| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Small sample binomial probability |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard binomial distribution concepts and normal approximation. Parts (a)-(c) require basic binomial probability calculations with small n=6, while part (d) involves routine application of normal approximation with continuity correction. The multi-step nature and need to find P(X≥2) first adds slight complexity, but all techniques are standard textbook exercises requiring no novel insight. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(X \sim B(6, 0.25)\) | B1 | B1 for a completely specified distribution. Condone \(B(6, 25\%)\) must be in (a)(i) |
| (a)(ii) Prizes are randomly placed in packets. Each packet has a 25% chance of containing a prize. Each packet contains a prize independently of others | B1 | B1 for a contextualised reason involving randomness, independence or constant probability. Must mention "prize" and "packet" and for constant prob "0.25" in correct statement. |
| (b) \(P(X = 1) = \binom{6}{1}(0.25)(1 - 0.25)^5 = [0.355957\ldots]\) or \(0.5339 - 0.1780 [= 0.3559]\) | M1 | 1st M1 for a correct expression for \(P(X = 1)\) may use \(P(X \leq 1) - P(X = 0)\) from tables with \(X \sim B(6, 0.25)\) (May be implied by sight of awrt 0.356 or answer in range) |
| \(P(\text{only 1 box contains exactly 1 prize}) = 2P(X = 1)(1 - P(X = 1))\) = answer in the range 0.458–0.459 (inc) | M1, A1 | 2nd M1 for writing or using \(2P(X = 1)(1 - P(X = 1))\) NB M0M1A0 is possible. Allow just \(2P(X = 1)(1 - P(X = 1))\) or a numerical expression with any \(p = P(X = 1)\) except \(p = 0.25\) provided \(0 < p < 1\) |
| (c) \(P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5339 = 0.4661\) awrt 0.466 | M1, A1 | M1 for writing or using \(1 - P(X \leq 1)\). A1 for awrt 0.466 (calc: 0.46606445...) |
| (d) \(Y \sim B(80, '0.4661') \to N(\text{awrt } 37.3, \text{ awrt } 19.9)\) | B1ft, M1, dM1A1ft | [Calc: 37.285..., 19.9078...]. 1st B1ft for mean = \(np\) and variance = \(np(1-p)\) where \(p = \text{'their (c)'} \approx 0.25\). Any ft values must be correct to at least 3sf. 1st M1 for a correct standardized expression (\(\pm\)) correct standardized expression (ft their \(\mu\) and \(\sigma\)) or \(z = \text{awrt} \pm 1.52\). 2nd M1 dependent on 1st M1 for using a continuity correction \(30 \pm 0.5\) |
| \(P(Y \leq 30) \approx P\left(Z < \frac{30.5 - '37.3'}{\sqrt{19.9'}}\right)\) | ||
| \(P(Z < -1.52) = 1 - 0.9357 = 0.0643\) (calc: 0.064165...) awrt 0.064 | A1 |
**(a)(i)** $X \sim B(6, 0.25)$ | B1 | B1 for a completely specified distribution. Condone $B(6, 25\%)$ must be in (a)(i)
**(a)(ii)** Prizes are randomly placed in packets. Each packet has a 25% chance of containing a prize. Each packet contains a prize independently of others | B1 | B1 for a contextualised reason involving randomness, independence or constant probability. Must mention "prize" and "packet" and for constant prob "0.25" in correct statement.
**(b)** $P(X = 1) = \binom{6}{1}(0.25)(1 - 0.25)^5 = [0.355957\ldots]$ or $0.5339 - 0.1780 [= 0.3559]$ | M1 | 1st M1 for a correct expression for $P(X = 1)$ may use $P(X \leq 1) - P(X = 0)$ from tables with $X \sim B(6, 0.25)$ (May be implied by sight of awrt 0.356 or answer in range)
| $P(\text{only 1 box contains exactly 1 prize}) = 2P(X = 1)(1 - P(X = 1))$ = answer in the range **0.458–0.459** (inc) | M1, A1 | 2nd M1 for writing or using $2P(X = 1)(1 - P(X = 1))$ NB M0M1A0 is possible. Allow just $2P(X = 1)(1 - P(X = 1))$ or a numerical expression with any $p = P(X = 1)$ except $p = 0.25$ provided $0 < p < 1$
**(c)** $P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5339 = 0.4661$ awrt **0.466** | M1, A1 | M1 for writing or using $1 - P(X \leq 1)$. A1 for awrt 0.466 (calc: 0.46606445...)
**(d)** $Y \sim B(80, '0.4661') \to N(\text{awrt } 37.3, \text{ awrt } 19.9)$ | B1ft, M1, dM1A1ft | [Calc: 37.285..., 19.9078...]. 1st B1ft for mean = $np$ and variance = $np(1-p)$ where $p = \text{'their (c)'} \approx 0.25$. Any ft values must be correct to at least 3sf. 1st M1 for a correct standardized expression ($\pm$) correct standardized expression (ft their $\mu$ and $\sigma$) or $z = \text{awrt} \pm 1.52$. 2nd M1 dependent on 1st M1 for using a continuity correction $30 \pm 0.5$
| $P(Y \leq 30) \approx P\left(Z < \frac{30.5 - '37.3'}{\sqrt{19.9'}}\right)$ | |
| $P(Z < -1.52) = 1 - 0.9357 = 0.0643$ (calc: 0.064165...) awrt **0.064** | A1 |
**Total: 12**
---2. Crispy-crisps produces packets of crisps. During a promotion, a prize is placed in $25 \%$ of the packets. No more than 1 prize is placed in any packet. A box contains 6 packets of crisps.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down a suitable distribution to model the number of prizes found in a box.
\item Write down one assumption required for the model.
\end{enumerate}\item Find the probability that in 2 randomly selected boxes, only 1 box contains exactly 1 prize.
\item Find the probability that a randomly selected box contains at least 2 prizes.
Neha buys 80 boxes of crisps.
\item Using a normal approximation, find the probability that no more than 30 of the boxes contain at least 2 prizes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q2 [12]}}