| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution problem requiring inverse normal calculations to find μ and σ from two given probabilities, then a forward calculation for the warranty proportion. While it involves multiple steps and careful z-score work, it follows a well-practiced textbook pattern 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 transformation2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Bell shaped curve; symmetrical about the mean; 95% of data lies within 2sd of mean; asymptotic (any 2) | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 3500) = 0.01 \Rightarrow \mu - 3500 = 2.3263\sigma\) | M1, A1 | |
| \(P(X < 5500) = 0.025 \Rightarrow 5500 - \mu = 1.96\sigma\) | A1 | |
| Solving for \(\mu\) and \(\sigma\) | M1 | |
| \(\sigma = 466.6028\ldots\) accept \(466.6/467\) | A1 | |
| \(\mu = 4585.4583\ldots\) accept \(4585.5/4590\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 4000) = P\!\left(Z < \dfrac{4000 - 4585.4583\ldots}{466.6028\ldots}\right)\) | M1, A1ft | Follow-through on \(\mu\), \(\sigma\) |
| \(= P(Z < -1.25)\) | A1 | |
| \(= 0.1056\) | A1 |
## Question 5:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Bell shaped curve; symmetrical about the mean; 95% of data lies within 2sd of mean; asymptotic (any 2) | B1, B1 | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 3500) = 0.01 \Rightarrow \mu - 3500 = 2.3263\sigma$ | M1, A1 | |
| $P(X < 5500) = 0.025 \Rightarrow 5500 - \mu = 1.96\sigma$ | A1 | |
| Solving for $\mu$ and $\sigma$ | M1 | |
| $\sigma = 466.6028\ldots$ accept $466.6/467$ | A1 | |
| $\mu = 4585.4583\ldots$ accept $4585.5/4590$ | A1 | |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 4000) = P\!\left(Z < \dfrac{4000 - 4585.4583\ldots}{466.6028\ldots}\right)$ | M1, A1ft | Follow-through on $\mu$, $\sigma$ |
| $= P(Z < -1.25)$ | A1 | |
| $= 0.1056$ | A1 | |
---5. A random variable $X$ has a normal distribution.
\begin{enumerate}[label=(\alph*)]
\item Describe two features of the distribution of $X$.
A company produces electronic components which have life spans that are normally distributed. Only $1 \%$ of the components have a life span less than 3500 hours and $2.5 \%$ have a life span greater than 5500 hours.
\item Determine the mean and standard deviation of the life spans of the components.
The company gives warranty of 4000 hours on the components.
\item Find the proportion of components that the company can expect to replace under the warranty.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2002 Q5 [12]}}