| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | State general binomial conditions |
| Difficulty | Moderate -0.8 Part (i) is pure recall of standard binomial conditions. Parts (ii) and (iii) are routine binomial probability calculations using standard formulas or tables, with (iii) likely requiring normal approximation. No problem-solving insight needed, just direct application of learned techniques. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| constant/given prob, independent trials, fixed/given no. of trials, only two outcomes | B1 B1 [2] | One option correct, Three options correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(8, 9, 0, 1) =\) | M1 | One term seen involving \((0.3)^i(0.7)^j\) \(^j(^jC_i)\) |
| \(^9C_8(0.3)^7(0.7) + (0.3)^9 + (0.7)^9 + ^4C_1(0.3)(0.7)^8\) | A1 | Correct unsimplified expression |
| \(= 0.196\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| mean \(= 90 \times 0.3 = 27\), var \(= 18.9\) | B1 | Expressions for 27 and 18.9 (4.347) seen |
| \(P(X > 35) = 1 - \Phi\left(\frac{35.5-27}{\sqrt{18.9}}\right)\) | M1 | Standardising one expression, must have sq rt in denom, cc not necessary |
| \(= 1 - \Phi(1.955) = 0.0253\) | M1 | Continuity correction applied at least once, \((1-\Phi_1) + (1-\Phi_2)\) accept (0.0329 + 0.5) if no cc |
| \(P(X < 27) = \Phi\left(\frac{26.5-27}{\sqrt{18.9}}\right) = 1 - \Phi(0.115)\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total prob \(= 0.480\) accept 0.48 | A1 [5] | Rounding to correct answer |
**(i)**
constant/given prob, independent trials, fixed/given no. of trials, only two outcomes | B1 B1 [2] | One option correct, Three options correct
**(ii)**
$P(8, 9, 0, 1) =$ | M1 | One term seen involving $(0.3)^i(0.7)^j$ $^j(^jC_i)$
$^9C_8(0.3)^7(0.7) + (0.3)^9 + (0.7)^9 + ^4C_1(0.3)(0.7)^8$ | A1 | Correct unsimplified expression
$= 0.196$ | A1 [3] | Correct answer
**(iii)**
mean $= 90 \times 0.3 = 27$, var $= 18.9$ | B1 | Expressions for 27 and 18.9 (4.347) seen
$P(X > 35) = 1 - \Phi\left(\frac{35.5-27}{\sqrt{18.9}}\right)$ | M1 | Standardising one expression, must have sq rt in denom, cc not necessary
$= 1 - \Phi(1.955) = 0.0253$ | M1 | Continuity correction applied at least once, $(1-\Phi_1) + (1-\Phi_2)$ accept (0.0329 + 0.5) if no cc
$P(X < 27) = \Phi\left(\frac{26.5-27}{\sqrt{18.9}}\right) = 1 - \Phi(0.115)$ | M1 |
$= 0.4542$
Total prob $= 0.480$ accept 0.48 | A1 [5] | Rounding to correct answer6 (i) State three conditions that must be satisfied for a situation to be modelled by a binomial distribution.
On any day, there is a probability of 0.3 that Julie's train is late.\\
(ii) Nine days are chosen at random. Find the probability that Julie's train is late on more than 7 days or fewer than 2 days.\\
(iii) 90 days are chosen at random. Find the probability that Julie's train is late on more than 35 days or fewer than 27 days.
\hfill \mbox{\textit{CAIE S1 2010 Q6 [10]}}