| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Single probability inequality |
| Difficulty | Standard +0.3 This is a straightforward application of normal approximation to binomial distribution. Part (i) requires simple probability subtraction (0.88 - 0.16), and part (ii) involves standard steps: identify p=0.72, calculate mean and variance for n=180, apply continuity correction, and use normal tables. The question is slightly above average difficulty only because it requires recognizing the setup and applying continuity correction correctly, but follows a very standard template with no conceptual challenges. |
| 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 |
| \(0.72\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 180 \times 0.72\), \(npq = 180 \times 0.72 \times 0.28\) | B1\(\checkmark\) | \(180 \times 0.72\), \(180 \times 0.72 \times 0.28\) seen, their values or correct |
| \(X \sim N(129.6,\ 36.288)\) | ||
| \(P(x > 115) = P\!\left(z > \dfrac{115.5 - 129.6}{\sqrt{36.288}}\right)\) | M1 | Standardising (\(\pm\)) must have sq rt |
| M1 | cc either 115.5 or 114.5 seen | |
| \(= P(z > -2.341)\) | M1 | Correct area, \(\Phi\) from final answer attempt fully correct method |
| \(= 0.990\) | A1 [5] |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.72$ | B1 [1] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 180 \times 0.72$, $npq = 180 \times 0.72 \times 0.28$ | B1$\checkmark$ | $180 \times 0.72$, $180 \times 0.72 \times 0.28$ seen, their values or correct |
| $X \sim N(129.6,\ 36.288)$ | | |
| $P(x > 115) = P\!\left(z > \dfrac{115.5 - 129.6}{\sqrt{36.288}}\right)$ | M1 | Standardising ($\pm$) must have sq rt |
| | M1 | cc either 115.5 or 114.5 seen |
| $= P(z > -2.341)$ | M1 | Correct area, $\Phi$ from final answer attempt fully correct method |
| $= 0.990$ | A1 [5] | |
---2 When visiting the dentist the probability of waiting less than 5 minutes is 0.16 , and the probability of waiting less than 10 minutes is 0.88 .\\
(i) Find the probability of waiting between 5 and 10 minutes.
A random sample of 180 people who visit the dentist is chosen.\\
(ii) Use a suitable approximation to find the probability that more than 115 of these people wait between 5 and 10 minutes.
\hfill \mbox{\textit{CAIE S1 2016 Q2 [6]}}