| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution (part i) and normal approximation to binomial (parts ii-iii). Part (i) requires direct calculation of P(X=5)+P(X=6)+P(X=7) using binomial formula. Part (ii) is a standard normal approximation with continuity correction. Part (iii) simply asks to verify np>5 and nq>5. All steps are routine textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(5,6,7) = \binom{8}{5}(0.68)^5(0.32)^3 + \binom{8}{6}(0.68)^6(0.32)^2 + \binom{8}{7}(0.68)^7(0.32)\) | M1 | Binomial term \(\binom{8}{x} p^x (1-p)^{8-x}\) seen, \(0 < p < 1\) |
| M1 | Summing 3 binomial terms | |
| A1 | Correct unsimplified answer | |
| \(= 0.722\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(np = 340\), \(npq = 108.8\) | B1 | Correct (unsimplified) mean and variance |
| \(P(x > 337) = P\!\left(z > \frac{337.5 - 340}{\sqrt{108.8}}\right)\) | M1 | Standardising with square root; must have used 500 |
| M1 | cc either 337.5 or 336.5 | |
| \(= P(z > -0.2396)\) | M1 | Correct area \((> 0.5)\); must have used 500 |
| \(= 0.595\) | A1 [5] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(np(340) > 5\) and \(nq(160) > 5\) | B1 [1] | Must have both, or at least the smaller; need numerical justification |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(5,6,7) = \binom{8}{5}(0.68)^5(0.32)^3 + \binom{8}{6}(0.68)^6(0.32)^2 + \binom{8}{7}(0.68)^7(0.32)$ | M1 | Binomial term $\binom{8}{x} p^x (1-p)^{8-x}$ seen, $0 < p < 1$ |
| | M1 | Summing 3 binomial terms |
| | A1 | Correct unsimplified answer |
| $= 0.722$ | A1 [4] | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $np = 340$, $npq = 108.8$ | B1 | Correct (unsimplified) mean and variance |
| $P(x > 337) = P\!\left(z > \frac{337.5 - 340}{\sqrt{108.8}}\right)$ | M1 | Standardising with square root; must have used 500 |
| | M1 | cc either 337.5 or 336.5 |
| $= P(z > -0.2396)$ | M1 | Correct area $(> 0.5)$; must have used 500 |
| $= 0.595$ | A1 [5] | Correct answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $np(340) > 5$ and $nq(160) > 5$ | B1 [1] | Must have both, or at least the smaller; need numerical justification |
---6 (i) In a certain country, $68 \%$ of households have a printer. Find the probability that, in a random sample of 8 households, 5, 6 or 7 households have a printer.\\
(ii) Use an approximation to find the probability that, in a random sample of 500 households, more than 337 households have a printer.\\
(iii) Justify your use of the approximation in part (ii).
\hfill \mbox{\textit{CAIE S1 2015 Q6 [10]}}