| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| 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 S2 question testing standard concepts: (i) requires a definition, (ii) is a direct binomial calculation with p=5/8 and n=6, and (iii) is a routine normal approximation to binomial with continuity correction. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.01a Population and sample: terminology2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Each element equally likely to be selected (and selections independent) OR each possible sample equally likely | B1, B1 | One of these two. "Selections independent" alone is insufficient, but don't need this. An example is insufficient. |
| (ii) \(B(6, 5/8)\); \(\binom{C_4}{p^5}(1-p)^2 = 0.32187\) | M1, M1, A1 | \(B(6; 5/8)\) stated or implied, allow e.g. \(499/799\). Correct formula, any \(p\). Answer, a.r.t. 0.322, can allow from wrong \(p\) |
| (iii) \(N(37.5, 225/16)\); \(\frac{39.5 - 37.5}{3.75} = 0.5333\) | B1, B1, M1 dep | Normal, mean 37.5, or 37.47 from \(499/799\), \(499/800\). 14.0625 or 3.75 seen, allow 14.07/14.1 or 3.75 |
| \(1 - \Phi(0.5333) = 0.297\) | dep M1, A1 | Standardise, wrong or no cc, \(np\), \(npq\), no \(\sqrt{n}\). Correct cc, \(\sqrt{npq}\), signs can be reversed. Tables used, answer \(< 0.5\), \(p = 5/8\) |
(i) Each element equally likely to be selected (and selections independent) OR each possible sample equally likely | B1, B1 | One of these two. "Selections independent" alone is insufficient, but don't need this. An example is insufficient.
(ii) $B(6, 5/8)$; $\binom{C_4}{p^5}(1-p)^2 = 0.32187$ | M1, M1, A1 | $B(6; 5/8)$ stated or implied, allow e.g. $499/799$. Correct formula, any $p$. Answer, a.r.t. 0.322, can allow from wrong $p$
(iii) $N(37.5, 225/16)$; $\frac{39.5 - 37.5}{3.75} = 0.5333$ | B1, B1, M1 dep | Normal, mean 37.5, or 37.47 from $499/799$, $499/800$. 14.0625 or 3.75 seen, allow 14.07/14.1 or 3.75
$1 - \Phi(0.5333) = 0.297$ | dep M1, A1 | Standardise, wrong or no cc, $np$, $npq$, no $\sqrt{n}$. Correct cc, $\sqrt{npq}$, signs can be reversed. Tables used, answer $< 0.5$, $p = 5/8$
---4 (i) Explain briefly what is meant by a random sample.
Random numbers are used to select, with replacement, a sample of size $n$ from a population numbered 000, 001, 002, ..., 799.\\
(ii) If $n = 6$, find the probability that exactly 4 of the selected sample have numbers less than 500 .\\
(iii) If $n = 60$, use a suitable approximation to calculate the probability that at least 40 of the selected sample have numbers less than 500 .
\hfill \mbox{\textit{OCR S2 2006 Q4 [10]}}