| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2024 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum or total of normal variables |
| Difficulty | Standard +0.3 This is a standard S3 question on linear combinations of normal variables requiring straightforward application of variance rules (sum/difference of independent normals) and normal probability calculations. Part (c) requires rearranging the inequality and recognizing G - 2T as a linear combination, which adds mild complexity but remains routine for this specification. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(C+C+C) = 3.6\) oe | B1 | Correct value; may be seen as \(\frac{18}{5}\) or implied by later calculation |
| \(\text{Var}(C+C+C) = 0.03^2 + 0.03^2 + 0.03^2 [= 0.0027]\) | M1 | Correct method to find variance; condone \(0.03^4\) instead of \(0.03^2\); may be implied by calculation or awrt 0.973 |
| \(P(C+C+C>3.5) = P\left(Z > \pm\frac{3.5-\text{"3.6"}}{\sqrt{\text{"0.0027"}}}\right)[= -1.9245...]\) | M1 | Correct standardisation using their mean and sd; allow \(\pm\frac{3.5-\text{"3.6"}}{\sqrt{\text{"0.0027"}}}\); may be implied by \(P(Z > \text{awrt}-1.92)\) or \(P(Z < \text{awrt}1.92)\) or awrt 0.973 |
| \(= 0.9726\) (calc \(0.97285...\)) awrt 0.973 | A1 | Do not isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(R-R) = 0\) | M1 | 0 may be seen or implied by later calculation |
| \(\text{Var}(R-R) = 0.03^2 + 0.03^2 [= 0.0018]\) | M1 | Correct method; may be implied by 0.0018 or \(\frac{9}{5000}\) or a later calculation; must be numerical value or expression |
| \(P\big((R-R)>0.05\big) = P\left(Z > \frac{0.05-\text{"0"}}{\sqrt{\text{"0.0018"}}}\right)[= 1.1785...]\) \([= 0.119\) (calc \(0.119296...)]\) | M1 | Correct standardisation using their mean and Var; allow \(\pm\) |
| \(2 \times P\big((R-R)>0.05\big) = 2\times\text{"0.119"}\) | M1 | \(2 \times \text{"their 0.119"}\) |
| \(= 0.238\) table or \(0.23859...\) calc awrt 0.238/0.239 | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu_G = 2.5 + 10\times 2.3 [= 25.5]\) | M1 | Correct method for finding mean of \(G\); may be implied by 25.5 or later work or sight of 20.5 (or may subtract 20 so 0.5) |
| \(\sigma_G^2 = 0.1 + 10\times 0.03^2 [= 0.109]\) | M1 | Correct method for finding var of \(G\); may be implied by sight of 0.109 provided this is not to find the variance of \(10R-T\); may be implied by 0.509 |
| Let \(X = G - 2T\) | M1 | Realising they need to find \(\pm(G-2T)\); allow \(\pm(G-2T-20)\) may be seen as part of a probability expression or implied by their calculation |
| \(\mu_X = \text{"25.5"} - 2\times 2.5 [= 20.5]\) | M1 | Correct method for finding mean of \(X\); may be from using their mean of \(G\) (which must be correct if no method or value is seen); allow 0.5 |
| \(\sigma_X^2 = \text{"0.109"} + 4\times 0.1 [= 0.509]\) | M1 | Correct method for finding var of \(X\); may be from using their variance of \(G\) (which must be correct if no method or value is seen); may be implied by \(10\times0.03^2+0.5\) or 0.509 |
| \(P(G-2T<20) = P\left(Z < \frac{20-\text{"20.5"}}{\sqrt{\text{"0.509"}}}\right)[= -0.7008...]\) | M1 | Correct standardisation using their mean and variance for \(G-2T\) (condone \(G-T\)); leading to probability \(< 0.5\); allow \(0-\text{"0.5"}\) for numerator for correct use of 20 with their "20.5" and their "0.509" |
| \(= 0.242\) (table) or \(0.2417...\) (calc) awrt 0.242 | A1 | From a correct distribution |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(C+C+C) = 3.6$ oe | B1 | Correct value; may be seen as $\frac{18}{5}$ or implied by later calculation |
