.\\
(i) Find the probability that the volume of glass in a randomly chosen paperweight is less than $60 \mathrm {~cm} ^ { 3 }$.\\
(ii) Find the probability that the total volume of a randomly chosen paperweight is more than $100 \mathrm {~cm} ^ { 3 }$.
The glass has a mass of 3.1 grams per $\mathrm { cm } ^ { 3 }$ and the wood has a mass of 0.8 grams per $\mathrm { cm } ^ { 3 }$.\\
(iii) Find the probability that the total mass of a randomly chosen paperweight is between 200 and 220 grams.\\
(iv) The factory manager introduces some modifications intended to reduce the mean mass of the paperweights to 200 grams or less. The variance is also affected but not the Normality. Subsequently, for a random sample of 10 paperweights, the sample mean mass is 205.6 grams and the sample standard deviation is 8.51 grams. Is there evidence, at the $5 \%$ level of significance, that the intended reduction of the mean mass has not been achieved?
\hfill \mbox{\textit{OCR MEI S3 2009 Q2 [18]}}