| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Binomial with derived parameters |
| Difficulty | Moderate -0.3 This is a straightforward binomial distribution question testing standard formulas (mean = np, variance = np(1-p)) and normal approximation. Part (i) requires solving simultaneous equations from mean and variance, which is routine algebra. Part (ii) involves direct calculation or normal approximation with continuity correction - both standard S1 techniques with no novel problem-solving required. Slightly easier than average due to being purely procedural. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) \(np = 11\); \(np(1-p) = 4.95\); \(n = 20\) (\(p = 0.55\)) | B1, B1, M1, A1 4 | For solving, need find a value for n; For correct answer |
| (b) \(P(X = 12) = (0.55)^{12} \times (0.45)^8 \times _{20}C_{12} = 0.162\) | M1, A1 2 | For (their p)^12 x (their q)^w12 x k ≠ 1; For correct answer |
| (ii) \(\mu = 100 \times 0.3 = 30\), \(\sigma^2 = 100 \times 0.3 \times 0.7\) | B1, M1 | For both mean and variance correct, allow \(\sigma = 21\); For standardising with or without cc, allow their 21 or their \(\sqrt{21}\) in denom |
| \(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{21}}\right) = \Phi(0.9820) = 0.837\) (exact) | M1, A1 4 | For use of any continuity correction 34.5 or 35.5; For correct answer |
**(i) (a)** $np = 11$; $np(1-p) = 4.95$; $n = 20$ ($p = 0.55$) | B1, B1, M1, A1 4 | For solving, need find a value for n; For correct answer
**(b)** $P(X = 12) = (0.55)^{12} \times (0.45)^8 \times _{20}C_{12} = 0.162$ | M1, A1 2 | For (their p)^12 x (their q)^w12 x k ≠ 1; For correct answer
**(ii)** $\mu = 100 \times 0.3 = 30$, $\sigma^2 = 100 \times 0.3 \times 0.7$ | B1, M1 | For both mean and variance correct, allow $\sigma = 21$; For standardising with or without cc, allow their 21 or their $\sqrt{21}$ in denom
$P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{21}}\right) = \Phi(0.9820) = 0.837$ (exact) | M1, A1 4 | For use of any continuity correction 34.5 or 35.5; For correct answer\begin{enumerate}[label=(\roman*)]
\item A garden shop sells polyanthus plants in boxes, each box containing the same number of plants. The number of plants per box which produce yellow flowers has a binomial distribution with mean 11 and variance 4.95.
\begin{enumerate}[label=(\alph*)]
\item Find the number of plants per box. [4]
\item Find the probability that a box contains exactly 12 plants which produce yellow flowers. [2]
\end{enumerate}
\item Another garden shop sells polyanthus plants in boxes of 100. The shop's advertisement states that the probability of any polyanthus plant producing a pink flower is 0.3. Use a suitable approximation to find the probability that a box contains fewer than 35 plants which produce pink flowers. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2002 Q7 [10]}}