| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Independent binomial samples with compound probability |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution across multiple parts. Parts (a)-(c) involve direct use of binomial probability formulas with clear parameters, while (d) requires a standard normal approximation—a routine technique taught in S2. The calculations are computational rather than conceptually challenging, with no novel problem-solving required beyond recognizing when to apply the normal approximation. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(W\) represent the number of white plants. \(W \sim B(12, 0.45)\) | use of B1M1 | |
| \(P(W = 5) = P(W \leq 5) - P(W < 4)\) | ||
| \(= {}^{12}C_5 0.45^5 0.55^7\) or equivalent | awrt 0.222(5) | values from correct table implies B |
| \(= 0.5269 - 0.3044\) | A1 (3 marks) | |
| \(= 0.2225\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(W \geq 7) = 1 - P(W \leq 6)\) | or \(= 1 - P(W < 7)\) | M1 |
| \(= 1 - 0.7393\) | implies method | A1 (2 marks) |
| \(= 0.2607\) | awrt 0.261 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{3 contain more white than coloured}) = \binom{10}{3}(0.2607)^3(1 - 0.2607)^7\) | use of B, \(n=10\) | M1A1† |
| \(= 0.256654\ldots\) | awrt 0.257 | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| mean \(= np = 22.5\); var \(= npq = 12.375\) | B1B1 | |
| \(P(W > 25) \approx P\left(Z > \frac{25.5 - 22.5}{\sqrt{12.375}}\right)\) | \(\pm\) standardise with \(\sigma\) and \(\mu\); \(\pm 0.5\) c.c. | M1;M1 |
| \(\approx P(Z > 0.8528\ldots)\) | awrt 0.85 | A1 |
| \(\approx 1 - 0.8023\) | 'one minus' | M1 |
| \(\approx 0.1977\) | awrt 0.197 or 0.198 | A1 (7 marks) |
**Part (a)**
Let $W$ represent the number of white plants. $W \sim B(12, 0.45)$ | use of B1M1 |
$P(W = 5) = P(W \leq 5) - P(W < 4)$ |
$= {}^{12}C_5 0.45^5 0.55^7$ or equivalent | awrt 0.222(5) | values from correct table implies B |
$= 0.5269 - 0.3044$ | | A1 (3 marks) |
$= 0.2225$ | |
**Part (b)**
$P(W \geq 7) = 1 - P(W \leq 6)$ | or $= 1 - P(W < 7)$ | M1 |
$= 1 - 0.7393$ | implies method | A1 (2 marks) |
$= 0.2607$ | awrt 0.261 | |
**Part (c)**
$P(\text{3 contain more white than coloured}) = \binom{10}{3}(0.2607)^3(1 - 0.2607)^7$ | use of B, $n=10$ | M1A1† |
$= 0.256654\ldots$ | awrt 0.257 | A1 (3 marks) |
**Part (d)**
mean $= np = 22.5$; var $= npq = 12.375$ | | B1B1 |
$P(W > 25) \approx P\left(Z > \frac{25.5 - 22.5}{\sqrt{12.375}}\right)$ | $\pm$ standardise with $\sigma$ and $\mu$; $\pm 0.5$ c.c. | M1;M1 |
$\approx P(Z > 0.8528\ldots)$ | awrt 0.85 | A1 |
$\approx 1 - 0.8023$ | 'one minus' | M1 |
$\approx 0.1977$ | awrt 0.197 or 0.198 | A1 (7 marks) |
---3. For a particular type of plant $45 \%$ have white flowers and the remainder have coloured flowers. Gardenmania sells plants in batches of 12. A batch is selected at random.
Calculate the probability that this batch contains
\begin{enumerate}[label=(\alph*)]
\item exactly 5 plants with white flowers,
\item more plants with white flowers than coloured ones.
Gardenmania takes a random sample of 10 batches of plants.
\item Find the probability that exactly 3 of these batches contain more plants with white flowers than coloured ones.
Due to an increasing demand for these plants by large companies, Gardenmania decides to sell them in batches of 50 .
\item Use a suitable approximation to calculate the probability that a batch of 50 plants contains more than 25 plants with white flowers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2007 Q3 [15]}}