# Question 8(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Inter-quartile range $=18.1-17.8=0.3$ | B1 | |
| Lower limit $17.8-(1.5\times 0.3)=17.35$; No outliers at lower end | M1 | FT their IQR for M marks only |
| | A1 | dep on 17.35; allow 'No values below 17.35 for first A1'; allow 'Lower limit $=17.35$ so no outliers (at lower end)' |
| Upper limit $18.1+(1.5\times 0.3)=18.55$; (Max is 18.6) so at least one outlier at upper end | M1 | |
| | A1 | dep on 18.55; allow 'At least one value above 18.55' or 'any value above 18.55 is an outlier' |
| **[5]** | | Do not allow 'There MAY be one outlier'; condone 'one outlier'; watch for use of median giving M0A0 |

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# Question 8(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A)=P(\text{All 3 have orange centres})=\frac{7}{20}\times\frac{6}{19}\times\frac{5}{18}=\frac{7}{228}$ | M1 | For $\frac{7}{20}\times$ |
| | M1 | For product of correct three fractions without extra terms |
| $=0.0307$ (0.030702) | A1 | CAO; allow full marks for fully simplified fractional answers |
| Alternative: $\frac{^7C_3}{^{20}C_3}=\frac{35}{1140}=\frac{7}{228}=0.0307$ | M1,M1,A1 | |
| $P(B)=P(\text{All 3 have same centres})=$ | | |
| $\left(\frac{7}{20}\times\frac{6}{19}\times\frac{5}{18}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\right)+\left(\frac{4}{20}\times\frac{3}{19}\times\frac{2}{18}\right)+\left(\frac{3}{20}\times\frac{2}{19}\times\frac{1}{18}\right)$ | M1 | For at least two correct triple products or fractions or decimals |
| $=0.0307+0.0175+0.0035+0.0009$ | M1 | For sum of all four correct |
| $=0.0526=\frac{1}{19}$ (0.052632) $\left(=\frac{7}{228}+\frac{1}{57}+\frac{1}{285}+\frac{1}{1140}\right)$ | A1 | CAO; allow 0.053 or anything which rounds up to 0.053 with working |
| **[6]** | | Please check all of the answer space for this part |

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# Question 8(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A\mid B)=\frac{0.0307\ldots}{0.0526\ldots}$ | M1 | For their $A$ divided by their $B$; allow 0.584 from $\frac{0.0307}{0.0526}$ |

## Question 8(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| P(All have orange centres) $= 0.0307^2 = 0.00094$ or $= \frac{49}{51984}$ $= (0.00094260)$ | M1 | For their $0.0307^2$; Allow $9.4 \times 10^{-4}$, condone 0.0009 or $9 \times 10^{-4}$ |
| | A1 **[2]** | FT |

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## Question 8(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Has to select} > 2) = 1 - P(\text{Has to select} \leq 2)$ $= 1 - \left(\frac{14}{20} + \left(\frac{6}{20} \times \frac{14}{19}\right)\right) = 1-(0.7+0.221) = 1-0.921$ | M1 | For $\left(\frac{6}{20} \times \frac{14}{19}\right)$; For any of the methods below allow SC2 for $1 - 0.079 = 0.921$ or $1 - 3/38 = 35/38$ o.e. as final answer |
| $= 0.079 \quad (=0.078947)$ | M1, A1 **[3]** | For $1 -$ sum of both CAO; This is $1 - P(C' + CC')$ |
| **OR** $P(\text{Has to select} > 2) = P(\text{First 2 both cherry}) = \left(\frac{6}{20} \times \frac{5}{19}\right)$ | M2 | For whole product; Without extra terms added. M1 if multiplied by $k/18$ only where $0 < k < 18$ (seen as a triple product only) |
| $= 0.079 \quad = \frac{3}{38}$ | A1 | CAO; This is $P(CC)$ |
| **OR** $1-(P(0 \text{ cherries}) + P(1 \text{ cherry}))=$ $1-\left(\frac{14}{20}\times\frac{13}{19}+\left(\frac{6}{20}\times\frac{14}{19}\right)+\left(\frac{14}{20}\times\frac{6}{19}\right)\right)$ $= 1-(0.4789+0.2211+0.2211) = 1-0.9209$ | M1, M1 | For any term; For $1 -$ sum of all three; This is $1 - P(C'C' + CC' + C'C)$ |
| $= 0.079$ | A1 | CAO |
| **OR** $\left(\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{14}{17}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{3}{17}\times\frac{14}{16}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{3}{17}\times\frac{2}{16}\times\frac{14}{15}\right)+\left(\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{3}{17}\times\frac{2}{16}\times\frac{1}{15}\times\frac{14}{14}\right)$ $= \frac{7}{114}+\frac{14}{969}+\frac{7}{2584}+\frac{7}{19380}+\frac{1}{38760}$ $= 0.079$ | M1, A1 | For any term (all correct); For sum of all five terms (all correct); CAO |

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*Note: This is $P(CCC' + CCCC' + CCCCC' + CCCCCC' + CCCCCCC')$*