| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Mixed sum threshold probability |
| Difficulty | Standard +0.3 This is a straightforward application of standard normal distribution properties for linear combinations. Parts (i)-(iii) require routine standardization and table lookup, (iv) uses basic variance formulas for linear combinations, and (v) involves reverse-engineering a confidence interval. All techniques are standard S3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Mean | Standard deviation | |
| Polymer 1 | 2025 | 44.6 |
| Polymer 2 | 1565 | 21.8 |
| Impact modifier | 1410 | 33.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P1 \sim N(2025,\ 44.6^2)\) | When a candidate's answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables, penalise the first occurrence only. | |
| \(P2 \sim N(1565,\ 21.8^2)\) | ||
| \(I \sim N(1410,\ 33.8^2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(P1 < 2100) = P\!\left(Z < \dfrac{2100-2025}{44.6} = 1.681(6)\right)\) | M1, A1 | For standardising. Award once, here or elsewhere. |
| \(= 0.9536/7\) | A1 | c.a.o. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Require \(P(P1 - P2 > 400)\) | M1 | |
| \(P1 - P2 \sim N(2025 - 1565 = 460,\ 44.6^2 + 21.8^2 = 2464.4)\) | B1, B1 | Mean. Variance. Accept sd \((= 49.64)\). |
| \(P(\text{this} > 400) = P\!\left(Z > \dfrac{400-460}{\sqrt{2464.4}} = -1.208(6)\right) = 0.8864/5\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T = P1 + P2 + I \sim N(5000,\ \sigma^2 = 44.6^2 + 21.8^2 + 33.8^2 = 3606.84)\) | B1, B1 | Mean. Variance. Accept sd \((= 60.056\ldots)\). |
| Require \(b\) s.t. \(P(T > b) = 0.95\) | ||
| \(\therefore \dfrac{b - 5000}{\sqrt{3606.84}} = -1.645\) | B1 | \(-1.645\) seen. |
| \(\therefore b = 5000 - 1.645 \times \sqrt{3606.84} = 4901.2\ldots\) | A1 | c.a.o. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= (1.2 \times 2025) + (1.3 \times 1565) + (0.8 \times 1410) = \pounds5592.50\) | B1 | Condone absence of £. |
| \(\text{Var} = (1.2^2 \times 44.6^2) + (1.3^2 \times 21.8^2) + (0.8^2 \times 33.8^2) = 4398.7076 \approx \pounds^2 4399\) | M1, A1 | Use of at least one of \((1.2^2 \times 44.6^2)\) etc. Condone absence of \(\pounds^2\). |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= (123.72 + 127.38)/2 = 125.55\) | B1 | Cao |
| \(s = \dfrac{127.38 - 125.55}{2.576/\sqrt{50}} = 5.02(3)\) | B1 | Sight of 2.576. |
| M1 | Or equivalent. | |
| A1 | cao | |
| [4] |
# Question 3:
## Distributions stated
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P1 \sim N(2025,\ 44.6^2)$ | | When a candidate's answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables, penalise the first occurrence only. |
| $P2 \sim N(1565,\ 21.8^2)$ | | |
| $I \sim N(1410,\ 33.8^2)$ | | |
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(P1 < 2100) = P\!\left(Z < \dfrac{2100-2025}{44.6} = 1.681(6)\right)$ | M1, A1 | For standardising. Award once, here or elsewhere. |
| $= 0.9536/7$ | A1 | c.a.o. |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Require $P(P1 - P2 > 400)$ | M1 | |
| $P1 - P2 \sim N(2025 - 1565 = 460,\ 44.6^2 + 21.8^2 = 2464.4)$ | B1, B1 | Mean. Variance. Accept sd $(= 49.64)$. |
| $P(\text{this} > 400) = P\!\left(Z > \dfrac{400-460}{\sqrt{2464.4}} = -1.208(6)\right) = 0.8864/5$ | A1 | cao |
| **[4]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = P1 + P2 + I \sim N(5000,\ \sigma^2 = 44.6^2 + 21.8^2 + 33.8^2 = 3606.84)$ | B1, B1 | Mean. Variance. Accept sd $(= 60.056\ldots)$. |
| Require $b$ s.t. $P(T > b) = 0.95$ | | |
| $\therefore \dfrac{b - 5000}{\sqrt{3606.84}} = -1.645$ | B1 | $-1.645$ seen. |
| $\therefore b = 5000 - 1.645 \times \sqrt{3606.84} = 4901.2\ldots$ | A1 | c.a.o. |
| **[4]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= (1.2 \times 2025) + (1.3 \times 1565) + (0.8 \times 1410) = \pounds5592.50$ | B1 | Condone absence of £. |
| $\text{Var} = (1.2^2 \times 44.6^2) + (1.3^2 \times 21.8^2) + (0.8^2 \times 33.8^2) = 4398.7076 \approx \pounds^2 4399$ | M1, A1 | Use of at least one of $(1.2^2 \times 44.6^2)$ etc. Condone absence of $\pounds^2$. |
| **[3]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= (123.72 + 127.38)/2 = 125.55$ | B1 | Cao |
| $s = \dfrac{127.38 - 125.55}{2.576/\sqrt{50}} = 5.02(3)$ | B1 | Sight of 2.576. |
| | M1 | Or equivalent. |
| | A1 | cao |
| **[4]** | | |
---3 In the manufacture of child car seats, a resin made up of three ingredients is used. The ingredients are two polymers and an impact modifier. The resin is prepared in batches. Each ingredient is supplied by a separate feeder and the amount supplied to each batch, in kg, is assumed to be Normally distributed with mean and standard deviation as shown in the table below. The three feeders are also assumed to operate independently of each other.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
& Mean & Standard deviation \\
\hline
Polymer 1 & 2025 & 44.6 \\
\hline
Polymer 2 & 1565 & 21.8 \\
\hline
Impact modifier & 1410 & 33.8 \\
\hline
\end{tabular}
\end{center}
(i) Find the probability that, in a randomly chosen batch of resin, there is no more than 2100 kg of polymer 1.\\
(ii) Find the probability that, in a randomly chosen batch of resin, the amount of polymer 1 exceeds the amount of polymer 2 by at least 400 kg .\\
(iii) Find the value of $b$ such that the total amount of the ingredients in a randomly chosen batch exceeds $b \mathrm {~kg} 95 \%$ of the time.\\
(iv) Polymer 1 costs $\pounds 1.20$ per kg, polymer 2 costs $\pounds 1.30$ per kg and the impact modifier costs $\pounds 0.80$ per kg. Find the mean and variance of the total cost of a batch of resin.\\
(v) Each batch of resin is used to make a large number of car seats from which a random sample of 50 seats is selected in order that the tensile strength (in suitable units) of the resin can be measured. From one such sample, the $99 \%$ confidence interval for the true mean tensile strength of the resin in that batch was calculated as $( 123.72,127.38 )$. Find the mean and standard deviation of the sample.
\hfill \mbox{\textit{OCR MEI S3 2013 Q3 [18]}}