| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This question combines standard normal distribution calculations with binomial probability (part i), solving simultaneous equations using z-scores (part ii), and independence of events (part iii). While multi-part, each component uses routine S1 techniques without requiring novel insight—slightly easier than average due to straightforward application of standard methods. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(> 42) = P\left(z > \frac{42 - 41.1}{3.4}\right)\) | M1 | Standardising no cc no sq rt no sq |
| \(= P(z \geq 0.2647) = 1 - 0.6045 = 0.3955\) | A1 | Correct prob rounding to 0.395 or 0.396 |
| \(\text{Prob} = (0.3955)(0.6045)^2{}_3C_1 = 0.433\) or \(0.434\) | M1 | Binomial \(_3C_r\) powers summing to 3, any \(p\), \(\Sigma p = 1\) |
| A1 [4] | Rounding to correct answer | |
| (ii) \(-1.282 = \frac{26.5 - \mu}{\sigma}\) | B1 | ±1.282 seen |
| \(1.645 = \frac{34.6 - \mu}{\sigma}\) | B1 | ±1.645 seen |
| M1 | An eqn with a z-value, \(\mu\) and \(\sigma\), no \(\sqrt{\sigma}\) no \(\sigma^2\) | |
| M1 | Sensible attempt to eliminate \(\mu\) or \(\sigma\) by substitution or subtraction | |
| \(\mu = 30.0\), \(\sigma = 2.77\) | A1 [5] | Correct answers, accept 30.1, accept 30, rounding to 2.77 |
| (iii) \(P(B6 < 34.6) = P\left(z < \frac{34.6 - 41.1}{3.4}\right)\) | M1 | Standardising for B6 no cc no sq rt no sq |
| \(= P(z < -1.912) = 1 - 0.9720 = 0.0280\) | A1 | Correct answer rounding to |
| \(P(B5 < 34.6) = 0.95\) | M1 | Mult by 0.95 or their regurgitated 0.95 |
| \(P(\text{both} < 34.6) = 0.028 \times 0.95 = 0.0266\) | A1 [4] | Correct answer rounding to 0.027, accept 0.027 |
(i) $P(> 42) = P\left(z > \frac{42 - 41.1}{3.4}\right)$ | M1 | Standardising no cc no sq rt no sq
$= P(z \geq 0.2647) = 1 - 0.6045 = 0.3955$ | A1 | Correct prob rounding to 0.395 or 0.396
$\text{Prob} = (0.3955)(0.6045)^2{}_3C_1 = 0.433$ or $0.434$ | M1 | Binomial $_3C_r$ powers summing to 3, any $p$, $\Sigma p = 1$
| A1 [4] | Rounding to correct answer
(ii) $-1.282 = \frac{26.5 - \mu}{\sigma}$ | B1 | ±1.282 seen
$1.645 = \frac{34.6 - \mu}{\sigma}$ | B1 | ±1.645 seen
| M1 | An eqn with a z-value, $\mu$ and $\sigma$, no $\sqrt{\sigma}$ no $\sigma^2$
| M1 | Sensible attempt to eliminate $\mu$ or $\sigma$ by substitution or subtraction
$\mu = 30.0$, $\sigma = 2.77$ | A1 [5] | Correct answers, accept 30.1, accept 30, rounding to 2.77
(iii) $P(B6 < 34.6) = P\left(z < \frac{34.6 - 41.1}{3.4}\right)$ | M1 | Standardising for B6 no cc no sq rt no sq
$= P(z < -1.912) = 1 - 0.9720 = 0.0280$ | A1 | Correct answer rounding to
$P(B5 < 34.6) = 0.95$ | M1 | Mult by 0.95 or their regurgitated 0.95
$P(\text{both} < 34.6) = 0.028 \times 0.95 = 0.0266$ | A1 [4] | Correct answer rounding to 0.027, accept 0.0277 The times taken to play Beethoven's Sixth Symphony can be assumed to have a normal distribution with mean 41.1 minutes and standard deviation 3.4 minutes. Three occasions on which this symphony is played are chosen at random.\\
(i) Find the probability that the symphony takes longer than 42 minutes to play on exactly 1 of these occasions.
The times taken to play Beethoven's Fifth Symphony can also be assumed to have a normal distribution. The probability that the time is less than 26.5 minutes is 0.1 , and the probability that the time is more than 34.6 minutes is 0.05 .\\
(ii) Find the mean and standard deviation of the times to play this symphony.\\
(iii) Assuming that the times to play the two symphonies are independent of each other, find the probability that, when both symphonies are played, both of the times are less than 34.6 minutes.
\hfill \mbox{\textit{CAIE S1 2012 Q7 [13]}}