| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2021 |
| Session | October |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.3 This is a standard S3 question on linear combinations of normal variables requiring routine application of formulas: (a) difference of two observations uses variance sum rule, (b) sample mean distribution with inverse normal lookup, (c) linear combination of independent normals. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(B_1 - B_2) = 0\) | B1 | For expected value being 0 written or used |
| \(\text{Var}(B_1 - B_2) = 0.006\) | B1 | For 0.006 being written or used for Variance |
| \(P(\ | B_1 - B_2\ | > 0.1) = 2P(B_1 - B_2 > 0.1)\) |
| \(= 2 \times P\left(Z > \dfrac{0.1}{\sqrt{0.006}}\right) \left[= 2 \times P(Z > 1.2909...)\right]\) | M1 | Correct standardisation using their 0.1 and 0.006. If expected value and/or standard deviation not stated then they must be correct |
| \(= 0.1967...\) awrt 0.197 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{B} \sim N\left(1.96, \dfrac{0.003}{n}\right)\) | B1 | The correct distribution written or used |
| \(P(\bar{B} > 2) = P\left(Z > \dfrac{2 - 1.96}{\sqrt{0.003/n}}\right)\ [< 0.01]\) | M1 | Correct standardisation. Allow using their distribution if stated but must contain \(\sqrt{n}\) for sd |
| \(\dfrac{2 - 1.96}{\sqrt{0.003/n}} > 2.3263\) | B1 | Using awrt 2.3263 |
| (dependent on previous M) using a \(z\) value, \(2 < z < 3\) | dM1 | |
| \(n = 11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu_M = 21.8 + 500 \times 1.96\ [= 1001.8]\) | M1 | Correct method for finding the mean of \(M\) |
| \(\sigma_M^2 = 0.6 + 500 \times 0.003\ [= 2.1]\) | M1 | Correct method for finding the var of \(M\) |
| Let \(X = 4T - 3M\) | M1 | Realising the need to find \(4T - 3M\) or \(4T - 3M - 100\) or \(100 + 3M - 4T\) |
| \(\mu_X = 4 \times 774 - 3 \times 1001.8\ [= 90.6]\) | M1 | Correct method for finding the mean of \(X\) (using \(4T - 3M - 100 = -9.4\) or \(100 + 3M - 4T = 9.4\)) |
| \(\sigma_X^2 = 16 \times 1.8 + 9 \times 2.1\ [= 47.7]\) | M1 | Correct method for finding the var of \(X\) |
| \(P(4T - 3M > 100) = P\left(Z > \dfrac{100 - 90.6}{\sqrt{47.7}}\right)\left[= P(Z > 1.361...)\right]\) | M1 | Correct standardisation using their mean of \(X\) and their standard deviation of \(X\). If not stated they must be correct |
| \(= 0.0869\) (table) or \(0.08675...\) (calc) | A1 | awrt 0.0869 or 0.0868 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(B_1 - B_2) = 0$ | B1 | For expected value being 0 written or used |
| $\text{Var}(B_1 - B_2) = 0.006$ | B1 | For 0.006 being written or used for Variance |
| $P(\|B_1 - B_2\| > 0.1) = 2P(B_1 - B_2 > 0.1)$ | M1 | Realising they need to consider both |
| $= 2 \times P\left(Z > \dfrac{0.1}{\sqrt{0.006}}\right) \left[= 2 \times P(Z > 1.2909...)\right]$ | M1 | Correct standardisation using their 0.1 and 0.006. If expected value and/or standard deviation not stated then they must be correct |
| $= 0.1967...$ awrt 0.197 | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{B} \sim N\left(1.96, \dfrac{0.003}{n}\right)$ | B1 | The correct distribution written or used |
| $P(\bar{B} > 2) = P\left(Z > \dfrac{2 - 1.96}{\sqrt{0.003/n}}\right)\ [< 0.01]$ | M1 | Correct standardisation. Allow using their distribution if stated but must contain $\sqrt{n}$ for sd |
| $\dfrac{2 - 1.96}{\sqrt{0.003/n}} > 2.3263$ | B1 | Using awrt 2.3263 |
| (dependent on previous M) using a $z$ value, $2 < z < 3$ | dM1 | |
| $n = 11$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu_M = 21.8 + 500 \times 1.96\ [= 1001.8]$ | M1 | Correct method for finding the mean of $M$ |
| $\sigma_M^2 = 0.6 + 500 \times 0.003\ [= 2.1]$ | M1 | Correct method for finding the var of $M$ |
| Let $X = 4T - 3M$ | M1 | Realising the need to find $4T - 3M$ or $4T - 3M - 100$ or $100 + 3M - 4T$ |
| $\mu_X = 4 \times 774 - 3 \times 1001.8\ [= 90.6]$ | M1 | Correct method for finding the mean of $X$ (using $4T - 3M - 100 = -9.4$ or $100 + 3M - 4T = 9.4$) |
| $\sigma_X^2 = 16 \times 1.8 + 9 \times 2.1\ [= 47.7]$ | M1 | Correct method for finding the var of $X$ |
| $P(4T - 3M > 100) = P\left(Z > \dfrac{100 - 90.6}{\sqrt{47.7}}\right)\left[= P(Z > 1.361...)\right]$ | M1 | Correct standardisation using their mean of $X$ and their standard deviation of $X$. If not stated they must be correct |
| $= 0.0869$ (table) or $0.08675...$ (calc) | A1 | awrt 0.0869 or 0.0868 |\begin{enumerate}
\item A company produces bricks.
\end{enumerate}
The weight of a brick, $B \mathrm {~kg}$, is such that $B \sim \mathrm {~N} \left( 1.96 , \sqrt { 0.003 } ^ { 2 } \right)$\\
Two bricks are chosen at random.\\
(a) Find the probability that the difference in weight of the 2 bricks is greater than 0.1 kg
A random sample of $n$ bricks is to be taken.\\
(b) Find the minimum sample size such that the probability of the sample mean being greater than 2 is less than 1\%
The bricks are randomly selected and stacked on pallets.\\
The weight of an empty pallet, $E \mathrm {~kg}$, is such that $E \sim \mathrm {~N} \left( 21.8 , \sqrt { 0.6 } ^ { 2 } \right)$\\
The random variable $M$ represents the total weight of a pallet stacked with 500 bricks.
The random variable $T$ represents the total weight of a container of cement.\\
Given that $T$ is independent of $M$ and that $T \sim \mathrm {~N} \left( 774 , \sqrt { 1.8 } ^ { 2 } \right)$\\
(c) calculate $\mathrm { P } ( 4 T > 100 + 3 M )$\\
\hfill \mbox{\textit{Edexcel S3 2021 Q7 [17]}}