| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution and normal approximation. Part (i) requires direct binomial calculation, part (ii) is a standard normal approximation with continuity correction, and part (iii) involves solving a simple inequality using complement probability. All techniques are routine for S1 level with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Method 1: \(P(<11) = 1 - P(11, 12, 13)\) | M1 | Binomial expression of form \({}^{13}C_x\ (p)^x(1-p)^{13-x}\), \(0 < x < 13\), \(0 < p < 1\) |
| \(= 1 - {}^{13}C_{11}(0.6)^{11}(0.4)^2 - {}^{13}C_{12}(0.6)^{12}(0.4) - (0.6)^{13}\) | M1 | Correct unsimplified answer |
| \(= 0.942\) | A1 | CAO |
| Method 2: \(P(<11) = P(0,1,2,3,4,5,6,7,8,9,10)\) | M1 | Binomial expression of form \({}^{13}C_x\ (p)^x(1-p)^{13-x}\), \(0 < x < 13\), \(0 < p < 1\) |
| \(= (0.4)^{13} + {}^{13}C_1(0.4)^{12}(0.6) + \ldots + {}^{13}C_{10}(0.4)^3(0.6)^{10}\) | M1 | Correct unsimplified answer |
| \(= 0.942\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 130\times0.35 = 45.5\), \(\text{var} = 130\times0.35\times0.65 = 29.575\) | B1 | Correct unsimplified mean and var (condone \(\sigma^2 = 29.6\), \(\sigma = 5.438\)) |
| \(P(\geqslant 50) = P\left(z > \frac{49.5-45.5}{\sqrt{29.575}}\right) = P(z > 0.7355)\) | M1 | Standardising, using \(\pm\left(\frac{x - their\ \text{mean}}{their\ \sigma}\right)\), \(x\) = value to standardise. 49.5 or 50.5 seen in \(\pm\) standardisation equation |
| \(= 1 - \Phi(0.7355)\) | M1 | Correct final area |
| \(= 1 - 0.7691\) | M1 | |
| \(= 0.231\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(1-(0.65)^n > 0.98\) or \(0.02 > (0.65)^n\) | M1 | Eqn or inequality involving \(0.65^n\) and \(0.02\) or \(0.35^n\) and \(0.98\) |
| \(n > 9.08\) | M1 | Attempt to solve their eqn or inequality by logs or trial and error |
| \(n = 10\) | A1 | CAO |
## Question 7(i):
**Method 1:** $P(<11) = 1 - P(11, 12, 13)$ | M1 | Binomial expression of form ${}^{13}C_x\ (p)^x(1-p)^{13-x}$, $0 < x < 13$, $0 < p < 1$
$= 1 - {}^{13}C_{11}(0.6)^{11}(0.4)^2 - {}^{13}C_{12}(0.6)^{12}(0.4) - (0.6)^{13}$ | M1 | Correct unsimplified answer
$= 0.942$ | A1 | CAO
**Method 2:** $P(<11) = P(0,1,2,3,4,5,6,7,8,9,10)$ | M1 | Binomial expression of form ${}^{13}C_x\ (p)^x(1-p)^{13-x}$, $0 < x < 13$, $0 < p < 1$
$= (0.4)^{13} + {}^{13}C_1(0.4)^{12}(0.6) + \ldots + {}^{13}C_{10}(0.4)^3(0.6)^{10}$ | M1 | Correct unsimplified answer
$= 0.942$ | A1 | CAO
---
## Question 7(ii):
$\mu = 130\times0.35 = 45.5$, $\text{var} = 130\times0.35\times0.65 = 29.575$ | B1 | Correct unsimplified mean and var (condone $\sigma^2 = 29.6$, $\sigma = 5.438$)
$P(\geqslant 50) = P\left(z > \frac{49.5-45.5}{\sqrt{29.575}}\right) = P(z > 0.7355)$ | M1 | Standardising, using $\pm\left(\frac{x - their\ \text{mean}}{their\ \sigma}\right)$, $x$ = value to standardise. 49.5 or 50.5 seen in $\pm$ standardisation equation
$= 1 - \Phi(0.7355)$ | M1 | Correct final area
$= 1 - 0.7691$ | M1 |
$= 0.231$ | A1 | Correct final answer
---
## Question 7(iii):
$1-(0.65)^n > 0.98$ or $0.02 > (0.65)^n$ | M1 | Eqn or inequality involving $0.65^n$ and $0.02$ **or** $0.35^n$ and $0.98$
$n > 9.08$ | M1 | Attempt to solve their eqn or inequality by logs **or** trial and error
$n = 10$ | A1 | CAO7 In a certain country, $60 \%$ of mobile phones sold are made by Company $A , 35 \%$ are made by Company $B$ and 5\% are made by other companies.\\
(i) Find the probability that, out of a random sample of 13 people who buy a mobile phone, fewer than 11 choose a mobile phone made by Company $A$.\\
(ii) Use a suitable approximation to find the probability that, out of a random sample of 130 people who buy a mobile phone, at least 50 choose a mobile phone made by Company $B$.\\
(iii) A random sample of $n$ mobile phones sold is chosen. The probability that at least one of these phones is made by Company $B$ is more than 0.98 . Find the least possible value of $n$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S1 2018 Q7 [11]}}