| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Distribution of linear combination |
| Difficulty | Standard +0.3 This is a standard S3 question on linear combinations of normal variables requiring routine application of formulas for mean and variance of linear combinations, followed by a normal probability calculation. Part (b) adds one extra step of working backwards from a probability to find variance, but still follows a predictable method with no novel insight required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{E(X)=4E(Y)-3E(W)=4(40)-3(50)\} \Rightarrow E(X)=10\) | B1 | \(E(X)=10\) seen or implied |
| \(\text{Var}(X)=16\text{Var}(Y)+9\text{Var}(W)\) | M1 | Either \((4^2)\text{Var}(Y)\) or \(+(3^2)\text{Var}(W)\) |
| M1 | For adding the variances | |
| \(\{\text{Var}(X)=16(9)+9(4)\} \Rightarrow \text{Var}(X)=180\) | A1 | \(\text{Var}(X)=180\) |
| \(\{P(X>25)=\}\ P\left(Z>\frac{25-10}{\sqrt{180}}\right)\) | M1 | Standardising \((\pm)\) with their mean and their standard deviation |
| \(= P(Z>1.11803...)\) | A1 | awrt \(\pm 1.12\) |
| \(= 1-0.8686 = 0.1314\) (or \(0.131777...\)) | A1 | awrt 0.131 or awrt 0.132 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{E(A-C)=3E(Y)-E(C)=3(40)-(115)\} \Rightarrow E(A-C)=5\) | B1 | \(E(A-C)=5\) |
| \(\text{Var}(A-C)=3\text{Var}(Y)+\text{Var}(C)\) | M1 | \(3\text{Var}(Y)\) and \(a+...\) |
| \(\{\text{Var}(A-C)=3(9)+\sigma^2\} \Rightarrow \text{Var}(A-C)=27+\sigma^2\) | A1 | \(\text{Var}(A-C)=27+\sigma^2\) |
| \(\{P(A-C<0)=0.2\} \Rightarrow P\left(Z<\frac{-5}{\sqrt{27+\sigma^2}}\right)=0.2\) | ||
| \(\frac{-5}{\sqrt{27+\sigma^2}}=k\ (=-0.8416)\) | M1 | Standardising \((\pm)\) with their mean and their standard deviation, in terms of \(\sigma^2\), and setting equal to \(k\) where \( |
| \(\pm 0.8416\) or awrt \(\pm 0.8416\) | B1 | |
| Correct equation. See notes. | A1 | E.g. Allow \(\frac{-5}{\sqrt{27+\sigma^2}}=[-0.85,-0.84]\) or \(\frac{5}{\sqrt{27+\sigma^2}}=[0.84,0.85]\) |
| \(\sigma^2=\left(\frac{-5}{-0.8416}\right)^2-27 \Rightarrow \sigma^2=...\) | dM1 | Squaring and rearranging leading to a positive value for \(\sigma^2\). Dependent on 2nd M1 |
| \(\sigma^2=8.2962...\) \((=8.4308...\) from using \(-0.84)\) | A1 cso | awrt 8.3 or awrt 8.4. \((=8.2945...\) from calculator, so need awrt 8.29 for full marks if no prior working shown) |
# Question 7:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{E(X)=4E(Y)-3E(W)=4(40)-3(50)\} \Rightarrow E(X)=10$ | B1 | $E(X)=10$ seen or implied |
| $\text{Var}(X)=16\text{Var}(Y)+9\text{Var}(W)$ | M1 | Either $(4^2)\text{Var}(Y)$ or $+(3^2)\text{Var}(W)$ |
| | M1 | For adding the variances |
| $\{\text{Var}(X)=16(9)+9(4)\} \Rightarrow \text{Var}(X)=180$ | A1 | $\text{Var}(X)=180$ |
| $\{P(X>25)=\}\ P\left(Z>\frac{25-10}{\sqrt{180}}\right)$ | M1 | Standardising $(\pm)$ with their mean and their standard deviation |
| $= P(Z>1.11803...)$ | A1 | awrt $\pm 1.12$ |
| $= 1-0.8686 = 0.1314$ (or $0.131777...$) | A1 | awrt 0.131 or awrt 0.132 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{E(A-C)=3E(Y)-E(C)=3(40)-(115)\} \Rightarrow E(A-C)=5$ | B1 | $E(A-C)=5$ |
| $\text{Var}(A-C)=3\text{Var}(Y)+\text{Var}(C)$ | M1 | $3\text{Var}(Y)$ and $a+...$ |
| $\{\text{Var}(A-C)=3(9)+\sigma^2\} \Rightarrow \text{Var}(A-C)=27+\sigma^2$ | A1 | $\text{Var}(A-C)=27+\sigma^2$ |
| $\{P(A-C<0)=0.2\} \Rightarrow P\left(Z<\frac{-5}{\sqrt{27+\sigma^2}}\right)=0.2$ | | |
| $\frac{-5}{\sqrt{27+\sigma^2}}=k\ (=-0.8416)$ | M1 | Standardising $(\pm)$ with their mean and their standard deviation, in terms of $\sigma^2$, and setting equal to $k$ where $|k|$ is in interval $[0.84, 0.85]$ |
| $\pm 0.8416$ or awrt $\pm 0.8416$ | B1 | |
| Correct equation. See notes. | A1 | E.g. Allow $\frac{-5}{\sqrt{27+\sigma^2}}=[-0.85,-0.84]$ or $\frac{5}{\sqrt{27+\sigma^2}}=[0.84,0.85]$ |
| $\sigma^2=\left(\frac{-5}{-0.8416}\right)^2-27 \Rightarrow \sigma^2=...$ | dM1 | Squaring and rearranging leading to a positive value for $\sigma^2$. **Dependent on 2nd M1** |
| $\sigma^2=8.2962...$ $(=8.4308...$ from using $-0.84)$ | A1 cso | awrt 8.3 or awrt 8.4. $(=8.2945...$ from calculator, so need awrt 8.29 for full marks if no prior working shown) |7. The random variable $X$ is defined as
$$X = 4 Y - 3 W$$
where $Y \sim \mathrm {~N} \left( 40,3 ^ { 2 } \right) , W \sim \mathrm {~N} \left( 50,2 ^ { 2 } \right)$ and $Y$ and $W$ are independent.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 25 )$
The random variables $Y _ { 1 } , Y _ { 2 }$ and $Y _ { 3 }$ are independent and each has the same distribution as $Y$. The random variable $A$ is defined as
$$A = \sum _ { i = 1 } ^ { 3 } Y _ { i }$$
The random variable $C$ is such that $C \sim \mathrm {~N} \left( 115 , \sigma ^ { 2 } \right)$
Given that $\mathrm { P } ( A - C < 0 ) = 0.2$ and that $A$ and $C$ are independent,
\item find the variance of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q7 [15]}}