| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 This is a standard S2 question combining normal distribution inverse problems with binomial probability. Part (a) requires using percentage points tables (routine), part (b) is direct binomial calculation with small n, and part (c) applies normal approximation to binomial—all well-practiced techniques with no novel insight required. Slightly above average due to the multi-part structure and need to carry forward the probability from part (a). |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(L > 100) = P\left(Z > \frac{100-\mu}{0.5}\right) = 0.3\) | M1 B1 | |
| \(\Rightarrow \frac{100 - \mu}{0.5} = 0.5244\) | A1 | (3) |
| \(\mu = 99.7378...\) cm | A1 | awrt 99.7 |
| \(X\) represents number more than 100cm. \(X \sim B(12, 0.3)\) | B1 | |
| \(P(X \leq 2) = 0.2528\) | M1A1 | awrt 0.253 |
| Normal approximation \(\mu = 400 \times 0.3 = 120, \sigma^2 = 84\) | M1, A1 | |
| \(P(X > 127) \approx 1 - P\left(Z < \frac{127.5 - 120}{\sqrt{84}}\right)\) | M1, M1, A1 | \(\pm 0.5\), standardise |
| \(\approx 1 - P(Z < 0.818)\) | ||
| \(= 1 - 0.7939\) | ||
| \(= 0.206\) or 0.207 | A1 | (6) |
| [12] |
| Answer | Marks |
|---|---|
| Item | Guidance |
| (a) | M1: standardising (\(\pm\)) with 100, \(\mu\) and 0.5 and setting equal to a z value. \(0.5 < z < 0.7\) |
| NB: Use of \(z = 0.7\) scores M0B0A0 | |
| B1: \(z = \pm 0.5244\) or better (Calc. Gives 0.5244005...). Must be used in an equation for \(\mu\). | |
| A1: awrt 99.7. Answer only is 0/3 | |
| NB: M1 + answer only of awrt 99.7 scores M1B0A1 but allow B1 for \(99.7376 \leq \mu \leq 99.7379\) | |
| (b) | B1: writing B(12, 0.3) |
| M1: writing \(P(X \leq 2)\). May be implied by sight of 0.252 or 0.253. | |
| NB: \(P(X < 3)\) alone is M0 unless they show that \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\) | |
| A1: awrt 0.253. Answer only scores 3/3 | |
| (c) | 1st M1: attempting to use a Normal approx. State \(N(\mu, \sigma^2)\) with \(\mu\) or \(\sigma\) correct |
| 1st A1: correct mean and var/sd | |
| 2nd M1: continuity correction used: either 127.5 or 126.5 seen | |
| 3rd M1: standardising with their \(\mu\) and \(\sigma\) and finding correct area. Must lead to \(P(Z > +\nu e)\) (o.e.) | |
| 2nd A1: \(\frac{127.5 - 120}{\sqrt{84}}\) or awrt 0.82 | |
| 3rd A1: for awrt 0.206 or 0.207 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(L > 100) = P\left(Z > \frac{100-\mu}{0.5}\right) = 0.3$ | M1 B1 | |
| $\Rightarrow \frac{100 - \mu}{0.5} = 0.5244$ | A1 | (3) |
| $\mu = 99.7378...$ cm | A1 | awrt 99.7 | |
| $X$ represents number more than 100cm. $X \sim B(12, 0.3)$ | B1 | |
| $P(X \leq 2) = 0.2528$ | M1A1 | awrt 0.253 | (3) |
| Normal approximation $\mu = 400 \times 0.3 = 120, \sigma^2 = 84$ | M1, A1 | |
| $P(X > 127) \approx 1 - P\left(Z < \frac{127.5 - 120}{\sqrt{84}}\right)$ | M1, M1, A1 | $\pm 0.5$, standardise | |
| $\approx 1 - P(Z < 0.818)$ | | |
| $= 1 - 0.7939$ | | |
| $= 0.206$ or 0.207 | A1 | (6) |
| | [12] | |
**Notes:**
| Item | Guidance |
|---|---|
| (a) | M1: standardising ($\pm$) with 100, $\mu$ and 0.5 and setting equal to a z value. $0.5 < z < 0.7$ |
| | NB: Use of $z = 0.7$ scores M0B0A0 |
| | B1: $z = \pm 0.5244$ or better (Calc. Gives 0.5244005...). Must be used in an equation for $\mu$. |
| | A1: awrt 99.7. Answer only is 0/3 |
| | NB: M1 + answer only of awrt 99.7 scores M1B0A1 but allow B1 for $99.7376 \leq \mu \leq 99.7379$ |
| (b) | B1: writing B(12, 0.3) |
| | M1: writing $P(X \leq 2)$. May be implied by sight of 0.252 or 0.253. |
| | NB: $P(X < 3)$ alone is M0 unless they show that $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ |
| | A1: awrt 0.253. Answer only scores 3/3 |
| (c) | 1st M1: attempting to use a Normal approx. State $N(\mu, \sigma^2)$ with $\mu$ or $\sigma$ correct |
| | 1st A1: correct mean and var/sd |
| | 2nd M1: continuity correction used: either 127.5 or 126.5 seen |
| | 3rd M1: standardising with their $\mu$ and $\sigma$ and finding correct area. Must lead to $P(Z > +\nu e)$ (o.e.) |
| | 2nd A1: $\frac{127.5 - 120}{\sqrt{84}}$ or awrt 0.82 |
| | 3rd A1: for awrt 0.206 or 0.207 |
---4. Pieces of ribbon are cut to length $L \mathrm {~cm}$ where $L \sim \mathrm {~N} \left( \mu , 0.5 ^ { 2 } \right)$
\begin{enumerate}[label=(\alph*)]
\item Given that $30 \%$ of the pieces of ribbon have length more than 100 cm , find the value of $\mu$ to the nearest 0.1 cm .
John selects 12 pieces of ribbon at random.
\item Find the probability that fewer than 3 of these pieces of ribbon have length more than 100 cm .
Aditi selects 400 pieces of ribbon at random.
\item Using a suitable approximation, find the probability that more than 127 of these pieces of ribbon will have length more than 100 cm .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q4 [12]}}