| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from percentile/frequency data |
| Difficulty | Standard +0.8 Part (a) is routine normal distribution probability calculation. Part (b) requires setting up and solving simultaneous equations from two percentiles (32/200 = 0.16 and 17/200 = 0.085), finding corresponding z-scores, then solving for μ and σ—this multi-step inverse problem with two unknowns is significantly harder than standard normal distribution questions. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P\!\left(\frac{50-54}{6.1} < z < \frac{60-54}{6.1}\right) = P(-0.6557 < Z < 0.9836)\) | M1 | |
| Both values correct | A1 | |
| \(\Phi(0.9836) - \Phi(-0.6557) = \Phi(0.9836) + \Phi(0.6557) - 1 = 0.8375 + 0.7441 - 1\) (Correct area) | M1 | |
| \(0.582\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{45-\mu}{\sigma} = -0.994\) | B1 | |
| \(\frac{56-\mu}{\sigma} = 1.372\) | B1 | |
| One appropriate standardisation equation with \(\mu\), \(\sigma\), z-value (not probability) and 45 or 56 | M1 | |
| \(11 = 2.366\sigma\) | M1 | Correct algebraic elimination of \(\mu\) or \(\sigma\) from *their* two simultaneous equations to form an equation in one variable |
| \(\sigma = 4.65\), \(\mu = 49.6\) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P\!\left(\frac{50-54}{6.1} < z < \frac{60-54}{6.1}\right) = P(-0.6557 < Z < 0.9836)$ | M1 | |
| Both values correct | A1 | |
| $\Phi(0.9836) - \Phi(-0.6557) = \Phi(0.9836) + \Phi(0.6557) - 1 = 0.8375 + 0.7441 - 1$ (Correct area) | M1 | |
| $0.582$ | A1 | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{45-\mu}{\sigma} = -0.994$ | B1 | |
| $\frac{56-\mu}{\sigma} = 1.372$ | B1 | |
| One appropriate standardisation equation with $\mu$, $\sigma$, z-value (not probability) and 45 or 56 | M1 | |
| $11 = 2.366\sigma$ | M1 | Correct algebraic elimination of $\mu$ or $\sigma$ from *their* two simultaneous equations to form an equation in one variable |
| $\sigma = 4.65$, $\mu = 49.6$ | A1 | |
---6 The lengths of female snakes of a particular species are normally distributed with mean 54 cm and standard deviation 6.1 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen female snake of this species has length between 50 cm and 60 cm .\\
The lengths of male snakes of this species also have a normal distribution. A scientist measures the lengths of a random sample of 200 male snakes of this species. He finds that 32 have lengths less than 45 cm and 17 have lengths more than 56 cm .
\item Find estimates for the mean and standard deviation of the lengths of male snakes of this species.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q6 [9]}}