| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single sum threshold probability |
| Difficulty | Standard +0.3 Part (a) is a straightforward application of summing independent normal variables (mean and variance scale predictably) followed by a standard normal probability calculation. Part (b) requires forming the linear combination L - 2S and recognizing this is normally distributed, then finding P(L - 2S ≥ 30). While this requires careful algebraic setup and understanding of variance rules for linear combinations, it's a standard Further Statistics 2 technique with no novel insight required. Slightly easier than average A-level due to being a direct application of taught methods. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(T = S_1 + S_2 + S_3\), then \(\text{E}(T) = 1500\) | M1 | Selecting and using the appropriate model and attempting \(3 \times 500\) |
| \(\text{Var}(T) = 75\) | M1 | For realising the need to use \(\text{Var}(S) + \text{Var}(S) + \text{Var}(S) = 3 \times 5^2\) |
| \(P(1490 < T < 1530) = 0.8756\ldots\) | A1 | awrt 0.876 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(W = \pm(L - 2S - 30)\) then \(\text{E}(W) = \pm(1020 - 2\times500 - 30)\) or Let \(X = \pm(L-2S)\) then \(\text{E}(X) = \pm(1020 - 2\times500)\) | M1 | Selecting and using the appropriate model \(\pm(L-2S-30)\) or \(\pm(L-2S)\) in an attempt to find the expected value |
| \(\text{E}(W) = -10\) (or 10) or \(\text{E}(X) = 20\) (or \(-20\)) | A1 | \(-10\) or 20 (or 10 or \(-20\)) |
| \(\text{Var}(\ldots) = 20^2 + 4 \times 5^2\) | M1 | For realising they need to use \(\text{Var}(L) + 4\text{Var}(S) = 20^2 + 4\times5^2\) |
| \(\text{Var}(\ldots) = 500\) | A1 | 500 only |
| \(P(W > 0)\) or \(P(X > 30)\) (or \(P(W<0)\) or \(P(X<30)\)) | M1 | Dependent on using an appropriate model and realising that \(P(W>0)\) or \(P(W<0)\) or \(P(X>30)\) or \(P(X<30)\) is required. May be implied by awrt 0.327. Using standardisation e.g. \(P\!\left(Z > \dfrac{0 - {-10}}{\sqrt{500}}\right)\) or \(P\!\left(Z > \dfrac{30 - {20}}{\sqrt{500}}\right)\) \((= P(Z > 0.4472\ldots))\) |
| \(= 0.3273\ldots\) | A1 | awrt 0.327 |
## Question 8:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $T = S_1 + S_2 + S_3$, then $\text{E}(T) = 1500$ | M1 | Selecting and using the appropriate model and attempting $3 \times 500$ |
| $\text{Var}(T) = 75$ | M1 | For realising the need to use $\text{Var}(S) + \text{Var}(S) + \text{Var}(S) = 3 \times 5^2$ |
| $P(1490 < T < 1530) = 0.8756\ldots$ | A1 | awrt 0.876 |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $W = \pm(L - 2S - 30)$ then $\text{E}(W) = \pm(1020 - 2\times500 - 30)$ **or** Let $X = \pm(L-2S)$ then $\text{E}(X) = \pm(1020 - 2\times500)$ | M1 | Selecting and using the appropriate model $\pm(L-2S-30)$ or $\pm(L-2S)$ in an attempt to find the expected value |
| $\text{E}(W) = -10$ (or 10) **or** $\text{E}(X) = 20$ (or $-20$) | A1 | $-10$ or 20 (or 10 or $-20$) |
| $\text{Var}(\ldots) = 20^2 + 4 \times 5^2$ | M1 | For realising they need to use $\text{Var}(L) + 4\text{Var}(S) = 20^2 + 4\times5^2$ |
| $\text{Var}(\ldots) = 500$ | A1 | 500 only |
| $P(W > 0)$ or $P(X > 30)$ (or $P(W<0)$ or $P(X<30)$) | M1 | Dependent on using an appropriate model and realising that $P(W>0)$ or $P(W<0)$ or $P(X>30)$ or $P(X<30)$ is required. May be implied by awrt 0.327. Using standardisation e.g. $P\!\left(Z > \dfrac{0 - {-10}}{\sqrt{500}}\right)$ or $P\!\left(Z > \dfrac{30 - {20}}{\sqrt{500}}\right)$ $(= P(Z > 0.4472\ldots))$ |
| $= 0.3273\ldots$ | A1 | awrt 0.327 |\begin{enumerate}
\item A company packs chickpeas into small bags and large bags.
\end{enumerate}
The weight of a small bag of chickpeas is normally distributed with mean 500 g and standard deviation 5 g
A random sample of 3 small bags of chickpeas is taken.\\
(a) Find the probability that the total weight of these 3 bags of chickpeas is between 1490 g and 1530 g
The weight of a large bag of chickpeas is normally distributed with mean 1020 g and standard deviation 20 g
One large bag and one small bag of chickpeas are chosen at random.\\
(b) Calculate the probability that the weight of the large bag of chickpeas is at least 30 g more than twice the weight of the small bag of chickpeas. Show your working clearly.
\hfill \mbox{\textit{Edexcel FS2 2024 Q8 [9]}}