| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: finding σ from a symmetric probability statement, calculating a single probability using tables, and applying binomial distribution. All steps are routine for S1 level with no novel problem-solving required, 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 1.882\) or \(1.881\) | B1 | \(\pm 1.882\) or \(\pm 1.881\) seen |
| \(1.882 = (32 - 20)/\sigma\) | M1 | Equation using their \(z\) (must be a \(z\)-value), 32, 20 and \(s\) |
| \(\sigma = 6.38\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(x > 13) = P\left(z > \frac{13-20}{6.376}\right)\) | M1 | Standardising |
| \(= P(z > -1.0978)\) | M1 | Correct area \(> 0.5\) |
| \(= 0.864\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{at least } 2) = 1 - P(0,1)\) | M1 | Using 0.03 and 0.97 or 0.06 and 0.94 in a binomial expression powers summing to 7 |
| \(= 1 - (0.97)^7 - (0.03)(0.97)^6 {}_7C_1\) | M1 | Correct unsimplified binomial expansion |
| \(= 0.0171\) | A1 [3] | Correct answer |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 1.882$ or $1.881$ | B1 | $\pm 1.882$ or $\pm 1.881$ seen |
| $1.882 = (32 - 20)/\sigma$ | M1 | Equation using their $z$ (must be a $z$-value), 32, 20 and $s$ |
| $\sigma = 6.38$ | A1 [3] | Correct answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(x > 13) = P\left(z > \frac{13-20}{6.376}\right)$ | M1 | Standardising |
| $= P(z > -1.0978)$ | M1 | Correct area $> 0.5$ |
| $= 0.864$ | A1 [3] | Correct answer |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{at least } 2) = 1 - P(0,1)$ | M1 | Using 0.03 and 0.97 or 0.06 and 0.94 in a binomial expression powers summing to 7 |
| $= 1 - (0.97)^7 - (0.03)(0.97)^6 {}_7C_1$ | M1 | Correct unsimplified binomial expansion |
| $= 0.0171$ | A1 [3] | Correct answer |
---5 The weights of letters posted by a certain business are normally distributed with mean 20 g . It is found that the weights of $94 \%$ of the letters are within 12 g of the mean.\\
(i) Find the standard deviation of the weights of the letters.\\
(ii) Find the probability that a randomly chosen letter weighs more than 13 g .\\
(iii) Find the probability that at least 2 of a random sample of 7 letters have weights which are more than 12 g above the mean.
\hfill \mbox{\textit{CAIE S1 2011 Q5 [9]}}