# Question 7:

## Part (a):
$\text{Geo}(0.25)$ | **M1** | Stated or implied. NB: $12/36 = \frac{1}{3}$ may be MR

$P(>8) = 0.75^8$ | **M1** | Or $1 - 0.25(1 + 0.75 + \ldots + 0.75^7)$. $0.75^9 = 0.075$ or equivalent addition: M1M1A0

$= 0.1(00113)$ or $\frac{6561}{65536}$ | **A1 [3]** | Exact or awrt $0.100$, allow "$0.1$"

## Part (b)(i):
**DR** — *Either*: Geometric stated or implied | **M1** | Identify geometric for $Z$. SR if M0: B1 for one clear correct comparison, B1 a second one; all correct B1; conclusion "agree/good match" B1

$E(Z) = 4$ which is close to $4.03$ | **A1** | One correct calculation interpreted

$\text{Var}(Z) = 12$ which is close to $11.57$ | **A1** | Another correct calculation interpreted

So second row is a good match for expected results for student $Z$ | **A1 [4]** | Needs two pieces of correct evidence, and a conclusion. Accept "The statement is true"

**Or:** Geometric stated or implied | **M1** |

Frequencies are (generally) decreasing | **A1** |

$4.03 \approx 4$ | **M1** |

So second row is a good match for expected results for student $Z$ | **A1** | Needs adequate correct evidence – conclusion needed! Or equivalent with variance

## Part (b)(ii):
**DR** — $X \sim B(20, 0.25)$ | **M1** | Stated or implied. $(B(30, 0.25)$: M1A0A0). SR if M0: B1 for one clear correct comparison, B1 a second one; all correct and conclusion "very likely but not definite" B1 [e.g. sampling without replacement reduces probabilities of higher results]

$E(X) = 5$ which is close to $4.97$ | **A1** | Two correct comparisons

$\text{Var}(X) = 3.75$ which is close to $3.7$

Therefore quite strong indication that the third row is $X$, but not definite | **A1 [3]** | Consistent conclusion, indicate uncertainty, must deny "definitely". Needs both previous marks. Needn't actually say "very likely" as denying "definitely" is enough!

# Mark Scheme Extracts - Y532/01 June 2022

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## Q3 Conclusions

*(In general, allow "Accept $H_0$" as a synonym for "Do not reject $H_0$")*

| Label | Answer/Working | Mark | Guidance |
|-------|---------------|------|----------|
| α | Wrong but validly obtained TS leading consistently to "Reject $H_0$. There is significant evidence that runners who do better in one race tend to do better in the other" | M1A1 | standard FT |
| β | Do not reject $H_0$. Runners who do better in the first race do not tend to do better in the second | M1A0 | too definite |
| γ | Do not reject $H_0$. There is insufficient evidence of association | M1A0 | no context |
| δ | Insufficient evidence to reject $H_0$. Runners who do better in the first race do not tend to do better in the second | M1A1 | "Evidence" used, albeit in wrong sentence, but BOD |
| ε | Do not reject $H_0$. There is evidence that runners who do well in the first race do not do any better than others in the second race | M1A0 | Non-rejection doesn't give positive evidence that $H_0$ is correct |
| ζ | (from correct TS and CV) Reject $H_0$. (+ anything) | M0A0 | inconsistent |
| η | If $H_0$, $H_1$ the wrong way round, but calculations right: Do not reject $H_0$. (+ anything) | M1A0 | |

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## Q4 Conclusions

*(Similar general principles apply here as in Q2)*

| Label | Answer/Working | Mark | Guidance |
|-------|---------------|------|----------|
| α | (from miscalculated comparison) Reject $H_0$. There is evidence of an association between direction of journey and timing | M1A0 | |
| β | Significant evidence to reject $H_0$. There is association between direction of journey and delays | M1A1 | as in Q2 ex δ |
| γ | Reject $H_0$. There is association between direction of journey and delays | M1A0 | too definite; as in Q2 ex β |
| δ | Reject $H_0$. There is association between them | M1A0 | no context |
| ε | (from correct comparison) Do not reject $H_0$. (+ anything) | M0A0 | |
| ζ | If $H_0$, $H_1$ the wrong way round but calculations right: Reject $H_0$. (+ anything) | M1A0 | |
| η | Reject $H_0$. There is significant evidence of association between direction of journey and delays | M1A1 | best answer |

