A2 June 2022 Q5
5. A random sample of 150 observations is taken from a geometric distribution with parameter 0.3
Estimate the probability that the mean of the sample is less than 3.45 (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{Geo}(0.3) \quad \mu = \dfrac{1}{0.3}\left[\text{or exact equivalent e.g. } \dfrac{10}{3}\right]\) | B1 | 1.1b |
| \(\sigma^2 = \dfrac{1 - 0.3}{0.3^2}\left[\text{or exact equivalent e.g. } \dfrac{70}{9}\right]\) | B1 | 1.1b |
| CLT \(\Rightarrow \overline{X} \approx \mathrm{N}\left(\dfrac{10}{3}, \ldots\right)\) oe | M1 | 2.1 |
| \(\Rightarrow \overline{X} \approx \mathrm{N}\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) and attempt (sight of) \(\mathrm{P}(\overline{X} \lt 3.45)\) | M1 | 3.4 |
| \(= 0.69579\ldots\) awrt 0.696 | A1 | 1.1b |
| (5) | ||
| Total 5 |
Notes
1st B1 correct mean
2nd B1 correct Var may be implied by sight of \(\frac{7}{135}\) in distribution of \(\overline{X}\)
1st M1 For use of CLT (must see \(\overline{X}\) and Normal with mean correct ft) or sight of \(\mathrm{N}\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) or \(\mathrm{N}\left(\text{“}\dfrac{10}{3}\text{”}, \dfrac{\text{“}70\text{”}}{\text{“}9\text{”} \times 150}\right)\) with any letter
Allow 3.33 or better for \(\frac{10}{3}\) and 7.78 or better for \(\frac{70}{9}\)
May be implied by 2nd M1
2nd M1 Using the normal distribution to find \(\mathrm{P}(\overline{X} \lt 3.45)\) ft their \(\text{“}\dfrac{10}{3}\text{”}\) and \(\dfrac{\text{“}\dfrac{70}{9}\text{”}}{150}\)
May be implied by correct answer.
A1 awrt 0.696
Correct answer with no incorrect working scores 5/5
Alternative (Use of \(Y = \sum X\))
| Scheme | Marks |
|---|---|
| \(\mu = \dfrac{150}{0.3}\ [= 500]\) | B1 |
| \(\sigma^2 = \dfrac{150 \times 0.7}{0.3^2}\left[\dfrac{3500}{3}\right] = 1166.\dot{6}\) | B1 |
| \(\Rightarrow Y \approx \mathrm{N}\left(500, \dfrac{3500}{3}\right)\) | M1 |
| \(\mathrm{P}(Y \lt 517.5)\) | M1 |
| \(= 0.69579\ldots\) | A1 |