| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Percentages or proportions given |
| Difficulty | Standard +0.3 This is a standard normal distribution problem requiring inverse normal calculations and z-score manipulation. Part (i) involves setting up two equations from given percentages (routine but multi-step), part (ii) requires understanding continuity correction for discrete rounding, and part (iii) involves symmetric properties of the normal distribution. All techniques are standard S1 content with no novel insight required, making it slightly easier than average A-level questions overall. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{34.1 - \mu}{\sigma} = 1.751\); \(\frac{26.7 - \mu}{\sigma} = -0.524\); \(\mu = 28.4\), \(\sigma = 3.25\) | B1 B1 M1 M1 A1 [5] | ±1.751 seen; ±0.524 seen; a standardising equation with a z-value, \(\mu\) and \(\sigma\); valid attempt to eliminate \(\mu\) or \(\sigma\); correct answers |
| (ii) \(\Phi\left(\frac{34.5 - 32.9}{2.4}\right) - \Phi\left(\frac{33.5 - 32.9}{2.4}\right) = \Phi(0.667) - \Phi(0.25) = 0.7477 - 0.5987 = 0.149\) | M1 M1 A1 [3] | one numerical standardising expression, no cc, no square root, can have 34; subtracting two areas; correct answer |
| (iii) \(\Phi\left(\frac{t - 32.9}{2.4}\right) - \Phi\left(\frac{31.8 - 32.9}{2.4}\right) = 0.5\); \(\Phi\left(\frac{t - 32.9}{2.4}\right) - (1 - 0.6765) = 0.5\); \(\Phi\left(\frac{t - 32.9}{2.4}\right) = 0.8235\); \(\frac{t - 32.9}{2.4} = 0.929\); \(t = 35.1\) | M1 M1 M1 M1 A1 [4] | using 2 standardising expressions to give an equation involving subtraction and 0.5, oe; adding their tail to 0.5 oe; solving a standardised equation, must be a z-value from their 0.8235; correct final answer |
(i) $\frac{34.1 - \mu}{\sigma} = 1.751$; $\frac{26.7 - \mu}{\sigma} = -0.524$; $\mu = 28.4$, $\sigma = 3.25$ | B1 B1 M1 M1 A1 [5] | ±1.751 seen; ±0.524 seen; a standardising equation with a z-value, $\mu$ and $\sigma$; valid attempt to eliminate $\mu$ or $\sigma$; correct answers
(ii) $\Phi\left(\frac{34.5 - 32.9}{2.4}\right) - \Phi\left(\frac{33.5 - 32.9}{2.4}\right) = \Phi(0.667) - \Phi(0.25) = 0.7477 - 0.5987 = 0.149$ | M1 M1 A1 [3] | one numerical standardising expression, no cc, no square root, can have 34; subtracting two areas; correct answer
(iii) $\Phi\left(\frac{t - 32.9}{2.4}\right) - \Phi\left(\frac{31.8 - 32.9}{2.4}\right) = 0.5$; $\Phi\left(\frac{t - 32.9}{2.4}\right) - (1 - 0.6765) = 0.5$; $\Phi\left(\frac{t - 32.9}{2.4}\right) = 0.8235$; $\frac{t - 32.9}{2.4} = 0.929$; $t = 35.1$ | M1 M1 M1 M1 A1 [4] | using 2 standardising expressions to give an equation involving subtraction and 0.5, oe; adding their tail to 0.5 oe; solving a standardised equation, must be a z-value from their 0.8235; correct final answer6 The lengths, in cm, of trout in a fish farm are normally distributed. 96\% of the lengths are less than 34.1 cm and 70\% of the lengths are more than 26.7 cm .\\
(i) Find the mean and the standard deviation of the lengths of the trout.
In another fish farm, the lengths of salmon, $X \mathrm {~cm}$, are normally distributed with mean 32.9 cm and standard deviation 2.4 cm .\\
(ii) Find the probability that a randomly chosen salmon is 34 cm long, correct to the nearest centimetre.\\
(iii) Find the value of $t$ such that $\mathrm { P } ( 31.8 < X < t ) = 0.5$.
\hfill \mbox{\textit{CAIE S1 2012 Q6 [12]}}