# Question 7:

## Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{a+b}{2} = 9$ and $\frac{13-5}{b-a} = \frac{1}{5}$ or $a = -11$ and $b = 29$ | M1 | Setting up 2 correct equations; may be implied by correct answers for $a$ and $b$ |
| $P\!\left(X > \frac{"29"-"11"}{3}\right) \left[= P(X>6)\right]$ or $P\!\left(X > \frac{9\times2}{3}\right) = \frac{23}{40}$ | M1A1 | Realising need to find $P(X>6)$; A1 for $\frac{23}{40}$ oe or 0.575 |

## Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{12}(c-1)^2 = 0.48 \Rightarrow c = 3.4$ | M1 | Correct equation to find $c$ with $c = \ldots$ |
| $E(Y) = \frac{1 + "3.4"}{2}$ | M1 | Correct method for finding $E(Y)$ using "their $c$" |
| $E(Y^2) = 0.48 + \left(\frac{1+3.4}{2}\right)^2 = 5.32$ | M1A1 | Correct method for $E(Y^2)$ using "their $E(Y)^2$"; A1 for 5.32 (allow $\frac{133}{25}$) |

## Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $W < 6$ or $W > 14$ or $X < 12$ or $Y > 8$ (any letter, ignore distribution) | M1 | For identifying a correct region |
| $2\times P(W<6) = 2\times\frac{6-0}{20}$ or $2\times P(W>14) = 2\times\frac{20-14}{20}$ or $P(W<6) = \frac{3}{10}$ **and** $P(W>14) = \frac{3}{10}$ or $P(X>14) = \frac{3}{5}$ or $P(Y<6) = \frac{3}{5}$ | M1 | Correct method for a required probability associated with correct distribution |
| $P(8 < W < 12) = \frac{12-8}{20}$ or $P(X<12) = \frac{12-10}{10}$ or $P(Y>8) = \frac{10-8}{10}$ | M1 | Correct method for second required probability |
| $P(\text{shortest side} < 6) = "\frac{3}{10}" + "\frac{3}{10}" + "\frac{1}{5}"$ or $"\frac{1}{5}" + "\frac{3}{5}"= \frac{4}{5}$ | dM1 | Dependent on previous 3 M marks |
| $\frac{4}{5}$ | A1 | cao |

**Alternative 1:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $6 < W < 8$ or $12 < W < 14$ (any letter, ignore distribution) | M1 | Identifying correct region |
| $P(6<W<8) = \frac{8-6}{20}$ or $P(12<W<14) = \frac{14-12}{20}$ each $= \frac{1}{10}$ | M1 | Correct method for required probability |
| $2\times P(6<W<8) = 2\times\frac{1}{10}$ or $2\times P(12<W<14) = 2\times\frac{1}{10}$ | M1 | Correct method for second probability |
| $P(\text{shortest side} < 6) = 1 - "\frac{1}{10}" - "\frac{1}{10}"$ or $1 - "\frac{2}{10}" = \frac{4}{5}$ | dM1A1 | Dependent on previous 3 M marks; cao |

**Alternative 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $W > 10$ or $12 < W < 14$ or $6 < W < 8$ or $W < 10$ (any letter) | M1 | Identifying correct region |
| $P(12<W<14) = \frac{1}{10}$ **and** $P(W>10) = \frac{1}{2}$ or $P(6<W<8) = \frac{1}{10}$ **and** $P(W<10) = \frac{1}{2}$ | M1 | Correct method for both required probabilities |
| $P(12<W<14 \mid W>10) = \frac{"\frac{1}{10}"}{"\frac{1}{2}"}\left[= \frac{1}{5}\right]$ or $P(6<W<8 \mid W<10) = \frac{"\frac{1}{10}"}{"\frac{1}{2}"}\left[= \frac{1}{5}\right]$ | M1 | Correct use of conditional probability |
| $P(\text{shortest side} < 6) = 1 - "\frac{1}{5}" = \frac{4}{5}$ | dM1A1 | Dependent on previous 3 M marks; cao |

> **NB**: Any answer of $\frac{4}{5}$ scores 5/5 provided it has not come from incorrect working.

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