| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine application of binomial distribution and its approximations. Part (a) is definition recall, (b) is direct binomial calculation, and (c) requires standard Poisson and Normal approximations with continuity correction—all textbook procedures with no novel problem-solving required. Slightly easier than average due to straightforward setup and mechanical application. |
| 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) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Binomial | B1 | |
| Let \(X\) represent the number of green mugs in a sample | B1 | |
| (b) \(X \sim B(10, 0.06)\) | may be implied or seen in part a | B1 |
| \(P(X = 3) = ^{10}C_3(0.06)^3(0.94)^7\) | \(^{10}C_3(p)^3(1-p)^7\) | M1 |
| \(= 0.016808 \ldots\) | awrt 0.0168 | A1 |
| (c) Let \(X\) represent number of green mugs in a sample of size 125 | B1 | |
| (i) \(X \sim P_0(125 \times 0.06 = 7.5)\) | may be implied | B1 |
| \(P(10 \leq X \leq 13) = P(X \leq 13) - P(X \leq 9)\) | M1 | |
| \(= 0.9784 - 0.7764\) | A1 | |
| \(= 0.2020\) | awrt 0.202 | A1 |
| (ii) \(P(10 \leq X \leq 13) = P(9.5 \leq Y \leq 13.5)\) where \(Y \sim N(7.5, 7.05)\) | 7.05; 9.5, 13.5 | B1; B1 |
| \(= P\left(\frac{9.5-7.5}{\sqrt{7.05}} \leq z \leq \frac{13.5-7.5}{\sqrt{7.05}}\right)\) | \(\pm 0.5\) stand. | M1; M1 |
| both values or both correct expressions, awrt 0.75 and 2.26 | A1 | |
| \(= P(0.75 \leq z \leq 2.26)\) | A1 | |
| \(= 0.2147\) | awrt 0.214 or 0.215 | A1 |
**(a)** Binomial | B1 |
Let $X$ represent the number of green mugs in a sample | B1 |
**(b)** $X \sim B(10, 0.06)$ | may be implied or seen in part a | B1
$P(X = 3) = ^{10}C_3(0.06)^3(0.94)^7$ | $^{10}C_3(p)^3(1-p)^7$ | M1
$= 0.016808 \ldots$ | awrt 0.0168 | A1 |
**(c)** Let $X$ represent number of green mugs in a sample of size 125 | B1 |
**(i)** $X \sim P_0(125 \times 0.06 = 7.5)$ | may be implied | B1
$P(10 \leq X \leq 13) = P(X \leq 13) - P(X \leq 9)$ | M1 |
$= 0.9784 - 0.7764$ | A1 |
$= 0.2020$ | awrt 0.202 | A1 |
**(ii)** $P(10 \leq X \leq 13) = P(9.5 \leq Y \leq 13.5)$ where $Y \sim N(7.5, 7.05)$ | 7.05; 9.5, 13.5 | B1; B1
$= P\left(\frac{9.5-7.5}{\sqrt{7.05}} \leq z \leq \frac{13.5-7.5}{\sqrt{7.05}}\right)$ | $\pm 0.5$ stand. | M1; M1
both values or both correct expressions, awrt 0.75 and 2.26 | A1 |
$= P(0.75 \leq z \leq 2.26)$ | A1 |
$= 0.2147$ | awrt 0.214 or 0.215 | A1 |A manufacturer produces large quantities of coloured mugs. It is known from previous records that 6\% of the production will be green.
A random sample of 10 mugs was taken from the production line.
\begin{enumerate}[label=(\alph*)]
\item Define a suitable distribution to model the number of green mugs in this sample. [1]
\item Find the probability that there were exactly 3 green mugs in the sample. [3]
\end{enumerate}
A random sample of 125 mugs was taken.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the probability that there were between 10 and 13 (inclusive) green mugs in this sample, using
\begin{enumerate}[label=(\roman*)]
\item a Poisson approximation, [3]
\item a Normal approximation. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2006 Q5 [13]}}