| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 This is a standard two-stage problem: first finding a probability from a normal distribution (routine), then applying normal approximation to binomial with continuity correction. The justification in part (iii) requires recalling np>5 and nq>5 criteria. Slightly above average due to the multi-step nature and need to apply approximation correctly, but follows a well-practiced template for S1. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(x < 3.0) = P\!\left(z < \frac{3.0-2.6}{0.25}\right) + P(z < 1.6) = 0.945\) | M1, M1, A1 [3] | Standardising no sq rt no cc; Correct area i.e. prob \(> 0.5\) legit |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \sim B(500,\ 0.9452) \sim N(472.6,\ 25.898)\) | M1, M1 | \(500\times0.9452\) and \(500\times0.9452\times(1-0.9452)\) seen |
| \(P\!\left(z > \frac{479.5 - 472.6}{\sqrt{25.89848}}\right) = P(z > 1.3558)\) | M1, M1 | Standardising must have sq rt; All M marks independent; cc either 479.5 or 480.5 seen |
| \(= 1 - 0.9125 = 0.0875\) | A1 [5] | Correct area i.e. \(< 0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(500\times0.9452\) and \(500\times(1-0.9452)\) are both \(> 5\) | B1\(\checkmark\) [1] | Must see at least \(500\times0.0548 > 5\) or ft their (i); accept \(np>5\), \(nq>5\) if both not \(npq>5\) |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(x < 3.0) = P\!\left(z < \frac{3.0-2.6}{0.25}\right) + P(z < 1.6) = 0.945$ | M1, M1, A1 [3] | Standardising no sq rt no cc; Correct area i.e. prob $> 0.5$ legit |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim B(500,\ 0.9452) \sim N(472.6,\ 25.898)$ | M1, M1 | $500\times0.9452$ and $500\times0.9452\times(1-0.9452)$ seen |
| $P\!\left(z > \frac{479.5 - 472.6}{\sqrt{25.89848}}\right) = P(z > 1.3558)$ | M1, M1 | Standardising must have sq rt; All M marks independent; cc either 479.5 or 480.5 seen |
| $= 1 - 0.9125 = 0.0875$ | A1 [5] | Correct area i.e. $< 0.5$ |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $500\times0.9452$ and $500\times(1-0.9452)$ are both $> 5$ | B1$\checkmark$ [1] | Must see at least $500\times0.0548 > 5$ or ft their (i); accept $np>5$, $nq>5$ if both not $npq>5$ |
---5 Plastic drinking straws are manufactured to fit into drinks cartons which have a hole in the top. A straw fits into the hole if the diameter of the straw is less than 3 mm . The diameters of the straws have a normal distribution with mean 2.6 mm and standard deviation 0.25 mm .\\
(i) A straw is chosen at random. Find the probability that it fits into the hole in a drinks carton.\\
(ii) 500 straws are chosen at random. Use a suitable approximation to find the probability that at least 480 straws fit into the holes in drinks cartons.\\
(iii) Justify the use of your approximation.
\hfill \mbox{\textit{CAIE S1 2016 Q5 [9]}}