| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard z-score calculations and inverse normal lookup. Part (i) involves three routine probability calculations using tables, while part (ii) requires working backwards from a probability to find a boundary value. All techniques are standard S1 material with no conceptual challenges beyond applying the normal distribution formula correctly. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(<1.2) = P\left(z < \frac{1.2-1.9}{0.55}\right) = P(z < -1.2727)\) | M1 | Standardising for wt 1.2 or 2.5, no cc, sq, sq rt. May be awarded in (ii) if not attempted in (i) |
| \(= 1 - \Phi(1.273) = 1 - 0.8986 = 0.1014\) | A1 | Accept 0.102. First correct proportion seen |
| \(P(>2.5) = P\left(z < \frac{2.5-1.9}{0.55}\right) = P(z > 1.0909)\) | A1 | Second correct proportion seen |
| \(= 1 - \Phi(1.0909) = 1 - 0.8623 = 0.138\) | ||
| \(P(1.2 < wt < 2.5) = 1 - 0.101 - 0.138\) | M1 | Third proportion \(1 -\) their previous 2 proportions or correct attempt for remaining proportion |
| \(= 0.761\) | A1√ 5 | Correct answer or \(1 -\) their 2 previous correct proportions |
| (ii) \(P(x > k) = 0.8 + 0.1377 = 0.9377\) | M1 | Valid method to obtain P(x > k) or P(x < k). \(\pm 1.536\) seen accept 3sf rounding to 1.53 or 1.54 |
| \(z = -1.536\) | A1 | |
| \(-1.536 = \frac{k - 1.9}{0.55}\) | M1 | Attempt to solve equation with their 'correct' area z value, k, 1.9 and 0.55 |
| \(k = 1.06\) | A1 4 | Correct answer or rounding to 1.05 |
**(i)** $P(<1.2) = P\left(z < \frac{1.2-1.9}{0.55}\right) = P(z < -1.2727)$ | M1 | Standardising for wt 1.2 or 2.5, no cc, sq, sq rt. May be awarded in (ii) if not attempted in (i)
$= 1 - \Phi(1.273) = 1 - 0.8986 = 0.1014$ | A1 | Accept 0.102. First correct proportion seen
$P(>2.5) = P\left(z < \frac{2.5-1.9}{0.55}\right) = P(z > 1.0909)$ | A1 | Second correct proportion seen
$= 1 - \Phi(1.0909) = 1 - 0.8623 = 0.138$ |
$P(1.2 < wt < 2.5) = 1 - 0.101 - 0.138$ | M1 | Third proportion $1 -$ their previous 2 proportions or correct attempt for remaining proportion
$= 0.761$ | A1√ 5 | Correct answer or $1 -$ their 2 previous correct proportions
**(ii)** $P(x > k) = 0.8 + 0.1377 = 0.9377$ | M1 | Valid method to obtain P(x > k) or P(x < k). $\pm 1.536$ seen accept 3sf rounding to 1.53 or 1.54
$z = -1.536$ | A1 |
$-1.536 = \frac{k - 1.9}{0.55}$ | M1 | Attempt to solve equation with their 'correct' area z value, k, 1.9 and 0.55
$k = 1.06$ | A1 4 | Correct answer or rounding to 1.055 Gem stones from a certain mine have weights, $X$ grams, which are normally distributed with mean 1.9 g and standard deviation 0.55 g . These gem stones are sorted into three categories for sale depending on their weights, as follows.
Small: under 1.2 g Medium: between 1.2 g and 2.5 g Large: over 2.5 g\\
(i) Find the proportion of gem stones in each of these three categories.\\
(ii) Find the value of $k$ such that $\mathrm { P } ( k < X < 2.5 ) = 0.8$.
\hfill \mbox{\textit{CAIE S1 2014 Q5 [9]}}