| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum versus sum comparison |
| Difficulty | Standard +0.8 This S3 question requires solid understanding of linear combinations of normal variables across multiple parts. Part (i) involves comparing 5B vs 5H (requiring variance of sums), part (ii) combines sample means with individual observations (Y₁ - X̄ requires careful variance calculation with n=5), and parts (c-d) test conceptual understanding of independence assumptions. The multi-step nature, careful variance tracking, and the conceptual trap in part (c) make this moderately challenging but still within standard S3 scope. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(R = B_1 + B_2 + B_3 + B_4 + B_5 - 5H\), so \(E(R) = -25\) (o.e.) | B1 | For \(E(R) = -25\) (or 25 if \(R\) defined the other way) |
| \(\text{Var}(R) = 5 \times 6^2 + 5^2 \times 4^2\), \(R \sim N(-25, \sqrt{580}^2)\) | M1A1 | M1 for attempt at \(\text{Var}(R) = 5\text{Var}(B) + 25\text{Var}(H)\). Condone swapping \(6^2\) and \(4^2\). A1 for normal and correct variance (ft their mean) |
| \(P(R > 0) = P\!\left(Z > \frac{0-(-25)}{\sqrt{580}}\right) = P(Z > 1.04) = 0.149619...\) (calc) or \(0.1492\) (tables) | dM1 A1 | dM1 for correct probability and standardising. A1 for answer in range \([0.149, 0.150]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{5}\right)\) | B1 | For correct distribution of \(\bar{X}\) (may be implied by correct answer for \(D\)) |
| \(\text{Var}(D) = \sigma^2 + \text{"}\frac{\sigma^2}{5}\text{"} \left[= \frac{6\sigma^2}{5}\right]\), so \(D \sim N\!\left(0, \frac{6\sigma^2}{5}\right)\) | M1, A1 | M1 for correct attempt at \(\text{Var}(D)\) (ft their \(\text{Var}(\bar{X})\)). A1 needs to be fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(Y_1 > \bar{X} + \sigma) = P(D > \sigma) = P\!\left(Z > \frac{\sigma}{\sqrt{\frac{6}{5}}\sigma}\right)\) | M1 | For expressing probability in terms of \(D\) and standardising |
| \(= P(Z > 0.912...) = 0.181\) (3 dp) | A1cso | For seeing \(P(Z > 0.912...)\) or prob of \(1 - 0.8186\) (tables) or \(0.180655...\) (calc) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Since \(U_1\) and \(\bar{U}\) are not independent (so variance formula cannot be used). Can be implied e.g. \(U_1\) used to calculate \(\bar{U}\), \(U_1\) and \(\bar{U}\) from same sample | B1 | Correct statement mentioning \(U_1\) and \(\bar{U}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(F = U_1 - \bar{U} = U_1 - \frac{U_1+U_2+U_3+U_4+U_5}{5} = \frac{4U_1-(U_2+U_3+U_4+U_5)}{5}\) | M1, A1 | M1 for forming expression in terms of \(U_1...U_5\) only. A1 for collecting \(U_1\) terms in form where \(\text{Var}(aX \pm bY)\) can be used |
| \(\text{Var}(F) = \frac{4^2\sigma^2 + 4\sigma^2}{5^2} = 0.8\sigma^2\), so \(F \sim N(0, 0.8\sigma^2)\) | dM1, A1 | dM1 for correct expression for \(\text{Var}(\text{their } F)\). A1 for correct distribution for \(F\) |
| \(P(F > \sigma) = P\!\left(Z > \frac{\sigma}{\sigma\sqrt{0.8}}\right) = P(Z > 1.118...)\) | M1 | Attempting correct prob and standardising using their \(\text{Var}(F)\); \(\sigma\) must cancel |
| \(= 0.1314\) (tables) or \(0.131776...\) (calc) awrt \(\mathbf{0.131}\)–\(\mathbf{0.132}\) | A1cso | For awrt 0.131 or 0.132 |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $R = B_1 + B_2 + B_3 + B_4 + B_5 - 5H$, so $E(R) = -25$ (o.e.) | B1 | For $E(R) = -25$ (or 25 if $R$ defined the other way) |
| $\text{Var}(R) = 5 \times 6^2 + 5^2 \times 4^2$, $R \sim N(-25, \sqrt{580}^2)$ | M1A1 | M1 for attempt at $\text{Var}(R) = 5\text{Var}(B) + 25\text{Var}(H)$. Condone swapping $6^2$ and $4^2$. A1 for normal and correct variance (ft their mean) |
