| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics 1 question testing standard geometric distribution knowledge. Part (a) is simple identification, (b) is routine probability calculation, (c) applies CLT in a standard way, and (d) is a textbook hypothesis test. All parts follow predictable patterns with no novel problem-solving required, though the multi-part structure and hypothesis testing elevate it slightly above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(X \sim \text{Geo}(\frac{1}{6})\) accept in words: geometric distribution with \(p = \frac{1}{6}\) | B1 |
| (b) | \(P(X < 3) = 1 - P(X > 3)\) or \(P(X = 1) + P(X = 2) + P(X = 3)\) | M1 |
| \(= 1 - \left(\frac{5}{6}\right)^3\) or \(\frac{1}{6} + \frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{91}{216}\) | A1cso | 1.1b |
| (c) | \(E(X) = 6\) \(\quad \text{Var}(X) = \frac{5}{(\frac{1}{6})^2} [= 30]\) | M1, A1 |
| \(\overline{X} \approx N\left(6, \sqrt{\left(\frac{30}{64}\right)^2}\right)\) | M1, A1 | 3.3, 1.1b |
| \(P(5.6 < \overline{X} < 7.2) = 0.68064\ldots\) | A1 | 1.1b |
| (d) | \(H_0: p = \frac{1}{6}\) \(\quad H_1: p < \frac{1}{6}\) | B1 |
| Probability approach: \(P(X \geq 16)\) | M1 | 3.4 |
| \(= P(X > 15) = \left(\frac{5}{6}\right)^{15} = 0.0649\ldots\) | A1 | 1.1b |
| (Not significant) insufficient evidence that probability is < \(\frac{1}{6}\) | A1 | 2.2b |
| OR CR approach: \(\left(\frac{5}{6}\right)^{r-1} < 0.05\) or \(\left(\frac{5}{6}\right)^d < 0.05\) with use of logs | M1 | 3.4 |
| CR: \(X \geq 18\) or \(X > 17\) | A1 | 1.1b |
| (Not significant) insufficient evidence that dice is biased | A1 | 2.2b |
(a) | $X \sim \text{Geo}(\frac{1}{6})$ accept in words: geometric distribution with $p = \frac{1}{6}$ | B1 | 3.3 |
(b) | $P(X < 3) = 1 - P(X > 3)$ or $P(X = 1) + P(X = 2) + P(X = 3)$ | M1 | 1.1b |
| $= 1 - \left(\frac{5}{6}\right)^3$ or $\frac{1}{6} + \frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{91}{216}$ | A1cso | 1.1b |
(c) | $E(X) = 6$ $\quad \text{Var}(X) = \frac{5}{(\frac{1}{6})^2} [= 30]$ | M1, A1 | 3.4, 1.1b |
| $\overline{X} \approx N\left(6, \sqrt{\left(\frac{30}{64}\right)^2}\right)$ | M1, A1 | 3.3, 1.1b |
| $P(5.6 < \overline{X} < 7.2) = 0.68064\ldots$ | A1 | 1.1b |
(d) | $H_0: p = \frac{1}{6}$ $\quad H_1: p < \frac{1}{6}$ | B1 | 2.5 |
| **Probability approach:** $P(X \geq 16)$ | M1 | 3.4 |
| $= P(X > 15) = \left(\frac{5}{6}\right)^{15} = 0.0649\ldots$ | A1 | 1.1b |
| **(Not significant) insufficient evidence that probability is < $\frac{1}{6}$** | A1 | 2.2b |
| **OR CR approach:** $\left(\frac{5}{6}\right)^{r-1} < 0.05$ or $\left(\frac{5}{6}\right)^d < 0.05$ with use of logs | M1 | 3.4 |
| **CR: $X \geq 18$ or $X > 17$** | A1 | 1.1b |
| **(Not significant) insufficient evidence that dice is biased** | A1 | 2.2b |
**Notes:**
- (a) B1 for both "geometric" or "Geo" and correct parameter of $\frac{1}{6}$. Probability must be seen in (a)
- (b) M1 for a correct method, may be implied by a correct expression
- (b) A1cso for a correct solution leading to printed answer with no incorrect working seen
- (c) 1st M1 for correct use of formula for $E(X)$ or Var$(X)$
- (c) 1st A1 for both correct and see value of 6 for $E(X)$ and at least correct formula used for Var$(X)$ the value of 30 may be implied by 2nd A1
- (c) 2nd M1 for use of CLT to get a normal distribution with correct mean and variance $\neq$ their 30
- (c) 2nd A1 for a correct normal distribution stated or used. May be implied by correct answer.
- (c) 3rd A1 for awrt 0.681 but allow 0.68 or 0.680 if correct distribution is seen
- (d) B1 for both hypotheses correct in terms of $p$
- (d) M1 for sight or use of $P(X \geq 16)$ (may be implied)
- (d) or correct expression with logs (use of logs may be implied by 17.43...)
- (d) 1st A1 for 0.065 or better but accept 0.06 if a correct expression is seen, or correct CR
- (d) 2nd A1 for a correct conclusion mentioning probability or biased
- (d) Condone 'evidence suggests probability is equal to $\frac{1}{6}$'
- **NB: Must have correct probability or correct CR to gain final A1.**
- **NB: Send responses comparing 0.935 with 0.95 to review**
---\begin{enumerate}
\item Every morning Geethaka repeatedly rolls a fair, six-sided die until he rolls a 3 and then he stops. The random variable $X$ represents the number of times he rolls the die each morning.\\
(a) Suggest a suitable model for the random variable $X$\\
(b) Show that $\mathrm { P } ( X \leqslant 3 ) = \frac { 91 } { 216 }$
\end{enumerate}
After 64 mornings Geethaka will calculate the mean number of times he rolled the die.\\
(c) Estimate the probability that the mean number of rolls is between 5.6 and 7.2
Nira wants to check Geethaka's die to decide whether or not the probability of rolling a 3 with his die is less than $\frac { 1 } { 6 }$
Nira rolls the die repeatedly until she rolls a 3\\
She obtains $x = 16$\\
(d) By carrying out a suitable test, determine what Nira's conclusion should be. You should state your hypotheses clearly and use a $5 \%$ level of significance.
\hfill \mbox{\textit{Edexcel FS1 2024 Q4 [12]}}