| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct expected frequency calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard z-score calculations and basic binomial probability. Part (i) requires a simple z-score calculation and multiplication by sample size. Part (ii) involves recognizing the 5% tail probability (1.645 standard deviations) and applying binomial distribution with clearly defined parameters. Both parts are routine procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(X > 20) = P(z > -6.4/3.7) = P(z > -1.730) = 0.9582\) | M1 | Standardising no cc no sq rt |
| A1 | Prob rounding to 0.958 | |
| Number of students \(= 335 \text{ or } 336\) | A1ft | Correct answer ft their prob, must be integer |
| [3] | ||
| (ii) \(P(\text{very slow}) = 0.05\) | B1 | 0.05 or 0.95 seen |
| \(P(0, 1, 2) = (0.95)^3 + {^3}C_1(0.05)(0.95)^2 + {^3}C_2(0.05)^2(0.95)^6\) | M1 | Binomial term with \(^nC_rp^r(1-p)^{n-r}\) seen any \(p\) |
| \(= 0.6634 + 0.2793 + 0.0515 = 0.994\) | M1 | Correct expression for P(0, 1, 2), \(p\) close to 0.05 |
| A1 | Answer rounding to 0.994 | |
| [4] |
**(i)** $P(X > 20) = P(z > -6.4/3.7) = P(z > -1.730) = 0.9582$ | M1 | Standardising no cc no sq rt
| A1 | Prob rounding to 0.958
Number of students $= 335 \text{ or } 336$ | A1ft | Correct answer ft their prob, must be integer
| [3] |
**(ii)** $P(\text{very slow}) = 0.05$ | B1 | 0.05 or 0.95 seen
$P(0, 1, 2) = (0.95)^3 + {^3}C_1(0.05)(0.95)^2 + {^3}C_2(0.05)^2(0.95)^6$ | M1 | Binomial term with $^nC_rp^r(1-p)^{n-r}$ seen any $p$
$= 0.6634 + 0.2793 + 0.0515 = 0.994$ | M1 | Correct expression for P(0, 1, 2), $p$ close to 0.05
| A1 | Answer rounding to 0.994
| [4] |3 The times taken by students to get up in the morning can be modelled by a normal distribution with mean 26.4 minutes and standard deviation 3.7 minutes.\\
(i) For a random sample of 350 students, find the number who would be expected to take longer than 20 minutes to get up in the morning.\\
(ii) 'Very slow' students are students whose time to get up is more than 1.645 standard deviations above the mean. Find the probability that fewer than 3 students from a random sample of 8 students are 'very slow'.
\hfill \mbox{\textit{CAIE S1 2010 Q3 [7]}}