| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.3 Part (a) requires standardizing with μ and σ expressed algebraically (σ = μ/2), then using normal tables—straightforward but slightly above routine. Part (b) is a reverse normal distribution problem requiring finding σ from a probability, which is standard S1 fare but involves an extra step of working backwards from tables. Both parts are typical textbook exercises with no novel insight required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(y < 0) = P\left(z < \frac{0 - \mu}{\mu/2}\right)\) | M1 | Standardising containing 0 (can be implied) and \(\mu\) only |
| \(= P(z < -2)\) | A1 | \(z < -2\) seen |
| \(= 1 - 0.9772 = 0.0228\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(x > 2.1) = 253/8000 = 0.031625\) | M1 | \(1 -\) their \(253/8000\) used to obtain a \(z\)-value |
| \(P(x < 2.1) = 0.968375 = \Phi(z)\) | ||
| \(z = 1.857\) or \(1.858\) or \(1.859 = \frac{2.1 - 2.04}{\sigma}\) | A1 | Rounded to 1.86 seen |
| \(\sigma = 0.0323\) | M1 | Solving for \(\sigma\) using their \(z\) val; must be a \(z\) val |
| A1 [4] | Correct answer |
## Question 4:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(y < 0) = P\left(z < \frac{0 - \mu}{\mu/2}\right)$ | M1 | Standardising containing 0 (can be implied) and $\mu$ only |
| $= P(z < -2)$ | A1 | $z < -2$ seen |
| $= 1 - 0.9772 = 0.0228$ | A1 **[3]** | Correct answer |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(x > 2.1) = 253/8000 = 0.031625$ | M1 | $1 -$ their $253/8000$ used to obtain a $z$-value |
| $P(x < 2.1) = 0.968375 = \Phi(z)$ | | |
| $z = 1.857$ or $1.858$ or $1.859 = \frac{2.1 - 2.04}{\sigma}$ | A1 | Rounded to 1.86 seen |
| $\sigma = 0.0323$ | M1 | Solving for $\sigma$ using their $z$ val; must be a $z$ val |
| | A1 **[4]** | Correct answer |
---4
\begin{enumerate}[label=(\alph*)]
\item The random variable $Y$ is normally distributed with positive mean $\mu$ and standard deviation $\frac { 1 } { 2 } \mu$. Find the probability that a randomly chosen value of $Y$ is negative.
\item The weights of bags of rice are normally distributed with mean 2.04 kg and standard deviation $\sigma \mathrm { kg }$. In a random sample of 8000 such bags, 253 weighed over 2.1 kg . Find the value of $\sigma$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2013 Q4 [7]}}