A2 June 2019 Q3
3. A biased spinner can land on the numbers 1, 2, 3, 4 or 5 with the following probabilities.
| Number on spinner | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Probability | 0.3 | 0.1 | 0.2 | 0.1 | 0.3 |
The spinner will be spun 80 times and the mean of the numbers it lands on will be calculated.
Find an estimate of the probability that this mean will be greater than 3.25 (6)
| Scheme | Marks | AO |
|---|---|---|
| {Let \(X\) = the number when the spinner is spun} \(\mu = \underline{3}\) | B1 | 1.1b |
| \([\mathrm{E}(X^2) =]\ 0.3 + 4 \times 0.1 + 9 \times 0.2 + 16 \times 0.1 + 25 \times 0.3 \quad \left[= 11.6 \text{ or } \tfrac{58}{5}\right]\) | M1 | 1.1b |
| \(\sigma^2\ [= 11.6 - 3^2 =]\ \underline{2.6}\) | A1 | 1.1b |
| \(\overline{X} \mathrel{\approx\!\sim} \mathrm{N}\left(\text{“}3\text{”}, \sqrt{\dfrac{\text{“}2.6\text{”}}{80}}^{\,2}\right)\) | M1 A1ft | 2.1 1.1b |
| \(\mathrm{P}(\overline{X} \gt 3.25) = [\mathrm{P}(Z \gt 1.3867\ldots) =]\ 0.0827589\ldots\) (calc) awrt 0.0828 | A1 | 3.4 |
| (6 marks) |
Notes
B1 for stating or using mean = 3
1st M1 for using the given model to attempt \(\mathrm{E}(X^2)\) with at least 3 correct products seen
1st A1 for \(\mathrm{Var}(X) = 2.6\) or \(\sigma = \sqrt{2.6} = 1.6124\ldots\) (awrt 1.61)
ALT Use of pgf (B1 when mean = 3 seen) (M1 when correct \(\mathrm{G}^{\prime\prime}(t)\) seen with attempt at \(\mathrm{G}^{\prime\prime}(1)\))
\[\begin{aligned}\mathrm{G}(t) &= 0.3t + 0.1t^2 + 0.2t^3 + 0.1t^4 + 0.3t^5\\ \mathrm{G}^{\prime}(t) &= 0.3 + 0.2t + 0.6t^2 + 0.4t^3 + 1.5t^4\\ \mathrm{G}^{\prime\prime}(t) &= 0.2 + 1.2t + 1.2t^2 + 6t^3 \quad \text{leading to } \mathrm{G}^{\prime\prime}(1) = 8.6\end{aligned}\]2nd M1 for use of CLT – must use \(\overline{X}\) and normal or sight of \(\mathrm{N}\left(\text{“}3\text{”}, \sqrt{\dfrac{\text{“}2.6\text{”}}{80}}^{\,2}\right)\) with any letter
2nd A1ft for a correct mean and variance, ft their 3 and their 2.6
This M1A1ft may be implied by sight of correct st. dev. used in a standardisation leading to \(\mathrm{P}(Z \gt 1.39)\) Must see correct use of \(Z\)
NB \(\dfrac{2.6}{80} = 0.0325\) and \(\sqrt{\dfrac{2.6}{80}} = 0.18027\ldots\) so allow e.g. \(\mathrm{N}(3, \text{awrt } (0.180)^2)\)
3rd A1 for using the normal model to find probability awrt 0.0828
ALT Use of \(\sum X\) (If see clear attempt at \(\mathrm{P}(\Sigma X \gt 260)\) condone \(\mathrm{P}(\Sigma X \gt 260.5)\)) then:
2nd M1 for \(\Sigma X \sim \mathrm{N}(\ldots)\) or any letter \(\sim \mathrm{N}\left(\text{“}240\text{”}, \sqrt{\text{“}2.6\text{”} \times 80}^{\,2}\right)\)
2nd A1ft for mean \(= \text{“}3\text{”} \times 80 = 240\) and variance \(= \text{“}2.6\text{”} \times 80 = 208\)
May see \(\mathrm{P}(\Sigma X \gt 260.5) = 0.077597\ldots\) but it will only score 2nd M1 2nd A1ft and 3rd A0