| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.3 Part (i) requires reverse lookup from normal tables to find standard deviation (routine but slightly beyond basic recall), and part (ii) applies binomial probability with complement rule (P(at least 1) = 1 - P(none)). Both are standard textbook techniques with no novel insight required, making this slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{190 - 160}{\sigma} = 0.878\) | B1, M1 [3] | \(\pm 0.878, 0.88\), rounding to 0.88 seen \((190 - 160)/\sigma = \) something |
| \(\sigma = 34.2\) | A1 | Correct answer |
| (ii) \(P(\text{at least } 1) = 1 - P(0)\) | M1 | Using \(1 - P(0)\), \(1 - P(0,1)\), \(P(1,2 \ldots 12)\) or \(P(2, \ldots 12)\) with \(p = 0.19\) or 0.81, terms must be evaluated to get the M1 |
| \(= 1 - (0.81)^{12} = 0.920\) | A1 [2] | Correct answer accept 0.92 |
**(i)** $z = 0.878$
$\frac{190 - 160}{\sigma} = 0.878$ | B1, M1 [3] | $\pm 0.878, 0.88$, rounding to 0.88 seen $(190 - 160)/\sigma = $ something
$\sigma = 34.2$ | A1 | Correct answer
**(ii)** $P(\text{at least } 1) = 1 - P(0)$ | M1 | Using $1 - P(0)$, $1 - P(0,1)$, $P(1,2 \ldots 12)$ or $P(2, \ldots 12)$ with $p = 0.19$ or 0.81, terms must be evaluated to get the M1
$= 1 - (0.81)^{12} = 0.920$ | A1 [2] | Correct answer accept 0.923 The amount of fibre in a packet of a certain brand of cereal is normally distributed with mean 160 grams. 19\% of packets of cereal contain more than 190 grams of fibre.\\
(i) Find the standard deviation of the amount of fibre in a packet.\\
(ii) Kate buys 12 packets of cereal. Find the probability that at least 1 of the packets contains more than 190 grams of fibre.
\hfill \mbox{\textit{CAIE S1 2013 Q3 [5]}}