| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Repeated binomial experiments |
| Difficulty | Moderate -0.3 This is a standard A-level statistics question on binomial-normal approximation with routine calculations. Part (a) is trivial reasoning, parts (b-d) involve standard binomial probability and normal approximation techniques taught in S2/Stats modules, and part (e) requires basic interpretation. The multi-step nature and repeated binomial aspect (part b) add slight complexity, but all techniques are textbook applications with no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| The seeds would be destroyed in the process so they would have none to sell | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \([S =\) no. of seeds out of 24 that germinate, \(S \sim B(24, 0.55)]\) \(T =\) no. of trays with at least 15 germinating, \(T \sim B(10, p)\) | M1 | For selection of appropriate model for \(T\) |
| \(p = P(S \geqslant 15) = 0.299126...\) | A1 | For correct value of parameter \(p\) (accept 0.3 or better) |
| \(P(T \geqslant 5) = 0.1487...\) awrt \(\mathbf{0.149}\) | A1 | For awrt 0.149 |
| Answer | Marks | Guidance |
|---|---|---|
| \(n\) is large and \(p\) close to 0.5 | B1 | Both correct conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim N(132, 59.4)\) | B1 | For correct normal distribution |
| \(P(X \geqslant 149.5) = P\!\left(Z \geqslant \dfrac{149.5 - 132}{\sqrt{59.4}}\right)\) | M1 | For correct use of continuity correction |
| \(= 0.01158...\) awrt \(\mathbf{0.0116}\) | A1cso | cso |
| Answer | Marks | Guidance |
|---|---|---|
| The probability is very small therefore there is evidence that the company's claim is incorrect | B1 | Correct statement |
## Question 5:
### Part (a):
| The seeds would be destroyed in the process so they would have none to sell | B1 | cao |
### Part (b):
| $[S =$ no. of seeds out of 24 that germinate, $S \sim B(24, 0.55)]$ $T =$ no. of trays with at least 15 germinating, $T \sim B(10, p)$ | M1 | For selection of appropriate model for $T$ |
| $p = P(S \geqslant 15) = 0.299126...$ | A1 | For correct value of parameter $p$ (accept 0.3 or better) |
| $P(T \geqslant 5) = 0.1487...$ awrt $\mathbf{0.149}$ | A1 | For awrt 0.149 |
### Part (c):
| $n$ is large and $p$ close to 0.5 | B1 | Both correct conditions |
### Part (d):
| $X \sim N(132, 59.4)$ | B1 | For correct normal distribution |
| $P(X \geqslant 149.5) = P\!\left(Z \geqslant \dfrac{149.5 - 132}{\sqrt{59.4}}\right)$ | M1 | For correct use of continuity correction |
| $= 0.01158...$ awrt $\mathbf{0.0116}$ | A1cso | cso |
### Part (e):
| The probability is very small therefore there is evidence that the company's claim is incorrect | B1 | Correct statement |
---\begin{enumerate}
\item A company sells seeds and claims that $55 \%$ of its pea seeds germinate.\\
(a) Write down a reason why the company should not justify their claim by testing all the pea seeds they produce.
\end{enumerate}
A random selection of the pea seeds is planted in 10 trays with 24 seeds in each tray.\\
(b) Assuming that the company's claim is correct, calculate the probability that in at least half of the trays 15 or more of the seeds germinate.\\
(c) Write down two conditions under which the normal distribution may be used as an approximation to the binomial distribution.
A random sample of 240 pea seeds was planted and 150 of these seeds germinated.\\
(d) Assuming that the company's claim is correct, use a normal approximation to find the probability that at least 150 pea seeds germinate.\\
(e) Using your answer to part (d), comment on whether or not the proportion of the company's pea seeds that germinate is different from the company's claim of $55 \%$
\hfill \mbox{\textit{Edexcel Paper 3 Q5 [9]}}