| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Expectation and variance with context application |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of normal random variables. Part (i) tests basic understanding of variance properties (textbook exercise), while parts (ii)-(v) require routine standardization and normal table lookups with no novel problem-solving. Slightly above average difficulty only due to the multi-part nature and part (v) requiring setup of an inequality involving two variables. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Mean | Standard deviation | |
| Mathematical ability, \(X\) | 30.1 | 5.1 |
| Spatial awareness, \(Y\) | 25.4 | 4.2 |
| Communication, \(W\) | 28.2 | 3.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}(X_1 + X_2) = 2\sigma^2\) | B1 | Allow \(2\text{Var}(X)\) and \(4\text{Var}(X)\) |
| \(\text{Var}(2X) = 4\sigma^2\) | B1 | |
| \(X_1 + X_2\) means two independent values from \(X\) are added together. \(2X\) means that one value from \(X\) is multiplied by 2. | E1 | Any comment explaining why \(X_1 + X_2\) is different from \(2X\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 22) = P\left(Z < \dfrac{22 - 30.1}{5.1}\right)\) | M1 | For standardising. Award once, here or elsewhere |
| \(= P(Z < -1.5882)\) | A1 | Correct z value |
| \(= 0.0561\) | A1 | cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X + Y + W \sim N(83.7, 58.86)\) | B1 | Mean |
| B1 | Variance (or sd \(= 7.67\)) | |
| \(P(X + Y + W > 100) = P\left(Z > \dfrac{100 - 83.7}{\sqrt{58.86}}\right)\) | M1 | Correct set up |
| \(P(Z > 2.1246) = 0.0168\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2X \sim N(60.2, 104.04)\) | ||
| \(\rightarrow 2X + Y \sim N(85.6, 121.68)\) | B1 | Variance |
| \(P(2X + Y > b) = 0.02\) | ||
| \(\rightarrow \dfrac{b - 85.6}{\sqrt{121.68}} = 2.054\) | B1 | 2.054 seen |
| M1 | Correct set up | |
| \(\rightarrow b = 108.26\) | ||
| Score exceeded by 2% is 108.3 | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(W < 0.6X) = P(W - 0.6X < 0)\) | M1 | Either way round |
| \(W - 0.6X \sim N(0.14, 24.5736)\) | B1 | Mean and variance |
| \(P(W - 0.6X < 0) = P(Z < -2.0455)\) | ||
| \(= 0.0204\) | A1 | cao. Allow convincing recovery |
| [3] |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}(X_1 + X_2) = 2\sigma^2$ | B1 | Allow $2\text{Var}(X)$ and $4\text{Var}(X)$ |
| $\text{Var}(2X) = 4\sigma^2$ | B1 | |
| $X_1 + X_2$ means two independent values from $X$ are added together. $2X$ means that one value from $X$ is multiplied by 2. | E1 | Any comment explaining why $X_1 + X_2$ is different from $2X$ |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 22) = P\left(Z < \dfrac{22 - 30.1}{5.1}\right)$ | M1 | For standardising. Award once, here or elsewhere |
| $= P(Z < -1.5882)$ | A1 | Correct z value |
| $= 0.0561$ | A1 | cao |
| **[3]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X + Y + W \sim N(83.7, 58.86)$ | B1 | Mean |
| | B1 | Variance (or sd $= 7.67$) |
| $P(X + Y + W > 100) = P\left(Z > \dfrac{100 - 83.7}{\sqrt{58.86}}\right)$ | M1 | Correct set up |
| $P(Z > 2.1246) = 0.0168$ | A1 | cao |
| **[4]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2X \sim N(60.2, 104.04)$ | | |
| $\rightarrow 2X + Y \sim N(85.6, 121.68)$ | B1 | Variance |
| $P(2X + Y > b) = 0.02$ | | |
| $\rightarrow \dfrac{b - 85.6}{\sqrt{121.68}} = 2.054$ | B1 | 2.054 seen |
| | M1 | Correct set up |
| $\rightarrow b = 108.26$ | | |
| Score exceeded by 2% is 108.3 | A1 | cao |
| **[4]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(W < 0.6X) = P(W - 0.6X < 0)$ | M1 | Either way round |
| $W - 0.6X \sim N(0.14, 24.5736)$ | B1 | Mean and variance |
| $P(W - 0.6X < 0) = P(Z < -2.0455)$ | | |
| $= 0.0204$ | A1 | cao. Allow convincing recovery |
| **[3]** | | |
---1 (i) Let $X$ be a random variable with variance $\sigma ^ { 2 }$. The independent random variables $X _ { 1 }$ and $X _ { 2 }$ are both distributed as $X$. Write down the variances of $X _ { 1 } + X _ { 2 }$ and $2 X$; explain why they are different.
A large company has produced an aptitude test which consists of three parts. The parts are called mathematical ability, spatial awareness and communication. The scores obtained by candidates in the three parts are continuous random variables $X , Y$ and $W$ which have been found to have independent Normal distributions with means and standard deviations as shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Mean & Standard deviation \\
\hline
Mathematical ability, $X$ & 30.1 & 5.1 \\
\hline
Spatial awareness, $Y$ & 25.4 & 4.2 \\
\hline
Communication, $W$ & 28.2 & 3.9 \\
\hline
\end{tabular}
\end{center}
(ii) Find the probability that a randomly selected candidate obtains a score of less than 22 in the mathematical ability part of the test.\\
(iii) Find the probability that a randomly selected candidate obtains a total score of at least 100 in the whole test.\\
(iv) For a particular role in the company, the score $2 X + Y$ is calculated. Find the score that is exceeded by only $2 \%$ of candidates.\\
(v) For a different role, a candidate must achieve a score in communication which is at least $60 \%$ of the score obtained in mathematical ability. What proportion of candidates do not achieve this?
\hfill \mbox{\textit{OCR MEI S3 2014 Q1 [17]}}