| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Moderate -0.8 This is a straightforward binomial probability question requiring standard calculations: part (i) uses direct binomial probability with small n, part (ii) applies normal approximation with continuity correction, and part (iii) asks for routine verification of np>5 and nq>5. All techniques are textbook applications with no problem-solving insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - (P(7) + P(8) + P(9))\) \(= 1 - (^9C_7\ 0.8^7 \times 0.2^2 + ^9C_8\ 0.8^8 \times 0.2^1 + ^9C_9\ 0.8^9 \times 0.2^0)\) | M1 | Any binomial term of form \(^9C_x p^x (1-p)^{9-x}\), \(x \neq 0\) |
| M1 | Correct unsimplified expression | |
| \(= 1 - (0.3019899 + 0.3019899 + 0.1342177) = 0.262\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 200 \times 0.8 = 160\); var \(= 200 \times 0.8 \times 0.2 = 32\) | B1 | Both unsimplified |
| \(P(X > 166) = P\!\left(Z > \dfrac{166.5 - 160}{\sqrt{32}}\right)\) | M1 | Standardise, \(z = \pm\dfrac{x - \text{their } 160}{\sqrt{\text{their } 32}}\) with square root |
| M1 | \(166.5\) or \(165.5\) seen in attempted standardisation expression | |
| \(= P(Z > 1.149) = 1 - 0.8747\) | M1 | \(1 - \Phi\)-value, correct area expression, linked to final answer |
| \(= 0.125\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 160,\ nq = 40\): both \(> 5\) (so normal approx. holds) | B1 | Both parts required |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - (P(7) + P(8) + P(9))$ $= 1 - (^9C_7\ 0.8^7 \times 0.2^2 + ^9C_8\ 0.8^8 \times 0.2^1 + ^9C_9\ 0.8^9 \times 0.2^0)$ | M1 | Any binomial term of form $^9C_x p^x (1-p)^{9-x}$, $x \neq 0$ |
| | M1 | Correct unsimplified expression |
| $= 1 - (0.3019899 + 0.3019899 + 0.1342177) = 0.262$ | A1 | Correct answer |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 200 \times 0.8 = 160$; var $= 200 \times 0.8 \times 0.2 = 32$ | B1 | Both unsimplified |
| $P(X > 166) = P\!\left(Z > \dfrac{166.5 - 160}{\sqrt{32}}\right)$ | M1 | Standardise, $z = \pm\dfrac{x - \text{their } 160}{\sqrt{\text{their } 32}}$ with square root |
| | M1 | $166.5$ or $165.5$ seen in attempted standardisation expression |
| $= P(Z > 1.149) = 1 - 0.8747$ | M1 | $1 - \Phi$-value, correct area expression, linked to final answer |
| $= 0.125$ | A1 | Correct final answer |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 160,\ nq = 40$: both $> 5$ (so normal approx. holds) | B1 | Both parts required |5 At the Nonland Business College, all students sit an accountancy examination at the end of their first year of study. On average, $80 \%$ of the students pass this examination.\\
(i) A random sample of 9 students who will take this examination is chosen. Find the probability that at most 6 of these students will pass the examination.\\
(ii) A random sample of 200 students who will take this examination is chosen. Use a suitable approximate distribution to find the probability that more than 166 of them will pass the examination.\\
(iii) Justify the use of your approximate distribution in part (ii).\\
\hfill \mbox{\textit{CAIE S1 2018 Q5 [9]}}