| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| 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 conditions. Parts (ii) and (iii) are routine applications: direct binomial calculation and standard normal approximation with continuity correction. All techniques are textbook exercises with no 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Constant \(p\), independent trials, fixed number of trials, only two outcomes | B1 | For an option |
| B1 2 | For a second option |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X<4) = 0.72^{14} + {}_{14}C_1 \times 0.28 \times 0.72^{13} + {}_{14}C_2 \times 0.28^2 \times 0.72^{12} + {}_{14}C_3 \times 0.28^3 \times 0.72^{11}\) | M1 | For adding with some \(C\) in \(P(0+1+2+3)\) or \(P(1+2+3)\) or \(P(0+1+2+3+4)\) or \(P(1+2+3+4)\) |
| M1 | For \(0.28\) and \(0.72\) to powers which sum to 14; need 2 or more terms | |
| \(= 0.0101 + 0.0548 + 0.1385 + 0.2154\) | A1 | For completely correct unsimplified form |
| \(= 0.419\) | A1 4 | For correct final answer; NB 0.418 is A0 if PA #1 or A1 if PA #2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mu = 50 \times 0.28 = 14\) | B1 | For 14 and 10.08 seen, can be implied |
| \(\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08\) | M1 | For standardising with or without cc, must have sq root |
| \(P(\text{more than } 18) = 1 - \Phi\!\left(\dfrac{18.5 - 14}{\sqrt{10.08}}\right)\) | M1 | For continuity correction 17.5 or 18.5 AND a final answer \(< 0.5\) |
| \(= 1 - \Phi(1.417)\) | ||
| \(= 1 - 0.9218\) or \(0.9217\) | ||
| \(= 0.0782\) or \(0.0783\) | A1 4 | For correct answer; NB 0.078 is A0 if RE #1 or A1 if RE #2 |
# Question 7:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Constant $p$, independent trials, fixed number of trials, only two outcomes | B1 | For an option |
| | B1 **2** | For a second option |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X<4) = 0.72^{14} + {}_{14}C_1 \times 0.28 \times 0.72^{13} + {}_{14}C_2 \times 0.28^2 \times 0.72^{12} + {}_{14}C_3 \times 0.28^3 \times 0.72^{11}$ | M1 | For adding with some $C$ in $P(0+1+2+3)$ or $P(1+2+3)$ or $P(0+1+2+3+4)$ or $P(1+2+3+4)$ |
| | M1 | For $0.28$ and $0.72$ to powers which sum to 14; need 2 or more terms |
| $= 0.0101 + 0.0548 + 0.1385 + 0.2154$ | A1 | For completely correct unsimplified form |
| $= 0.419$ | A1 **4** | For correct final answer; **NB** 0.418 is A0 if PA #1 or A1 if PA #2 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu = 50 \times 0.28 = 14$ | B1 | For 14 and 10.08 seen, can be implied |
| $\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08$ | M1 | For standardising with or without cc, must have sq root |
| $P(\text{more than } 18) = 1 - \Phi\!\left(\dfrac{18.5 - 14}{\sqrt{10.08}}\right)$ | M1 | For continuity correction 17.5 or 18.5 AND a final answer $< 0.5$ |
| $= 1 - \Phi(1.417)$ | | |
| $= 1 - 0.9218$ or $0.9217$ | | |
| $= 0.0782$ or $0.0783$ | A1 **4** | For correct answer; NB 0.078 is A0 if RE #1 or A1 if RE #2 |7 (i) State two conditions which must be satisfied for a situation to be modelled by a binomial distribution.
In a certain village 28\% of all cars are made by Ford.\\
(ii) 14 cars are chosen randomly in this village. Find the probability that fewer than 4 of these cars are made by Ford.\\
(iii) A random sample of 50 cars in the village is taken. Estimate, using a normal approximation, the probability that more than 18 cars are made by Ford.
\hfill \mbox{\textit{CAIE S1 2004 Q7 [10]}}