| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Standard +0.3 This is a multi-part normal distribution question with standard techniques: finding σ from a percentile (using z = -1.96), calculating probabilities between values, computing expected profit, and applying binomial approximation. Part (a) is routine inverse normal, (b) is standard probability calculation, (c) requires careful arithmetic with three cases, and (d) uses binomial distribution. All techniques are standard A-level fare with no novel insights required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{7.902-8}{x} = -1.96\) oe | M1 | Using normal distribution to set up equation. Allow \(\sigma\) for \(x\) and awrt \(\pm 1.96\) |
| \([x=]0.05*\) | A1cso* | cso. For correct expression for \(x\) followed by 0.05 or 0.05000... No incorrect working seen |
| SC B1 for \(\frac{7.902-8}{0.05} = -1.96 \Rightarrow 0.024998\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(7.94 \leqslant L \leqslant 8.09) = 0.8490...\) awrt 0.849 | B1 | awrt 0.849 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(L<7.94)=]\ 0.115069...\) (awrt 0.115) or \([P(L>8.09)=]\ 0.03593...\) (awrt 0.036) | B1 | awrt 0.115 (implied by awrt 57.5 for number of rods) or awrt 0.036 (implied by awrt 18 for number of rods) |
| \([P(L<7.94)=]\ 0.115069...\) & \([P(L>8.09)=]\ 0.03593...\) | B1 | awrt 0.115 and awrt 0.036 |
| Expected income per 500 rods \(= \sum(\text{Income} \times \text{probability} \times 500)\) \((500\times"0.849"\times0.5)+(500\times"0.1150..."\times0.05)+(500\times"0.03593..."\times0.4)\) or Expected profit per rod \(= \sum(\text{Profit}\times\text{probability})\) \(0.30\times"0.849"+-0.15\times"0.1150..."+0.20\times"0.03593..."\ [=0.2446..]\) | M1 | Correct method to find total income of 500 rods. Attempt at all 3 with at least two correct and no extras or correct method to find sum of all three profits with at least two of 30, \(-15\) or 20 correct. May work in pence but need to be consistent. Allow awrt 24.5 or 0.245 |
| \(500\times\sum(\text{Profit}\times\text{probability})\) or \(\sum(\text{Income}\times\text{probability}\times500)-500\times0.2\) \(= 500\times"0.2446..."\) or \(= "222.3"-500\times0.2\) \(=[\pounds]122.3...\) awrt [£]122 | M1d | Dep on previous method. May work in pence but need to be consistent. Allow \("0.2446..."\times500\) or "their income" for \(500\text{ rods}-500\times0.2\) (accept 499 or 501) |
| A1 | All previous marks must be awarded for awrt 122, awrt 12200p. NB if uses any integer values for numbers of rods then A0 other than for 18 for \(L>8.09\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(X \sim B(200, 0.015)\) | M1 | Selecting appropriate model. May be seen or used. Allow \(B(200,0.985)\) or \(Po(3)\). Condone \(B(0.015,200)\) or \(B(0.985,200)\) |
| \(P(X\leqslant 5)=\) | M1 | Writing or using \(P(X\leqslant 5)\). Do not accept \(P(X<6)\) unless found \(P(X\leqslant 5)\) |
| \(P(X\geqslant 6)=\) | M1 | Writing or using \(P(X\geqslant 6)\). Do not accept \(P(X>5)\) unless found \(P(X\geqslant 6)\) |
| \(0.9176...\) | A1 | 0.92 (Poisson 0.916...) |
| \(0.0824\) | A1 | 0.08 or better |
| Manufacturer is unlikely to achieve their aim since \(0.9176 < 0.95\) | A1ft | Need at least one method mark awarded. Correct conclusion with comparison. Ft "their \(p=0.9176...\)" as long as \(p>0.9\). If "their \(0.9176...\)"\(<0.95\) must be unlikely... If "their \(0.9176...\)"\(>0.95\) they must say...be likely... To ft the alternative then \(p<0.1\) |
| Manufacturer is unlikely to achieve their aim since \(0.0824 > 0.05\) | A1ft |
# Question 2:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{7.902-8}{x} = -1.96$ oe | M1 | Using normal distribution to set up equation. Allow $\sigma$ for $x$ and awrt $\pm 1.96$ |
| $[x=]0.05*$ | A1cso* | cso. For correct expression for $x$ followed by 0.05 or 0.05000... No incorrect working seen |
