| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Challenging +1.2 This question tests standard linear combinations of normal variables with multiple parts requiring careful setup of distributions and probability calculations. Part (a) and (b) involve forming new distributions from sums/differences/multiples of normals (routine S3 technique), while part (c) requires understanding of sampling distributions and working backwards from a probability equation. The multi-step nature and need to correctly handle variance rules elevates this above average difficulty, but it remains a standard S3 question type without requiring novel insight. |
| 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 Notes |
| \(W = L - (M_1 + M_2 + M_3)\) | B1 | Allow \(L-(M+M+M)\) but not \(L-3M\). Can be implied by correct \(\text{Var}(W)\). May use \(W = L-(M_1+M_2+M_3)-15\) for B1. |
| \(E(W) = 1510 - 3 \times 520 = -50\) | B1 | Accept 50 if definition reversed. Accept \(E(W) = 1510 - 3\times520 - 15 = -65\) |
| \(\text{Var}(W) = 20^2 + 10^2 + 10^2 + 10^2 = 700\) | M1, A1 | Attempt \(\text{Var}(W) = \text{Var}(L) + 3\text{Var}(M)\). Do not condone missing squares, cao. |
| \(P(W>15) = P\!\left(Z > \dfrac{15--50}{\sqrt{700}}\right)\) | dM1 | Attempting correct probability and standardising with their mean and sd, dependent on 1st M1. If values for \(W\) not used or wrong variance score M0. Must use 15. Accept \(P(W>0) = P\!\left(Z > \dfrac{0--65}{\sqrt{700}}\right)\) |
| \(= P(Z > 2.456769...)\) | ||
| \(= 0.0069\) | A1 | 0.0071 by calc. awrt 0.007 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(X = 3M - L\) | Can be implied by correct variance. | |
| \(E(X) = 3\times520 - 1510 = 50\) | B1 | Accept \(-50\) if reversed. |
| \(\text{Var}(X) = 3^2 \times 10^2 + 20^2 = 1300\) | M1, A1 | Attempt \(\text{Var}(W) = 3^2\text{Var}(M) + \text{Var}(S)\). Do not condone missing squares, cao. Condone \(10^2 + 3^2\times20^2\) for M1A0. |
| \(P(X>0) = P\!\left(Z > \dfrac{-50}{\sqrt{1300}}\right)\) | dM1 | Attempting correct probability and standardising with their mean and sd. |
| \(= P(Z > -1.38675...) = 0.9177\) | A1 | 0.9172 by calc. awrt 0.917–0.918 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(P(\text{all 5 bags weigh more than 520g}) = \left(\dfrac{1}{2}\right)^5 = \dfrac{1}{32} = 0.03125\) | B1 | 0.03125 |
| \(\bar{M} \sim N\!\left(520, \dfrac{10^2}{5}\right)\) or \(\displaystyle\sum_{i=1}^{5}M_i \sim N(2600, 500)\) | B1 | Both mean and variance required in either case. Can be implied below. |
| \(P(\bar{M}>d) = P\!\left(Z > \dfrac{d-520}{\frac{10}{\sqrt{5}}}\right) = 0.03125\) or \(P(T>5d) = P\!\left(Z > \dfrac{5d-2600}{\sqrt{500}}\right) = 0.03125\) | M1 | Standardise using \(d\), 520 and 10 or \(5d\), 2600 and \(\sqrt{500}\). |
| \(\dfrac{d-520}{\frac{10}{\sqrt{5}}} = 1.86(27...)\) or \(\dfrac{5d-2600}{\sqrt{500}} = 1.86(27...)\) | M1 | Equate to \(z\) value |
| \(d = 528.3\) | A1 | awrt 528.3 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $W = L - (M_1 + M_2 + M_3)$ | B1 | Allow $L-(M+M+M)$ but **not** $L-3M$. Can be implied by correct $\text{Var}(W)$. May use $W = L-(M_1+M_2+M_3)-15$ for B1. |
| $E(W) = 1510 - 3 \times 520 = -50$ | B1 | Accept 50 if definition reversed. Accept $E(W) = 1510 - 3\times520 - 15 = -65$ |
| $\text{Var}(W) = 20^2 + 10^2 + 10^2 + 10^2 = 700$ | M1, A1 | Attempt $\text{Var}(W) = \text{Var}(L) + 3\text{Var}(M)$. Do not condone missing squares, cao. |
| $P(W>15) = P\!\left(Z > \dfrac{15--50}{\sqrt{700}}\right)$ | dM1 | Attempting correct probability and standardising with their mean and sd, dependent on 1st M1. If values for $W$ not used or wrong variance score M0. Must use 15. Accept $P(W>0) = P\!\left(Z > \dfrac{0--65}{\sqrt{700}}\right)$ |
| $= P(Z > 2.456769...)$ | | |
| $= 0.0069$ | A1 | 0.0071 by calc. awrt 0.007 |
**(6 marks)**
---
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $X = 3M - L$ | | Can be implied by correct variance. |
| $E(X) = 3\times520 - 1510 = 50$ | B1 | Accept $-50$ if reversed. |
| $\text{Var}(X) = 3^2 \times 10^2 + 20^2 = 1300$ | M1, A1 | Attempt $\text{Var}(W) = 3^2\text{Var}(M) + \text{Var}(S)$. Do not condone missing squares, cao. Condone $10^2 + 3^2\times20^2$ for M1A0. |
| $P(X>0) = P\!\left(Z > \dfrac{-50}{\sqrt{1300}}\right)$ | dM1 | Attempting correct probability and standardising with their mean and sd. |
| $= P(Z > -1.38675...) = 0.9177$ | A1 | 0.9172 by calc. awrt 0.917–0.918 |
**(5 marks)**
---
## Part (c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $P(\text{all 5 bags weigh more than 520g}) = \left(\dfrac{1}{2}\right)^5 = \dfrac{1}{32} = 0.03125$ | B1 | 0.03125 |
| $\bar{M} \sim N\!\left(520, \dfrac{10^2}{5}\right)$ or $\displaystyle\sum_{i=1}^{5}M_i \sim N(2600, 500)$ | B1 | Both mean and variance required in either case. Can be implied below. |
| $P(\bar{M}>d) = P\!\left(Z > \dfrac{d-520}{\frac{10}{\sqrt{5}}}\right) = 0.03125$ or $P(T>5d) = P\!\left(Z > \dfrac{5d-2600}{\sqrt{500}}\right) = 0.03125$ | M1 | Standardise using $d$, 520 and 10 or $5d$, 2600 and $\sqrt{500}$. |
| $\dfrac{d-520}{\frac{10}{\sqrt{5}}} = 1.86(27...)$ or $\dfrac{5d-2600}{\sqrt{500}} = 1.86(27...)$ | M1 | Equate to $z$ value |
| $d = 528.3$ | A1 | awrt 528.3 |
**(5 marks)**
**Total: 16 marks**7. Sugar is packed into medium bags and large bags. The weights of the medium bags of sugar are normally distributed with mean 520 grams and standard deviation 10 grams. The weights of the large bags of sugar are normally distributed with mean 1510 grams and standard deviation 20 grams.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen large bag of sugar weighs at least 15 grams more than the combined weight of 3 randomly chosen medium bags of sugar.
\item Find the probability that a randomly chosen large bag of sugar weighs less than 3 times the weight of a randomly chosen medium bag of sugar.
A random sample of 5 medium bags of sugar is taken.
\item Find the value of $d$ so that the probability that all 5 bags of sugar each weigh more than 520 grams is equal to the probability that the mean weight of the 5 bags of sugar is more than $d$ grams.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q7 [16]}}