| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question requiring standard cumulative probability calculations and a basic check of the variance formula np(1-p). Part (a) involves routine use of tables or calculator functions, while part (b) requires simple comparison of theoretical variance (8) with claimed variance (20.41), making it slightly above average due to the multi-part nature but still standard S1 material. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a)(i) | ||
| \(B(50, 0.2)\) | M1 | Use of in (a) |
| \(P(R \leq 15) = 0.969\) to \(0.97\) | A1 | AWFW, 0.9692 |
| 2 | ||
| 6(a)(ii) | ||
| \(P(R = 10) = P(R \leq 10) - P(R \leq 9)\) | M1 | Stated or implied |
| or \(P(R = 10) = \binom{50}{10}(0.2)^{10}(0.8)^{40}\) | M1 | Stated or implied |
| \(= 0.5836 - 0.4437 = 0.139\) to \(0.141\) | A1 | AWFW, 0.1399 |
| 2 | ||
| 6(a)(iii) | ||
| \(P(5 < R < 15) = P(R \leq 14\) or \(15) = 0.9393\) or \(0.9692\) | M1 | Accept values to 3 dp |
| minus \(P(R \leq 5\) or \(4) = 0.0480\) or \(0.0185\) | M1 | Accept values to 3 dp |
| \(= 0.89\) to \(0.893\) | A1 | AWFW, 0.8913 |
| or \(B(50, 0.2)\) expressions stated for at least 3 of \(5 \leq R \leq 15\) | M1 | Or implied by a correct answer |
| A2 | ||
| 3 | ||
| 6(b) | ||
| Mean, \(\mu = np = 50 \times 0.2 = 10\) | B1 | Either; CAO |
| or Estimate of \(p\), \(\hat{p} = 0.21\) | ||
| Variance, \(\sigma^2 = np(1 - p) = 10 \times 0.8 = 8\) | B1 | CAO |
| Mean or Estimate of \(p\) is similar to that expected but Variance (standard deviation) is different from that expected | B1 | 10.5 and 10 or 0.21 and 0.2. Either point. 20.41 and 8 or 4.5 and 2.8 |
| Reason to doubt validity of Sly's claim | B1 | Must be based on both 10 or 0.2 and 8 or on both 10 or 0.2 and 2.8 correctly |
| 4 |
| **6(a)(i)** |
|----------|
| $B(50, 0.2)$ | M1 | Use of in (a) |
| $P(R \leq 15) = 0.969$ to $0.97$ | A1 | AWFW, 0.9692 |
| | | 2 |
| **6(a)(ii)** |
|----------|
| $P(R = 10) = P(R \leq 10) - P(R \leq 9)$ | M1 | Stated or implied |
| or $P(R = 10) = \binom{50}{10}(0.2)^{10}(0.8)^{40}$ | M1 | Stated or implied |
| $= 0.5836 - 0.4437 = 0.139$ to $0.141$ | A1 | AWFW, 0.1399 |
| | | 2 |
| **6(a)(iii)** |
|----------|
| $P(5 < R < 15) = P(R \leq 14$ or $15) = 0.9393$ or $0.9692$ | M1 | Accept values to 3 dp |
| minus $P(R \leq 5$ or $4) = 0.0480$ or $0.0185$ | M1 | Accept values to 3 dp |
| $= 0.89$ to $0.893$ | A1 | AWFW, 0.8913 |
| or $B(50, 0.2)$ expressions stated for at least 3 of $5 \leq R \leq 15$ | M1 | Or implied by a correct answer |
| | A2 | |
| | | 3 |
| **6(b)** |
|----------|
| Mean, $\mu = np = 50 \times 0.2 = 10$ | B1 | Either; CAO |
| or Estimate of $p$, $\hat{p} = 0.21$ | | |
| Variance, $\sigma^2 = np(1 - p) = 10 \times 0.8 = 8$ | B1 | CAO |
| Mean or Estimate of $p$ is similar to that expected but Variance (standard deviation) is different from that expected | B1 | 10.5 and 10 or 0.21 and 0.2. Either point. 20.41 and 8 or 4.5 and 2.8 |
| Reason to doubt validity of Sly's claim | B1 | Must be based on both 10 or 0.2 and 8 or on both 10 or 0.2 and 2.8 correctly |
| | | 4 |
**Total for Q6: 11**
---6 Plastic clothes pegs are made in various colours.\\
The number of red pegs may be modelled by a binomial distribution with parameter $p$ equal to 0.2 .
The contents of packets of 50 pegs of mixed colours may be considered to be random samples.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that a packet contains:
\begin{enumerate}[label=(\roman*)]
\item less than or equal to 15 red pegs;
\item exactly 10 red pegs;
\item more than 5 but fewer than 15 red pegs.
\end{enumerate}\item Sly, a student, claims to have counted the number of red pegs in each of 100 packets of 50 pegs. From his results the following values are calculated.
Mean number of red pegs per packet $= 10.5$\\
Variance of number of red pegs per packet $= 20.41$\\
Comment on the validity of Sly's claim.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2006 Q6 [11]}}