# Question 3:

## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $[F = \text{no. of faults in } A \text{ m}^2]$ $F \sim Po(A \times 0.4)$ | M1 | For selecting the correct model $Po(0.4A)$ |
| $[P(F=0) \Rightarrow]$ $0.0907 = e^{-0.4A}$ | M1 | For a correct equation; may be implied by correct answer with no incorrect working |
| $A = 6$ | A1 | For $A = 6$ (or awrt 6.0) |

## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $[T = \text{no. of tablecloths with no faults}]$ $T \sim B(20, 0.0907)$ | M1 | For selecting a correct model $B(20, 0.0907)$ |
| $P(T > 1) = 1 - P(T \leq 1)$ | M1 | For correctly interpreting "more than 1" to reach $1 - P(T \leq 1)$; may be implied by A1 |
| $= 0.55276\ldots =$ awrt $0.553$ | A1 | For awrt 0.553 |

## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $[X \sim B(100, 0.0907)]$ | M1 | For $X \sim B(100, 0.0907)$ used; may be implied by correct $E(X)$ or $Var(X)$ |
| $E(X) = 100 \times 0.0907 = 9.07$ | A1 | For 9.07 |

## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $Var(X) = 100 \times 0.0907 \times (1 - 0.0907) = 8.247351\ldots$ awrt $8.25$ | A1 | For awrt 8.25. SC - award M0A1A0 for: using $X \sim B(100, 0.4)$ leading to $E(X)=40$, $Var(X)=24$; or using $X \sim B(100,p)$, $0<p<1$ and $E(X)=100p$, $Var(X)=100p(1-p)$ |

## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $X \approx Po(9.07)$ | M1 | For selecting the correct Poisson model — ft their answer to (c)(i) |
| $P(X = 10) \approx 0.11947\ldots$ $\mathbf{0.1195}$ or awrt $0.119$ | A1 | For 0.1195 or awrt 0.119 |

## Part (e):
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0: \lambda = 0.4$ (or $\lambda = 12$) $\quad H_1: \lambda < 0.4$ (or $\lambda < 12$) | B1 | For both hypotheses correct in terms of $\lambda$ or $\mu$ |
| $[Y = \text{no. of faults from new machine}]$ $Y \sim Po(12)$ | M1 | For selecting a suitable model; sight or use of $Po(12)$; may be implied by 1st A1 |
| $P(Y \leq 6) = 0.04582\ldots$ | A1 | For a correct probability; must be 0.046 or better |
| [Significant] there is evidence to support the claim | A1 | For a correct conclusion in context using "claim" or "rate of faults" |

## Part (f):
| Working | Mark | Guidance |
|---------|------|----------|
| $p\text{-value} = 0.04582\ldots$ awrt $\mathbf{0.0458}$ | B1ft | B1ft for awrt 0.0458 o.e. e.g. 4.58%, or ft their answer to 1st A1 in (e) |

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