October 2020 Paper 2 Q12
12 In this question you must show detailed reasoning.
A 5-sided spinner can give scores of 1, 2, 3, 4 or 5. After observing a large number of spins, Elaine models the probability distribution of \(X\), the score on the spinner, as shown in Fig. 12.
| \(x\) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | 0.2 | 0.3 | \(p\) | \(p\) | \(q\) |
Fig. 12
When the spinner is spun twice, the probability of obtaining a total score of 9 is 0.06.
Elaine’s teacher believes that the probability that the spinner shows a 1 is greater than 0.2. The spinner is spun 100 times and gives a score of 1 on 28 occasions.
| Scheme | Marks | AO |
|---|---|---|
| \(2p + q + 0.2 + 0.3 = 1\) soi oe | B1 | 2.1 |
| \(2 \times p \times q = 0.06\) soi | M1 | 3.1a |
| eliminate \(p\) or \(q\) with a correct substitution from one of their equations | M1 | 1.1 |
| \(q^2 - 0.5q + 0.06 = 0\) or \(2p^2 - 0.5p + 0.03 = 0\) oe | A1 | 1.1 |
| \(q = 0.2\) or 0.3 and \(p = 0.15\) or 0.1 | A1 | 1.1 |
| (\(q < 2p\) so) \(q = 0.2\) and \(p = 0.15\) | A1 | 3.2a |
| [6] |
Notes
M1: allow M1 if 2 omitted
A1: eg \(2 \times \frac{0.03}{q} + q = 0.5\) or \(2p + \frac{0.03}{p} = 0.5\)
NB if 2 omitted, A0 for \(2p^2 - 0.5p + 0.06 = 0\) or \(2q^2 - q + 0.24 = 0\) which have no real roots (corrected from the printed mark scheme: printed as \(2p^2 - 0.5p + 0.6 = 0\) or \(2q^2 + q + 0.24 = 0\))
A1: may be implied by eg \(q = 0.2\) or 0.3 and \(2p = 0.3\) or 0.2
| Scheme | Marks | AO |
|---|---|---|
| \(10 \times q \times (1-q)^9\) soi | M1 | 1.1 |
| 0.27 or 0.268 or awrt 0.2684 isw | A1 | 1.1 |
| [2] |
Notes
A1: FT their \(q\) where \(0 < q < 1\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H_0}: p = 0.2\) \(\mathrm{H_1}: p > 0.2\) | B1 | 1.1 |
| \(p\) is the probability that the spinner shows a 1 (on any given spin) oe | B1 | 2.5 |
| use of \(X \sim \mathrm{B}(100, 0.2)\) where \(x\) is the number of 1s obtained in 100 spins to obtain \(\mathrm{P}(X \geqslant k)\) or \(\mathrm{P}(X \leqslant k)\) | M1 | 3.3 |
| \(\mathrm{P}(X \leqslant 27) =\) awrt 0.97 or \(\mathrm{P}(X \geqslant 28) =\) awrt 0.034 | A1 | 1.1 |
| \(0.034 < 0.05\) or \(0.97 > 0.95\) | M1 | 3.4 |
| significant or reject \(\mathrm{H_0}\) or accept \(\mathrm{H_1}\); may be embedded in conclusion in context | A1 | 1.1 |
| there is sufficient evidence to suggest (at 5% level) that the probability of a score of 1 is greater than 0.2 | A1 | 2.2b |
| [7] |
Notes
B1: both hypotheses; allow equivalent in words or eg \(\mathrm{P}(1) = 0.2\)
allow any parameter as long as clearly defined as probability
M1: \(k = 27, 28\) or 29
M0 for \(\mathrm{P}(X = k)\)
NB \(\mathrm{P}(X = 28) = 0.014\ldots\) \(\mathrm{P}(X = 27) = 0.020168\ldots\)
A1: or critical region is \(X \geqslant 28\)
M1: or 28 is in critical region
FT their probability, dependent on award of first M1
A1: must have the correct probability or correct critical region for the last two A marks
A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; A0 if answer spoiled