| $P(R > 0) = P\!\left(Z > \frac{0-(-25)}{\sqrt{580}}\right) = P(Z > 1.04) = 0.149619...$ (calc) or $0.1492$ (tables) | dM1 A1 | dM1 for correct probability and standardising. A1 for answer in range $[0.149, 0.150]$ |
## Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{5}\right)$ | B1 | For correct distribution of $\bar{X}$ (may be implied by correct answer for $D$) |
| $\text{Var}(D) = \sigma^2 + \text{"}\frac{\sigma^2}{5}\text{"} \left[= \frac{6\sigma^2}{5}\right]$, so $D \sim N\!\left(0, \frac{6\sigma^2}{5}\right)$ | M1, A1 | M1 for correct attempt at $\text{Var}(D)$ (ft their $\text{Var}(\bar{X})$). A1 needs to be fully correct |
## Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(Y_1 > \bar{X} + \sigma) = P(D > \sigma) = P\!\left(Z > \frac{\sigma}{\sqrt{\frac{6}{5}}\sigma}\right)$ | M1 | For expressing probability in terms of $D$ and standardising |
| $= P(Z > 0.912...) = 0.181$ (3 dp) | A1cso | For seeing $P(Z > 0.912...)$ or prob of $1 - 0.8186$ (tables) or $0.180655...$ (calc) |
## Part (ii)(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Since $U_1$ and $\bar{U}$ are not independent (so variance formula cannot be used). Can be implied e.g. $U_1$ used to calculate $\bar{U}$, $U_1$ and $\bar{U}$ from same sample | B1 | Correct statement mentioning $U_1$ and $\bar{U}$ |
## Part (ii)(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $F = U_1 - \bar{U} = U_1 - \frac{U_1+U_2+U_3+U_4+U_5}{5} = \frac{4U_1-(U_2+U_3+U_4+U_5)}{5}$ | M1, A1 | M1 for forming expression in terms of $U_1...U_5$ only. A1 for collecting $U_1$ terms in form where $\text{Var}(aX \pm bY)$ can be used |
| $\text{Var}(F) = \frac{4^2\sigma^2 + 4\sigma^2}{5^2} = 0.8\sigma^2$, so $F \sim N(0, 0.8\sigma^2)$ | dM1, A1 | dM1 for correct expression for $\text{Var}(\text{their } F)$. A1 for correct distribution for $F$ |
| $P(F > \sigma) = P\!\left(Z > \frac{\sigma}{\sigma\sqrt{0.8}}\right) = P(Z > 1.118...)$ | M1 | Attempting correct prob and standardising using their $\text{Var}(F)$; $\sigma$ must cancel |
| $= 0.1314$ (tables) or $0.131776...$ (calc) awrt $\mathbf{0.131}$–$\mathbf{0.132}$ | A1cso | For awrt 0.131 or 0.132 |
---\begin{enumerate}
\item (i) The volume, $B \mathrm { ml }$, in a bottle of Burxton's water has a normal distribution $B \sim \mathrm {~N} \left( 325,6 ^ { 2 } \right)$ and the volume, $H \mathrm { ml }$, in a bottle of Hargate's water has a normal distribution $H \sim \mathrm {~N} \left( 330,4 ^ { 2 } \right)$.\\
Rebecca buys 5 bottles of Burxton's water and one bottle of Hargate's water.\\
Find the probability that the total volume in the 5 bottles of Burxton's water is more than 5 times the volume in the bottle of Hargate's water.\\
(5)\\
(ii) Two independent random samples $X _ { 1 } , X _ { 2 } , X _ { 3 } , X _ { 4 } , X _ { 5 }$ and $Y _ { 1 } , Y _ { 2 } , Y _ { 3 } , Y _ { 4 } , Y _ { 5 }$ are each taken from a normal population with mean $\mu$ and standard deviation $\sigma$.\\
(a) Find the distribution of the random variable $D = Y _ { 1 } - \bar { X }$\\
(b) Hence show that $\mathrm { P } \left( Y _ { 1 } > \bar { X } + \sigma \right) = 0.181$ correct to 3 decimal places.
\end{enumerate}
Ankit believes that $\mathrm { P } \left( U _ { 1 } > \bar { U } + \sigma \right) = 0.181$ correct to 3 decimal places, for any random sample $U _ { 1 } , U _ { 2 } , U _ { 3 } , U _ { 4 } , U _ { 5 }$ taken from a normal population with mean $\mu$ and standard deviation $\sigma$.\\
(c) Explain briefly why the result from part (b) should not be used to confirm Ankit's belief.\\
(d) Find, correct to 3 decimal places, the actual value of $\mathrm { P } \left( U _ { 1 } > \bar { U } + \sigma \right)$.
\hfill \mbox{\textit{Edexcel S3 2015 Q5 [17]}}