| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| 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 to find parameters from given proportions, then forward calculations for probabilities. The binomial component in part (iii) is routine. While it involves multiple steps, all techniques are textbook-standard for S1 level with no novel insight required, making it slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.431 = \frac{135 - \mu}{\sigma}\) | B1 | One ±z-value correct, accept 0.430 |
| \(-0.842 = \frac{127 - \mu}{\sigma}\) | B1 | A second ±z-value correct |
| \(\sigma = 6.29\) \(\mu = 132\) | M1 | Solving two equations relating μ, σ, 135, 127 and their z-values (must be z-values) |
| A1 | Correct answer accept 6.28 | |
| A1 | Correct answer | |
| (ii) \(P(X < 145) = P\left(z < \frac{145-132.3}{6.284}\right)\) | M1 | Standardising no sq rt no cc |
| M1 | Correct use of normal tables | |
| \(= P(z < 2.023) = 0.978\) | A1 | Answer rounding to 0.978 or 0.979 |
| (iii) \(p = 1/3\) \(P(\text{at least } 2) = 1 - P(0, 1) = 1 - [(2/3)^8 + ^8C_1 \times (1/3)^1(2/3)^7]\) | M1 | Binomial expression with powers summing to 8 and \(^nC_{\text{something}}\) (any p) |
| A1 | Correct unsimplified expression | |
| \(= 0.805\) | A1 | Answer rounding to 0.805 |
**(i)** $0.431 = \frac{135 - \mu}{\sigma}$ | B1 | One ±z-value correct, accept 0.430 |
$-0.842 = \frac{127 - \mu}{\sigma}$ | B1 | A second ±z-value correct |
$\sigma = 6.29$ $\mu = 132$ | M1 | Solving two equations relating μ, σ, 135, 127 and their z-values (must be z-values) |
| A1 | Correct answer accept 6.28 |
| A1 | Correct answer |
**(ii)** $P(X < 145) = P\left(z < \frac{145-132.3}{6.284}\right)$ | M1 | Standardising no sq rt no cc |
| M1 | Correct use of normal tables |
$= P(z < 2.023) = 0.978$ | A1 | Answer rounding to 0.978 or 0.979 |
**(iii)** $p = 1/3$ $P(\text{at least } 2) = 1 - P(0, 1) = 1 - [(2/3)^8 + ^8C_1 \times (1/3)^1(2/3)^7]$ | M1 | Binomial expression with powers summing to 8 and $^nC_{\text{something}}$ (any p) |
| A1 | Correct unsimplified expression |
$= 0.805$ | A1 | Answer rounding to 0.805 |7 The heights that children of a particular age can jump have a normal distribution. On average, 8 children out of 10 can jump a height of more than 127 cm , and 1 child out of 3 can jump a height of more than 135 cm .\\
(i) Find the mean and standard deviation of the heights the children can jump.\\
(ii) Find the probability that a randomly chosen child will not be able to jump a height of 145 cm .\\
(iii) Find the probability that, of 8 randomly chosen children, at least 2 will be able to jump a height of more than 135 cm .
\hfill \mbox{\textit{CAIE S1 2010 Q7 [11]}}