| SC B1 for $\frac{7.902-8}{0.05} = -1.96 \Rightarrow 0.024998$ | | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(7.94 \leqslant L \leqslant 8.09) = 0.8490...$ awrt **0.849** | B1 | awrt 0.849 |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(L<7.94)=]\ 0.115069...$ (awrt 0.115) **or** $[P(L>8.09)=]\ 0.03593...$ (awrt 0.036) | B1 | awrt 0.115 (implied by awrt 57.5 for number of rods) **or** awrt 0.036 (implied by awrt 18 for number of rods) |
| $[P(L<7.94)=]\ 0.115069...$ **&** $[P(L>8.09)=]\ 0.03593...$ | B1 | awrt 0.115 **and** awrt 0.036 |
| Expected income per 500 rods $= \sum(\text{Income} \times \text{probability} \times 500)$ $(500\times"0.849"\times0.5)+(500\times"0.1150..."\times0.05)+(500\times"0.03593..."\times0.4)$ **or** Expected profit per rod $= \sum(\text{Profit}\times\text{probability})$ $0.30\times"0.849"+-0.15\times"0.1150..."+0.20\times"0.03593..."\ [=0.2446..]$ | M1 | Correct method to find total income of 500 rods. Attempt at all 3 with at least two correct and no extras **or** correct method to find sum of all three profits with at least two of 30, $-15$ or 20 correct. May work in pence but need to be consistent. Allow awrt 24.5 or 0.245 |
| $500\times\sum(\text{Profit}\times\text{probability})$ **or** $\sum(\text{Income}\times\text{probability}\times500)-500\times0.2$ $= 500\times"0.2446..."$ **or** $= "222.3"-500\times0.2$ $=[\pounds]122.3...$ awrt **[£]122** | M1d | Dep on previous method. May work in pence but need to be consistent. Allow $"0.2446..."\times500$ or "their income" for $500\text{ rods}-500\times0.2$ (accept 499 or 501) |
| | A1 | All previous marks must be awarded for awrt 122, awrt 12200p. **NB** if uses any integer values for numbers of rods then A0 other than for 18 for $L>8.09$ |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $X \sim B(200, 0.015)$ | M1 | Selecting appropriate model. May be seen or used. Allow $B(200,0.985)$ or $Po(3)$. Condone $B(0.015,200)$ or $B(0.985,200)$ |
| $P(X\leqslant 5)=$ | M1 | Writing or using $P(X\leqslant 5)$. Do not accept $P(X<6)$ unless found $P(X\leqslant 5)$ |
| $P(X\geqslant 6)=$ | M1 | Writing or using $P(X\geqslant 6)$. Do not accept $P(X>5)$ unless found $P(X\geqslant 6)$ |
| $0.9176...$ | A1 | 0.92 (Poisson 0.916...) |
| $0.0824$ | A1 | 0.08 or better |
| Manufacturer is unlikely to achieve their aim since $0.9176 < 0.95$ | A1ft | Need at least one method mark awarded. Correct conclusion with comparison. Ft "their $p=0.9176...$" as long as $p>0.9$. If "their $0.9176...$"$<0.95$ must be unlikely... If "their $0.9176...$"$>0.95$ they must say...be likely... To ft the alternative then $p<0.1$ |
| Manufacturer is unlikely to achieve their aim since $0.0824 > 0.05$ | A1ft | |
---\begin{enumerate}
\item A manufacturer uses a machine to make metal rods.
\end{enumerate}
The length of a metal rod, $L \mathrm {~cm}$, is normally distributed with
\begin{itemize}
\item a mean of 8 cm
\item a standard deviation of $x \mathrm {~cm}$
\end{itemize}
Given that the proportion of metal rods less than 7.902 cm in length is $2.5 \%$\\
(a) show that $x = 0.05$ to 2 decimal places.\\
(b) Calculate the proportion of metal rods that are between 7.94 cm and 8.09 cm in length.
The cost of producing a single metal rod is 20p\\
A metal rod
\begin{itemize}
\item where $L < 7.94$ is sold for scrap for 5 p
\item where $7.94 \leqslant L \leqslant 8.09$ is sold for 50 p
\item where $L > 8.09$ is shortened for an extra cost of 10 p and then sold for 50 p\\
(c) Calculate the expected profit per 500 of the metal rods.
\end{itemize}
Give your answer to the nearest pound.
The same manufacturer makes metal hinges in large batches.\\
The hinges each have a probability of 0.015 of having a fault.\\
A random sample of 200 hinges is taken from each batch and the batch is accepted if fewer than 6 hinges are faulty.
The manufacturer's aim is for 95\% of batches to be accepted.\\
(d) Explain whether the manufacturer is likely to achieve its aim.
\hfill \mbox{\textit{Edexcel Paper 3 2022 Q2 [12]}}