# Question 3:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cumulative frequency table: Upper Bounds 9.1, 9.3, 9.5, 9.7, 9.9, 10.1; CF: 0, 5, 12, 27, 43, 50 | B1 | May be implied from graph. Condone omission of 0 |
| Linear scales on both axes | G1 | Linear horizontal scale; linear vertical scale 0 to 50. No inequality scales |
| Correct labels on axes | G1 | Heating quality or $x$ and Cumulative frequency or CF. Not just frequency or fd |
| Points plotted at correct UCB positions | G1 | Ignore $(9.1, 0)$ at this stage. No midpoint or LCB plots. Plotted within $\frac{1}{2}$ small square |
| Points joined (line or smooth curve) including $(9.1, 0)$ | G1 | All G's dep on attempt at cumulative frequency but not cumulative $fx$'s or other spurious values |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median $= 9.67$ | B1 FT | Based on $25^{th}$ to $26^{th}$ value. Allow answers between 9.66 and 9.68 without checking curve. B0 for interpolation |

## Part (ii) continued – IQR
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Q_1 = 9.51$, $Q_3 = 9.83$, Interquartile range $= 9.83 - 9.51 = 0.32$ | B1 FT for $Q_3$ or $Q_1$; B1 FT for IQR providing both $Q_1$ and $Q_3$ correct | Based on $12^{th}$ to $13^{th}$ and $37^{th}$ to $38^{th}$ values. Allow answers between 9.50–9.52 and 9.82–9.84 without checking. B0 for interpolation. Allow correct IQR from graph if quartiles not stated |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Lower limit $9.51 - 1.5 \times 0.32 = 9.03$; Upper limit $9.83 + 1.5 \times 0.32 = 10.31$; Thus there are no outliers in the sample | B1 FT their $Q_1$, IQR; B1 FT their $Q_3$, IQR; E1 (NB E mark dep on both B marks) | Any use of median $\pm 1.5\times$ IQR scores B0B0E0. If FT leads to limits above 9.1 or below 10.1 then E0. No marks for $\pm 2$ or $3\times$ IQR |

## Part (iv)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{All 3 more than } 9.5) = \frac{38}{50}\times\frac{37}{49}\times\frac{36}{48} = 0.4304$ $(= \frac{50616}{117600} = \frac{2109}{4900})$ | M1 for $\frac{38}{50}\times$ (triple product); M1 for product of remaining fractions; A1 CAO | $\left(\frac{38}{50}\right)^3$ gives 0.4389 scores M1M0A0. $^3C_0 \times 0.24^0 \times 0.76^3$ still scores M1. Allow unsimplified fraction as final answer |

## Part (iv)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{at least 2 more than } 9.5) = 3\times\frac{38}{50}\times\frac{37}{49}\times\frac{12}{48} + 0.4304 = 3\times0.1435 + 0.4304 = 0.861$ $(= \frac{2109}{2450})$ | M1 for product of 3 correct fractions; M1 for $3\times$ a sensible triple or sum of 3 sensible triples; M1 indep for $+0.4304$ FT; A1 CAO | Or: $1-\left[\left(\frac{12}{50}\times\frac{11}{49}\times\frac{10}{48}\right)+\left(3\times\frac{12}{50}\times\frac{11}{49}\times\frac{38}{48}\right)\right] = 1-[0.01122+0.12796]=0.861$ |

---