| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.8 This question combines normal distribution with standardization, inverse normal lookup, and binomial approximation. Part (i) requires solving for μ using z-scores (non-trivial algebra with μ in both mean and SD), part (ii) is standard probability calculation, and part (iii) requires recognizing a binomial situation and applying normal approximation—multiple conceptual steps beyond routine exercises. |
| 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) \(z = \pm 1.751\) | B1 | Correct \(z\) |
| \(\pm \frac{20 - \mu}{\mu/4} = 1.751\) | M1 | Standardising no cc, no sqrt, must be a z-value |
| \(\mu = 13.9\) | A1 [3] | Correct answer |
| (ii) \(P(X < 10) = P(z < \frac{10 - 13.91}{13.91/4})\) | M1 | Standardising attempt with 10, their \(\mu\) and their \(\mu/4\), no cc, no sqrt |
| \(= P(z < -1.124)\) | M1 | "\(\Phi_1 + \Phi_2 - 1\)", fl their mean |
| \(= 1 - 0.8694 = 0.131\) | ||
| \(P(10 < X < 20) = 0.96 - 0.131 = 0.829\) or \(0.830\) | A1 [3] | Correct answer |
| (iii) \(\mu = 250 \times 0.96 = 240\) | B1 | 240 and 9.6 or sq rt 9.6 seen unsimplified |
| \(\sigma^2 = 250 \times 0.96 \times 0.04 = 9.6\) | ||
| \(P(\geq 235) = 1 - \Phi\left(\pm \frac{234.5 - 240}{\sqrt{9.6}}\right)\) | M1 | Standardising, with or without cc, must have sq rt in denom |
| M1 | Continuity correction 234.5 or 235.5 only | |
| \(= \Phi(1.775) = 0.962\) | A1 [5] | Correct region \(> 0.5\), fl their mean; Correct answer |
**(i)** $z = \pm 1.751$ | B1 | Correct $z$
$\pm \frac{20 - \mu}{\mu/4} = 1.751$ | M1 | Standardising no cc, no sqrt, must be a z-value
$\mu = 13.9$ | A1 [3] | Correct answer
**(ii)** $P(X < 10) = P(z < \frac{10 - 13.91}{13.91/4})$ | M1 | Standardising attempt with 10, their $\mu$ and their $\mu/4$, no cc, no sqrt
$= P(z < -1.124)$ | M1 | "$\Phi_1 + \Phi_2 - 1$", fl their mean
$= 1 - 0.8694 = 0.131$ | |
$P(10 < X < 20) = 0.96 - 0.131 = 0.829$ or $0.830$ | A1 [3] | Correct answer
**(iii)** $\mu = 250 \times 0.96 = 240$ | B1 | 240 and 9.6 or sq rt 9.6 seen unsimplified
$\sigma^2 = 250 \times 0.96 \times 0.04 = 9.6$ | |
$P(\geq 235) = 1 - \Phi\left(\pm \frac{234.5 - 240}{\sqrt{9.6}}\right)$ | M1 | Standardising, with or without cc, must have sq rt in denom
| M1 | Continuity correction 234.5 or 235.5 only
$= \Phi(1.775) = 0.962$ | A1 [5] | Correct region $> 0.5$, fl their mean; Correct answer
---5 The random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\frac { 1 } { 4 } \mu$. It is given that $\mathrm { P } ( X > 20 ) = 0.04$.\\
(i) Find $\mu$.\\
(ii) Find $\mathrm { P } ( 10 < X < 20 )$.\\
(iii) 250 independent observations of $X$ are taken. Find the probability that at least 235 of them are less than 20.
\hfill \mbox{\textit{CAIE S1 2011 Q5 [11]}}