$H_0: \lambda = 6.3 \text{ [or } \mu \text{]}$ | B2 | Both, B2. One error e.g. "$H_0 = 6.3$", or "$H_1: \lambda \ne 6.3$, B1, but $x, r$ etc: 0
$H_1: \lambda < 6.3 \text{ [or } \mu \text{]}$ | | 
$P(≤ 2) = e^{-6.3}(1 + 6.3 + 19.845)$ | M1 | Correct formula for at least 2 terms, can be implied by 0.0134
| A1 | Fully correct formula for $≤ 2$, can be implied by answer
$= 0.0498$ | A1 | Answer, a.r.t. 0.0498
$< 0.05$ | B1 | SR tables: B2 if a.r.t. 0.0506, else 0 [then can get B1M1A1]
| | Explicitly state $< 0.05$, not from $H_1$: $\lambda \ne 6.3$, or CR $≤ 2$ and explicitly state 2 in CR, needs essentially correct distribution
$\text{Therefore reject } H_0.$ | M1 | Not needed for final M1A1
$\text{Significant evidence that average number of pips has been reduced.}$ | A1√ | Correct method, comparison and first conclusion
| | Interpreted in context acknowledging uncertainty somewhere, $\sqrt{\text{on}}$ $p$ etc
| | SR: P(< 2) [0.0134] or Po(= 2) [0.0364]: B2 M1 A0 B0 M0 but allow "Po(= 2) = 0.0498" etc
| | SR: Normal: B2 M1 A0 B0
| [8] |

## Question 7(i)(a):

$\int_1^4 \frac{1}{2\sqrt{x}} xdx = \left[\frac{1}{3}x^3\right]_1 = 7/3 \text{ or } 2.333\ldots$ | M1 | Attempt to integrate $xf(x)$, correct limits
| B1 | Correct indefinite integral, a.e.f.
| A1 | Final answer 7/3 or equiv or a.r.t. 2.33
| [3] |

## Question 7(i)(b):

$\int_1^m \frac{1}{2\sqrt{x}} dx = 0.5$ | M1 | This or complementary integral, limits needed [not "$÷$∞"]
$\sqrt{m} - 1 = 0.5$ | A1 | Correct equation, any equivalent simplified form
$m = 2.25$ | A1 | Answer 9/4 or exact equivalent only
| [3] |

## Question 7(ii):

$1.5\int_1^\infty y^{y-2.5} y^2 dy = 1.5\left[\frac{y^{0.5}}{0.5}\right]_1$ | M1 | Attempt to integrate $y^2 f(y)$, limits 1 and $\infty$, allow any letter
| B1 | Correct indefinite integral [= 3y], ignore $\mu$ [= 3]
$\text{Upper limit gives infinite answer}$ | A1 | Give correct reason, c.w.o. apart from constant, allow "$= \infty$"
| [3] |

## Question 8(i):

Location of bacteria must be independent – the position of one does not affect that of another | M1 | "Found independently": M1. Allow "are independent", "singly". Context needed somewhere in answer.
| A1 | Correct explanation, not just of "singly", e.g. not "must not group together". No extra or wrong conditions, but allow both "singly" and "independently". Right explanation, not "independent": M1A0
| [2] |

**Examples:**

α | Number of bacteria occurring in a particular volume is independent of the number in another interval of the same volume. Number in one volume occurs randomly. | M1A0

β | Bacteria are distributed independently from one another. This means that they cannot be in groups. | M1A0

γ | Position of each bacterium must be independent of the position of other bacteria. Not well modelled by Poisson if they tended to form groups, they must not be influenced by the surrounding bacteria or certain conditions (e.g. heat). | M1A0

δ | Bacteria need to be independent. The results of one cannot influence the result of another. | M1A0

ε | Bacteria must occur independently, so the state of one bacterium has no effect on any other bacteria. | M1A0

ζ | Probability of bacteria must be independent, they cannot affect the probability of another bacterium occurring. | M1A1

η | Bacteria must occur independently, so if one occurs it can't cause more to appear. | M1A1

## Question 8(ii):

$1 - P(≤ 4) = [1 - 0.8912]$ | M1 | Allow M1 for 1 – .9580 [= 0.042] or wrong $\lambda$, 0.8912 etc: M0
$= 0.1088$ | A1 | 0.109 or 0.1088 or better
| [2] |

## Question 8(iii):

$\text{Po}(0.925)$ | M1 | Po(0.925) stated or implied | [37/40]
$e^{-0.925} \frac{0.925^2}{2!}$ | M1 | Correct Po formula for $r = 2$, any $\lambda$, can be implied by:
| A1 | Answer 0.17(0) or 0.1696 or better
| [3] |

## Question 8(iv):

$\text{Po}(250)$ | B1 | Po(250) stated or implied
$\lambda > 15 \text{ or } \lambda \text{ large } [\text{or } \mu \text{]}$ | B1 | Either of these
$N(250, 250)$ | M1* | N, mean their 100 × 2.5 …
| | …, variance (or SD) their mean
$\Phi\left(\frac{239.5-250}{\sqrt{250}}\right) = 1 - \Phi(0.664)$ | Dep*M1 | Standardise, allow wrong or no cc and/or no $\sqrt{\text{or}}$ or $\sigma^2$, needs A1
| A1√ | Continuity correction and $\sqrt{\text{correct}}$
$= 0.2533$ | A1 | Final answer a.r.t. 0.253, c.w.o.
| [7] |

## Question 9(i):

$H_0: \mu = 8; H_1: \mu \ne 8$ | B2 | Both, B2. One error, B1, allow $x/r/t$ here, but not $\bar{H}$
$\text{where } \mu \text{ is the population mean amount of sleep obtained by Year 11 pupils}$ | B1 | Need "population" or equivalent, but allow "average amount of sleep obtained by Year 11 pupils". Allow "$\mu$ is population mean".
| [3] |

## Question 9(ii):

$\Phi\left(\frac{0.28}{0.87/\sqrt{64}}\right) = \Phi(2.575)$ | M1 | Standardise, with $\sqrt{n}$ or $n$, allow cc, $\sqrt{\text{errors}}$
| A1 | $z = 2.575$ or 2.57 or 2.58, can be implied by, e.g., 0.005 or 0.995
$2 × (1 - \text{above})$ | M1 | Correct handling of tails
$= 0.01 \text{ or } 1\%$ | A1 | Answer 0.01 or 1% correct to 2 SF, c.w.o.
| [4] |

## Question 9(iii):

Rejecting $H_0$ when $\mu = 8$ | B1 | Or equivalent, some mention of context, not "probability of …"
| [1] |

## Question 9(iv):

$\Phi\left(\frac{8.28-7.9}{0.87/\sqrt{64}}\right) - \Phi\left(\frac{7.72-7.9}{0.87/\sqrt{64}}\right)$ | M1 | Find P(between 7.72 and 8.28 | $\mu = 7.9$), allow 1 – 2×P(1 tail) (need attempt to find correct region, not isw – i.e., not ans 0.049)
$= \Phi(3.494) - \Phi(-1.655) = [0.99976 - (1 - 0.951) \text{ or } 1 - \ldots]$ | M1 | Correct handling of tails, needn't attempt to evaluate, needs 64
$= 0.951$ | A1 | Final answer, a.r.t. 0.951.
| [3] | SR: One tail only used: M1M0A0. 0.951 from no working: B2