| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from percentile/frequency data |
| Difficulty | Standard +0.8 This question requires converting frequency data to proportions, then using inverse normal distribution to find z-scores, followed by solving simultaneous equations to find μ and σ. Part (ii) requires understanding of the empirical rule. The multi-step nature and need to work backwards from proportions to parameters makes this moderately challenging, though the techniques are standard for S1. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.04h Select appropriate distribution5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(-1.253 = \frac{6 - \mu}{\sigma}\) | B1 | \(Z = \pm 1.253\) |
| \(0.648 = \frac{12 - \mu}{\sigma}\) | B1 | \(Z = \pm 0.648\) |
| \(\mu = 9.9\) | M1 | Any equation with \(\mu\) and \(\sigma\) and a reasonable \(z\) value not a prob. Allow cc or −, not √o or \(\sigma^2\) |
| \(\sigma = 3.15\) or 3.16 | M1 [Indpt] | Att. to solve by substitution or elimination |
| A1 [5] | ||
| (ii) need \(P(z < -1\) or \(z > 1)\) | B1 | \(z = 1\) or − 1 seen |
| \(= 1 - \Phi(1) + \Phi(-1)\) | M1 | Correct area i.e. \(2 - 2\Phi\) |
| \(= 2 - 2 \times 0.8413\) | M1 | Mult their prob if sensible, by 1000 |
| \(= 0.3174\) | ||
| number = 317 | A1 [4] | Accept 317, 317.4, 318 |
**(i)** $-1.253 = \frac{6 - \mu}{\sigma}$ | B1 | $Z = \pm 1.253$
$0.648 = \frac{12 - \mu}{\sigma}$ | B1 | $Z = \pm 0.648$
$\mu = 9.9$ | M1 | Any equation with $\mu$ and $\sigma$ and a reasonable $z$ value not a prob. Allow cc or −, not √o or $\sigma^2$
$\sigma = 3.15$ or 3.16 | M1 [Indpt] | Att. to solve by substitution or elimination
| A1 [5]
**(ii)** need $P(z < -1$ or $z > 1)$ | B1 | $z = 1$ or − 1 seen
$= 1 - \Phi(1) + \Phi(-1)$ | M1 | Correct area i.e. $2 - 2\Phi$
$= 2 - 2 \times 0.8413$ | M1 | Mult their prob if sensible, by 1000
$= 0.3174$ |
number = 317 | A1 [4] | Accept 317, 317.4, 3186 The lengths of body feathers of a particular species of bird are modelled by a normal distribution. A researcher measures the lengths of a random sample of 600 body feathers from birds of this species and finds that 63 are less than 6 cm long and 155 are more than 12 cm long.\\
(i) Find estimates of the mean and standard deviation of the lengths of body feathers of birds of this species.\\
(ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean?
\hfill \mbox{\textit{CAIE S1 2012 Q6 [9]}}