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## Q5(c)

| Label | Answer/Working | Mark | Guidance |
|-------|---------------|------|----------|
| α | Not valid as it is based on a small sample | B0 | |
| β | 14 is not equal to 10 so it is unlikely to be valid | B0 | don't expect them to be exactly equal |
| γ | 14 is not close to 10 so it may not be valid | B1 | dubious about this as "may not be" is always true, but BOD |
| δ | Not too far from part (iii), so valid enough but not completely reliable | B1 | allow this judgement |
| ε | It is not very valid as the values should be closer | B1 | BOD. Don't want to penalise the idea of something being "partly valid" |
| ζ | Probably valid as 14 is not far from 10 | B1 | |
| η | Less valid as 14 is not close to 10 | B1 | |

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## Q7(b)(i)

| Label | Answer/Working | Mark | Guidance |
|-------|---------------|------|----------|
| α | Second row is a good match. Average number of cards chosen before red is 4, as only $\frac{1}{4}$ of the pack is red and student continuously picks up cards; 10% chance of picking up 8 cards before a red; 25% chance of all reds being close together | M1A0, A1A0 | enough to imply $\text{Geo}(\frac{1}{4})$, even though wrong, but no more |
| β | I agree. The expected number for $\text{Geo}(\frac{1}{4})$ is 4 which is very similar to the mean (4.03). The variance is much higher which is accurate for geometric distributions as it is not averaged over a certain number of trials | M1A1, A0A0 | I suspect candidate has forgotten variance formula! |
| γ | $4 \approx 4.03$ and $12 \approx 11.57$ so the student is correct | M1A1, A1A1 | "it is a good match". Allow this |
| δ | In geometric, prob of getting red is same on each attempt and since replacement is occurring the number of reds is likely to be less than number of reds for student Y since there is no replacement | M0, SR B1 | "geometric" confused, so use SR |

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## Q7(b)(ii)

| Label | Answer/Working | Mark | Guidance |
|-------|---------------|------|----------|
| α | Some differences, but very similar, and more similar than other rows, so highly likely that row 3 is student X | M0 | |
| β | I agree. $E(X) = 5$ which is close to 4.97. $\text{Var}(X) = 3.75$ which is close to 5 | M1A0 | 5 and 3.75 imply M1 but wrong number (5) then used |
| γ | Very likely that student X's results are the third row since the number of reds in the pack remains constant; $\frac{12}{48}$ chance that the student would pick a red. Therefore 4.97 is close to 5 (which is $\frac{1}{4}$ of 20) | M1A0 | |
| δ | I agree. $E(X) = 5$ which is close to 4.97; $\text{Var}(X) = 3.75$ which is close to 3.7 | M1A1A0 | correct apart from no rejection of "definite" |
| ε | It cannot be said that row 3 is definitely Student X's results, since the observed values of mean and variance may be unlikely results from Y or Z's distributions. It's however likely that row 3 is student X since the observed mean and variance are very similar to the expected mean and variance (5 and 3.75) of X's distribution | M1A1A1 | |
| ζ | I agree that it could be student X's results as the expected number of reds is 5 which is similar to the observed mean of 4.97. However, I disagree that it is definitely student X's results as student 1 has an observed mean of 5.03 which is just as close | M1A1A1 | I think this is a good answer and I am happy to treat the two means as two different facts |
| η | There are more 8s recorded for student 3 than for student 1, as expected as the cards are replaced, but it cannot be said for certain that X is student 3. Additionally student 3 has a lower mean than student 1, but this would not be expected for student X | B1B0B0 | SC: no dist implied |