| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Standard +0.8 This question requires working backwards from P(X≥1) to find p, then applying binomial properties and Poisson approximation. While the individual techniques are standard Further Stats 1 content, the reverse-engineering of the parameter from a complement probability and the multi-stage nature (find p, calculate mean/variance, apply approximation, evaluate validity) elevates it above routine exercises. It's moderately challenging but well within expected FS1 scope. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \geq 1) = 1 - P(X=0)\), \(1 - P(X=0) = 0.049\) | B1 | Realising P(at least 1 call) \(= 1 - P(X=0)\) |
| \(P(X=0) = 0.951\) | B1 | Calculating \(P(X=0) = 0.951\) |
| \(x^5 = 0.951\), \(x = 0.99\) | M1 | Forming equation \(x^5 = 0.951\); may be implied by \(p = 0.01\) |
| \(p = 0.01\) | A1 | 0.01 only |
| \(X \sim B(1000, 0.01)\) | M1 | Realising need to use \(B(1000, 0.01)\); may be stated or used |
| Mean \(= np = 10\) | A1ft | \(= 10\) or ft their \(p\), only if \(0 < p < 1\) |
| Variance \(= np(1-p) = 9.9\) | A1ft | \(= 9.9\) or ft their \(p\), only if \(0 < p < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim Po(``10")\); require \(P(X>6) = 1 - P(X \leq 6)\) | M1 | Using \(Po(\text{"their 10"})\) and \(1 - P(X \leq 6)\) |
| \(= 1 - 0.1301 = 0.870\) | A1 | awrt 0.870; award M1 A1 for awrt 0.870 with no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The approximation is valid as the number of calls is large | B1 | Need context of number of calls |
| The probability of connecting to the wrong agent is small | B1 | Need context of connecting to wrong agent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The answer is accurate to 2 decimal places | B1 | Evaluating accuracy of answer in (b); allow 2 significant figures |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \geq 1) = 1 - P(X=0)$, $1 - P(X=0) = 0.049$ | B1 | Realising P(at least 1 call) $= 1 - P(X=0)$ |
| $P(X=0) = 0.951$ | B1 | Calculating $P(X=0) = 0.951$ |
| $x^5 = 0.951$, $x = 0.99$ | M1 | Forming equation $x^5 = 0.951$; may be implied by $p = 0.01$ |
| $p = 0.01$ | A1 | 0.01 only |
| $X \sim B(1000, 0.01)$ | M1 | Realising need to use $B(1000, 0.01)$; may be stated or used |
| Mean $= np = 10$ | A1ft | $= 10$ or ft their $p$, only if $0 < p < 1$ |
| Variance $= np(1-p) = 9.9$ | A1ft | $= 9.9$ or ft their $p$, only if $0 < p < 1$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim Po(``10")$; require $P(X>6) = 1 - P(X \leq 6)$ | M1 | Using $Po(\text{"their 10"})$ and $1 - P(X \leq 6)$ |
| $= 1 - 0.1301 = 0.870$ | A1 | awrt 0.870; award M1 A1 for awrt 0.870 with no incorrect working |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The approximation is valid as the number of calls is large | B1 | Need context of number of calls |
| The probability of connecting to the wrong agent is small | B1 | Need context of connecting to wrong agent |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The answer is accurate to 2 decimal places | B1 | Evaluating accuracy of answer in (b); allow 2 significant figures |
---\begin{enumerate}
\item A call centre routes incoming telephone calls to agents who have specialist knowledge to deal with the call. The probability of a caller, chosen at random, being connected to the wrong agent is p.
\end{enumerate}
The probability of at least 1 call in 5 consecutive calls being connected to the wrong agent is 0.049
The call centre receives 1000 calls each day.\\
(a) Find the mean and variance of the number of wrongly connected calls a day.\\
(b) Use a Poisson approximation to find, to 3 decimal places, the probability that more than 6 calls each day are connected to the wrong agent.\\
(c) Explain why the approximation used in part (b) is valid.
The probability that more than 6 calls each day are connected to the wrong agent using the binomial distribution is 0.8711 to 4 decimal places.\\
(d) Comment on the accuracy of your answer in part (b).\\
\hfill \mbox{\textit{Edexcel FS1 Q2 [12]}}