| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Challenging +1.2 This is a standard Further Statistics 2 question on linear combinations of normal variables. Part (a) requires forming W - 2X and finding P(W - 2X > 0), which is routine application of the formula for variance of linear combinations. Part (b) involves setting up nW + 2nX and using inverse normal tables to find n from a given probability. While it requires careful algebraic manipulation and understanding of the topic, it follows predictable patterns for FS2 questions without requiring novel insight or particularly complex reasoning. |
| Spec | 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 \(T = W - 2X\), then \(E(T) = 2.5 - 2\times1.27\) | M1 | Selecting and using appropriate model \(\pm(W-2X)\); may be implied by \(-0.04\) |
| \(= -0.04\) | A1 | \(-0.04\) oe |
| \(\text{Var}(T) = 0.7^2 + 2^2 \times 0.4^2\) | M1 | Realising need to use \(\text{Var}(W) + 4\,\text{Var}(X)\); allow use of \(0.7\) for \(\text{Var}(W)\) instead of \(0.7^2\) and/or \(0.4\) instead of \(0.4^2\) |
| \(= 1.13\) | A1 | 1.13 only |
| \(P\!\left(Z > \frac{0 - \text{"}-0.04\text{"}}{\sqrt{\text{"1.13"}}}\right) = P(Z > 0.0376...)\) | M1 | Realising \(P(T>0)\) required; \(\frac{0-\text{"their }-0.04\text{"}}{\sqrt{\text{"their 1.13"}}}\); may be implied by correct answer |
| \(= \text{awrt } 0.484/0.485\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B = W_1+W_2+\cdots+W_n+X_1+X_2+\cdots+X_{2n}\) | M1 | Selecting and using appropriate model; may be implied by 0.81 |
| \(E(B) = 5.04n\) | B1 | \(5.04n\) only |
| \(\text{Var}(B) = n\times0.7^2 + 2n\times0.4^2 = 0.81n\) | A1 | \(0.81n\) |
| \(\pm\frac{252 - \text{"5.04n"}}{\sqrt{\text{"0.81n"}}}\) | M1 | Standardising using their mean and sd; if mean and sd not given must be correct here |
| \(\frac{252 - \text{"5.04n"}}{\sqrt{\text{"0.81n"}}} = 0.8\) | M1 | Constructing equation; equate standardisation to 0.8 or awrt 0.7998; must be of form \(\frac{252-an}{b\sqrt{n}}=0.8\) or \(\frac{252-an}{bn}=0.8\) |
| \(5.04n + 0.72\sqrt{n} - 252 = 0\) oe | ||
| \(\sqrt{n} = -7.14...\) or \(7\) | M1 | Correctly solving their 3-term quadratic; condone \(n=7\) |
| \(n = 7^2\) | M1 | Realising need to square their answer / squaring their quadratic equation |
| \(= 49\) | A1cso | 49 only |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $T = W - 2X$, then $E(T) = 2.5 - 2\times1.27$ | M1 | Selecting and using appropriate model $\pm(W-2X)$; may be implied by $-0.04$ |
| $= -0.04$ | A1 | $-0.04$ oe |
| $\text{Var}(T) = 0.7^2 + 2^2 \times 0.4^2$ | M1 | Realising need to use $\text{Var}(W) + 4\,\text{Var}(X)$; allow use of $0.7$ for $\text{Var}(W)$ instead of $0.7^2$ and/or $0.4$ instead of $0.4^2$ |
| $= 1.13$ | A1 | 1.13 only |
| $P\!\left(Z > \frac{0 - \text{"}-0.04\text{"}}{\sqrt{\text{"1.13"}}}\right) = P(Z > 0.0376...)$ | M1 | Realising $P(T>0)$ required; $\frac{0-\text{"their }-0.04\text{"}}{\sqrt{\text{"their 1.13"}}}$; may be implied by correct answer |
| $= \text{awrt } 0.484/0.485$ | A1 | |
# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B = W_1+W_2+\cdots+W_n+X_1+X_2+\cdots+X_{2n}$ | M1 | Selecting and using appropriate model; may be implied by 0.81 |
| $E(B) = 5.04n$ | B1 | $5.04n$ only |
| $\text{Var}(B) = n\times0.7^2 + 2n\times0.4^2 = 0.81n$ | A1 | $0.81n$ |
| $\pm\frac{252 - \text{"5.04n"}}{\sqrt{\text{"0.81n"}}}$ | M1 | Standardising using their mean and sd; if mean and sd not given must be correct here |
| $\frac{252 - \text{"5.04n"}}{\sqrt{\text{"0.81n"}}} = 0.8$ | M1 | Constructing equation; equate standardisation to 0.8 or awrt 0.7998; must be of form $\frac{252-an}{b\sqrt{n}}=0.8$ or $\frac{252-an}{bn}=0.8$ |
| $5.04n + 0.72\sqrt{n} - 252 = 0$ oe | | |
| $\sqrt{n} = -7.14...$ or $7$ | M1 | Correctly solving their 3-term quadratic; condone $n=7$ |
| $n = 7^2$ | M1 | Realising need to square their answer / squaring their quadratic equation |
| $= 49$ | A1cso | 49 only |7 A manufacturer makes two versions of a toy. One version is made out of wood and the other is made out of plastic.
The weights, $W \mathrm {~kg}$, of the wooden toys are normally distributed with mean 2.5 kg and standard deviation 0.7 kg . The weights, $X \mathrm {~kg}$, of the plastic toys are normally distributed with mean 1.27 kg and standard deviation 0.4 kg . The random variables $W$ and $X$ are independent.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the weight of a randomly chosen wooden toy is more than double the weight of a randomly chosen plastic toy.
The manufacturer packs $n$ of these wooden toys and $2 n$ of these plastic toys into the same container. The maximum weight the container can hold is 252 kg .
The probability of the contents of this container being overweight is 0.2119 to 4 decimal places.
\item Calculate the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2019 Q7 [14]}}