| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| 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 straightforward application of standard results for linear combinations of independent normal variables. Part (a) requires finding the mean and variance of a sum then dividing by 5 (routine calculation), and part (b) involves setting up W-X as a normal distribution and using inverse normal tables. Both parts are direct applications of learned formulas with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.05f Pearson correlation coefficient5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = \dfrac{X_1+X_2+X_3+Y_1+Y_2}{5}\), \(X \sim N(30,4.5^2)\), \(Y \sim N(20,3.5^2)\); \(X,Y\) independent | ||
| \(E(A) = \dfrac{3(30)+2(20)}{5}\) or \(\text{Var}(A) = \dfrac{3(4.5)^2+2(3.5)^2}{25}\) | M1 | A correct method for finding \(E(A)\) or \(\text{Var}(A)\) |
| \(E(A) = 26\) or \(\text{Var}(A) = 3.41\) | A1 | At least one of \(E(A)=26\) or \(\text{Var}(A)=3.41\) |
| Both \(E(A) = 26\) and \(\text{Var}(A) = 3.41\) | A1 | |
| \(A \sim N(26,\ 3.41)\) | ||
| \(P(A < 24) = P\!\left(Z < \dfrac{24-26}{\sqrt{3.41}}\right)\) | M1 | Standardising (\(\pm\)) with their mean and their standard deviation |
| \(= P(Z < -1.08306...)\) | ||
| \(= 1 - 0.8599\) | M1 | For a probability tail compatible with 24 and their mean |
| \(= 0.1401\) (or \(0.139391...\)); 0.14 or awrt 0.140 or awrt 0.139 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W \sim N(\mu, 2.8^2)\); \(P(W - X < 4) = 0.1\); \(W, X\) independent | ||
| \(E(W-X) = E(W) - E(X) = \mu - 30\) | B1 | |
| \(\text{Var}(W-X) = 2.8^2 + 4.5^2 = 28.09\) | M1 | \(2.8^2 + 4.5^2\) |
| \(W - X \sim N(\mu - 30,\ 28.09)\) | ||
| \(P(W-X < 4) = 0.1 \Rightarrow P\!\left(Z < \dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}}\right) = 0.1\) | ||
| \(\dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}} = k\ (= -1.2816)\) | M1 | Standardising (\(\pm\)) with mean in terms of \(\mu\) and their s.d., setting equal to \(k\) where \( |
| \(\pm 1.2816\) or awrt \(\pm 1.2816\) | B1 | For either \(-1.2816\) or \(1.2816\) |
| Correct equation | A1 | E.g. allow \(\dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}} \in [-1.29,-1.28]\) or \(\dfrac{(\mu-30)-4}{\sqrt{2.8^2+4.5^2}} \in [1.28,1.29]\) |
| \(\mu = 34 + 1.2816(5.3) \Rightarrow \mu = 40.792...(= 40.784\) from using \(-1.28)\); awrt 40.8 | A1 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \dfrac{X_1+X_2+X_3+Y_1+Y_2}{5}$, $X \sim N(30,4.5^2)$, $Y \sim N(20,3.5^2)$; $X,Y$ independent | | |
| $E(A) = \dfrac{3(30)+2(20)}{5}$ or $\text{Var}(A) = \dfrac{3(4.5)^2+2(3.5)^2}{25}$ | M1 | A correct method for finding $E(A)$ or $\text{Var}(A)$ |
| $E(A) = 26$ or $\text{Var}(A) = 3.41$ | A1 | At least one of $E(A)=26$ or $\text{Var}(A)=3.41$ |
| Both $E(A) = 26$ and $\text{Var}(A) = 3.41$ | A1 | |
| $A \sim N(26,\ 3.41)$ | | |
| $P(A < 24) = P\!\left(Z < \dfrac{24-26}{\sqrt{3.41}}\right)$ | M1 | Standardising ($\pm$) with their mean and their standard deviation |
| $= P(Z < -1.08306...)$ | | |
| $= 1 - 0.8599$ | M1 | For a probability tail compatible with 24 and their mean |
| $= 0.1401$ (or $0.139391...$); **0.14** or awrt **0.140** or awrt **0.139** | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W \sim N(\mu, 2.8^2)$; $P(W - X < 4) = 0.1$; $W, X$ independent | | |
| $E(W-X) = E(W) - E(X) = \mu - 30$ | B1 | |
| $\text{Var}(W-X) = 2.8^2 + 4.5^2 = 28.09$ | M1 | $2.8^2 + 4.5^2$ |
| $W - X \sim N(\mu - 30,\ 28.09)$ | | |
| $P(W-X < 4) = 0.1 \Rightarrow P\!\left(Z < \dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}}\right) = 0.1$ | | |
| $\dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}} = k\ (= -1.2816)$ | M1 | Standardising ($\pm$) with mean in terms of $\mu$ and their s.d., setting equal to $k$ where $|k| \in [1.28, 1.29]$ |
| $\pm 1.2816$ or awrt $\pm 1.2816$ | B1 | For either $-1.2816$ or $1.2816$ |
| Correct equation | A1 | E.g. allow $\dfrac{4-(\mu-30)}{\sqrt{2.8^2+4.5^2}} \in [-1.29,-1.28]$ or $\dfrac{(\mu-30)-4}{\sqrt{2.8^2+4.5^2}} \in [1.28,1.29]$ |
| $\mu = 34 + 1.2816(5.3) \Rightarrow \mu = 40.792...(= 40.784$ from using $-1.28)$; awrt **40.8** | A1 | |7. The independent random variables $X$ and $Y$ are such that
$$X \sim \mathrm {~N} \left( 30,4.5 ^ { 2 } \right) \text { and } Y \sim \mathrm {~N} \left( 20,3.5 ^ { 2 } \right)$$
The random variables $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ are independent and each has the same distribution as $X$. The random variables $Y _ { 1 }$ and $Y _ { 2 }$ are independent and each has the same distribution as $Y$.
Given that the random variable $A$ is defined as
$$A = \frac { X _ { 1 } + X _ { 2 } + X _ { 3 } + Y _ { 1 } + Y _ { 2 } } { 5 }$$
\begin{enumerate}[label=(\alph*)]
\item find $\mathrm { P } ( A < 24 )$
The random variable $W$ is such that $W \sim \mathrm {~N} \left( \mu , 2.8 ^ { 2 } \right)$
Given that $\mathrm { P } ( W - X < 4 ) = 0.1$ and that $W$ and $X$ are independent,
\item find the value of $\mu$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q7 [12]}}