| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Type I and Type II errors |
| Difficulty | Standard +0.3 This is a straightforward application of normal approximation to binomial distribution with clear parameters (n=90, p=0.05 and p=0.35). Students must recognize the binomial setup, apply continuity correction, and perform standard normal calculations. While it requires understanding of Type I/II errors and justifying the approximation conditions (np>5, nq>5), these are routine S2 skills with no novel problem-solving required. |
| Spec | 2.04d Normal approximation to binomial2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(90, 0.05)\) | M1 | \(B(90, any)\) stated or implied, but \(p\) can be algebraic or omitted |
| \(\approx \text{Po}(4.5)\) | A1 | Poisson (4.5) stated or implied |
| \(P(\leq 6)\) from \(\text{Po}(4.5) = \mathbf{0.8311}\) | M1, A1 | Allow M1A0 for 0.7029, 0.9134, 0.8436 or 0.8180 but nothing else |
| \(n\) large, \(p\) small, or \(n > 50\), \(np < 5\), therefore Poisson | B1 | Either pair asserted, \(n = 90\), \(p = 0.05\) |
| Total: 5 | Exact (0.836055): M1A0M0A0B0, 1/5. Normal (4.5, 4.275): M1A0M0A0B0, 1/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(90, 0.35) \approx N(31.5, 20.475)\) | M1 | Normal, attempt at \(90 \times 0.35\) |
| A1 | Both parameters correct, allow \(\sqrt{}\) | |
| \(P(\leq 28) = \Phi\!\left(\frac{28.5 - 31.5}{\sqrt{20.475}}\right)\) | M1 | Standardise "29" using \(np\), \(\sqrt{npq}\), allow wrong/no cc, \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(= \Phi(-0.6630) = \mathbf{0.2537}\) | A1, A1 | cc and \(\sqrt{}\) correct. Answer in range \([0.2535, 0.2545]\) |
| Total: 5 | \(s^2 = 819/40\). Variance \(np\) or \(nq\), or extra \(\sqrt{90}\) in SD: M0 so N(31.5, 58.5) is 1/5 |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(90, 0.05)$ | M1 | $B(90, any)$ stated or implied, but $p$ can be algebraic or omitted |
| $\approx \text{Po}(4.5)$ | A1 | Poisson (4.5) stated or implied |
| $P(\leq 6)$ from $\text{Po}(4.5) = \mathbf{0.8311}$ | M1, A1 | Allow M1A0 for 0.7029, 0.9134, 0.8436 or 0.8180 but nothing else |
| $n$ large, $p$ small, or $n > 50$, $np < 5$, therefore Poisson | B1 | Either pair asserted, $n = 90$, $p = 0.05$ |
| **Total: 5** | | Exact (0.836055): M1A0M0A0B0, 1/5. Normal (4.5, 4.275): M1A0M0A0B0, 1/5 |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(90, 0.35) \approx N(31.5, 20.475)$ | M1 | Normal, attempt at $90 \times 0.35$ |
| | A1 | Both parameters correct, allow $\sqrt{}$ |
| $P(\leq 28) = \Phi\!\left(\frac{28.5 - 31.5}{\sqrt{20.475}}\right)$ | M1 | Standardise "29" using $np$, $\sqrt{npq}$, allow wrong/no cc, $\sigma^2$ or $\sqrt{\sigma}$ |
| $= \Phi(-0.6630) = \mathbf{0.2537}$ | A1, A1 | cc and $\sqrt{}$ correct. Answer in range $[0.2535, 0.2545]$ |
| **Total: 5** | | $s^2 = 819/40$. Variance $np$ or $nq$, or extra $\sqrt{90}$ in SD: M0 so N(31.5, 58.5) is 1/5 |
---4 A continuous random variable is normally distributed with mean $\mu$. A significance test for $\mu$ is carried out, at the $5 \%$ significance level, on 90 independent occasions.\\
(i) Given that the null hypothesis is correct on all 90 occasions, use a suitable approximation to find the probability that on 6 or fewer occasions the test results in a Type I error. Justify your approximation.\\
(ii) Given instead that on all 90 occasions the probability of a Type II error is 0.35 , use a suitable approximation to find the probability that on fewer than 29 occasions the test results in a Type II error.
\hfill \mbox{\textit{OCR S2 2015 Q4 [10]}}