| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial distribution formulas with no conceptual challenges. Parts (a)-(c) involve direct substitution into standard formulas (binomial probability, mean, variance), part (b) requires summing probabilities, and part (d) uses routine normal approximation with continuity correction. All techniques are standard S2 textbook exercises requiring only recall and basic calculation. |
| 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 |
|---|---|---|
| \(P(X \leq 3) - P(X \leq 2) = 0.9144 - 0.7382\) or \((0.2)^3(0.8)^6\frac{9!}{3!6!}\) | B1; M1; A1 awrt 0.176 | (3) |
| \(= 0.1762\) | \(= 0.1762\) | |
| (b) \(P(X \leq 4) = 0.9804\) | M1A1 awrt 0.98 | (2) |
| (c) Mean = 3, variance = 2.85, \(\frac{57}{20}\) | B1 B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 4) = 1 - P(X \leq 4)\) | M1; M1 | (3) |
| \(= 1 - 0.8153\) | A1 awrt 0.185 | |
| \(= 0.1847\) |
**(a)** Let $X$ be the random variable the number of games Bhim loses. $X \sim B(9, 0.2)$
$P(X \leq 3) - P(X \leq 2) = 0.9144 - 0.7382$ or $(0.2)^3(0.8)^6\frac{9!}{3!6!}$ | B1; M1; A1 awrt 0.176 | (3) | B1 – writing or use of $B(9, 0.2)$; M1 for writing/using $P(X \leq 3) - P(X \leq 2)$ or $(\rho)^3(1-p)^6\frac{9!}{3!6!}$; A1 awrt 0.176
$= 0.1762$ | $= 0.1762$ | |
**(b)** $P(X \leq 4) = 0.9804$ | M1A1 awrt 0.98 | (2) | M1 for writing or using $P(X \leq 4)$; A1 awrt 0.98
**(c)** Mean = 3, variance = 2.85, $\frac{57}{20}$ | B1 B1 | (2) | B1 3; B1 2.85, or exact equivalent
**(d)** $Po(3)$
$P(X > 4) = 1 - P(X \leq 4)$ | M1; M1 | (3) | M1 for using Poisson; M1 for writing or using $1 - P(X \leq 4)$; NB $P(X \leq 4)$ is 0.7254 $Po(3.5)$ and 0.8912 $Po(2.5)$
$= 1 - 0.8153$ | A1 awrt 0.185 | |
$= 0.1847$ | | |
**Special case: Use of Po(1.8) in (a) and (b)**
(a) can get B1 M1 A0 – B1 if written $B(9, 0.2)$, M1 for $e^{-1.8}\frac{1.8^3}{3!}$ or awrt to 0.161
If B(9, 0.2) is not seen then the only mark available for using Poisson is M1.
(b) can get M1 A0 - M1 for writing or using $P(X \leq 4)$ or may be implied by awrt 0.964
**Use of Normal in (d)**
Can get M0 M1 A0.- for M1 they must write $1 - P(X \leq 4)$ or get awrt 0.187
---Bhim and Joe play each other at badminton and for each game, independently of all others, the probability that Bhim loses is 0.2
Find the probability that, in 9 games, Bhim loses
\begin{enumerate}[label=(\alph*)]
\item exactly 3 of the games, [3]
\item fewer than half of the games. [2]
\end{enumerate}
Bhim attends coaching sessions for 2 months. After completing the coaching, the probability that he loses each game, independently of all others, is 0.05
Bhim and Joe agree to play a further 60 games.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the mean and variance for the number of these 60 games that Bhim loses. [2]
\item Using a suitable approximation calculate the probability that Bhim loses more than 4 games. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2010 Q2 [10]}}