| $\text{Var}(C+C+C) = 0.03^2 + 0.03^2 + 0.03^2 [= 0.0027]$ | M1 | Correct method to find variance; condone $0.03^4$ instead of $0.03^2$; may be implied by calculation or awrt 0.973 |
| $P(C+C+C>3.5) = P\left(Z > \pm\frac{3.5-\text{"3.6"}}{\sqrt{\text{"0.0027"}}}\right)[= -1.9245...]$ | M1 | Correct standardisation using their mean and sd; allow $\pm\frac{3.5-\text{"3.6"}}{\sqrt{\text{"0.0027"}}}$; may be implied by $P(Z > \text{awrt}-1.92)$ or $P(Z < \text{awrt}1.92)$ or awrt 0.973 |
| $= 0.9726$ (calc $0.97285...$) **awrt 0.973** | A1 | Do not isw |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(R-R) = 0$ | M1 | 0 may be seen or implied by later calculation |
| $\text{Var}(R-R) = 0.03^2 + 0.03^2 [= 0.0018]$ | M1 | Correct method; may be implied by 0.0018 or $\frac{9}{5000}$ or a later calculation; must be numerical value or expression |
| $P\big((R-R)>0.05\big) = P\left(Z > \frac{0.05-\text{"0"}}{\sqrt{\text{"0.0018"}}}\right)[= 1.1785...]$ $[= 0.119$ (calc $0.119296...)]$ | M1 | Correct standardisation using their mean and Var; allow $\pm$ |
| $2 \times P\big((R-R)>0.05\big) = 2\times\text{"0.119"}$ | M1 | $2 \times \text{"their 0.119"}$ |
| $= 0.238$ table or $0.23859...$ calc **awrt 0.238/0.239** | A1cso | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu_G = 2.5 + 10\times 2.3 [= 25.5]$ | M1 | Correct method for finding mean of $G$; may be implied by 25.5 or later work or sight of 20.5 (or may subtract 20 so 0.5) |
| $\sigma_G^2 = 0.1 + 10\times 0.03^2 [= 0.109]$ | M1 | Correct method for finding var of $G$; may be implied by sight of 0.109 provided this is not to find the variance of $10R-T$; may be implied by 0.509 |
| Let $X = G - 2T$ | M1 | Realising they need to find $\pm(G-2T)$; allow $\pm(G-2T-20)$ may be seen as part of a probability expression or implied by their calculation |
| $\mu_X = \text{"25.5"} - 2\times 2.5 [= 20.5]$ | M1 | Correct method for finding mean of $X$; may be from using their mean of $G$ (which must be correct if no method or value is seen); allow 0.5 |
| $\sigma_X^2 = \text{"0.109"} + 4\times 0.1 [= 0.509]$ | M1 | Correct method for finding var of $X$; may be from using their variance of $G$ (which must be correct if no method or value is seen); may be implied by $10\times0.03^2+0.5$ or 0.509 |
| $P(G-2T<20) = P\left(Z < \frac{20-\text{"20.5"}}{\sqrt{\text{"0.509"}}}\right)[= -0.7008...]$ | M1 | Correct standardisation using their mean and variance for $G-2T$ (condone $G-T$); leading to probability $< 0.5$; allow $0-\text{"0.5"}$ for numerator for correct use of 20 with their "20.5" and their "0.509" |
| $= 0.242$ (table) or $0.2417...$ (calc) **awrt 0.242** | A1 | From a correct distribution |
*Note: candidates who attempt $10R+T < 2T+20 \Rightarrow 10R-T < 20$ can score maximum M1M0M0M0M1M0M1A0 (the first method mark is implied by the fourth method mark)*
---\begin{enumerate}
\item The weights of bags of carrots, $C \mathrm {~kg}$, are such that $C \sim \mathrm {~N} \left( 1.2,0.03 ^ { 2 } \right)$
\end{enumerate}
Three bags of carrots are selected at random.\\
(a) Calculate the probability that their total weight is more than 3.5 kg .
The weights of bags of potatoes, $R \mathrm {~kg}$, are such that $R \sim \mathrm {~N} \left( 2.3,0.03 ^ { 2 } \right)$\\
Two bags of potatoes are selected at random.\\
(b) Calculate the probability that the difference in their weights is more than 0.05 kg .
The weights of trays, $T \mathrm {~kg}$, are such that $T \sim \mathrm {~N} \left( 2.5 , \sqrt { 0.1 } ^ { 2 } \right)$\\
The random variable $G$ represents the total weight, in kg, of a single tray packed with 10 bags of potatoes where $G$ and $T$ are independent.\\
(c) Calculate $\mathrm { P } ( G < 2 T + 20 )$
\hfill \mbox{\textit{Edexcel S3 2024 Q6 [16]}}