# Question 1:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(30, 0.05)$ | B1 | Must include B(inomial), $n = 25$ and $p = 0.05$. Do not allow $p = 0.95$ in part (a) |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| The **probability** of an oyster surviving/not surviving is **constant** | B1 | For either correct assumption in context. Ignore extraneous non-contradicting comments |
| The survival of each oyster is **independent** of the others | B1 | |

## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{30}C_{24}(0.05)^6(0.95)^{24}$ oe | M1 | Allow $^{30}C_6$ oe or $P(X \leq 6) - P(X \leq 5)$ with one correct probability |
| $= 0.002708\ldots$ awrt $0.0027$ | A1 | awrt $0.0027$ (correct answer scores 2 out of 2) |

## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(Y \geq 3) = 1 - P(Y \leq 2)$ from $Y \sim B(30, 0.05)$ or $P(X \leq 27)$ from $X \sim B(30, 0.95)$ | M1 | Writing/using $1 - P(Y \leq 2)$ with $B(30, 0.05)$ or writing/using $P(X \leq 27)$ with $B(30, 0.95)$ |
| $= 1 - 0.8122$ | | |
| $= 0.1878$ awrt $0.188$ | A1 | awrt $0.188$ (correct answer scores 2 out of 2) |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $A \sim Po(10)$ | B1 | Writing or using $Po(10)$ — sight of $0.1301$ or $0.8699$ can imply this mark |
| $P(A \geq n) > 0.8$ | | |
| $P(A \leq n-1) < 0.2$ or $P(A \leq 6) = 0.1301\ldots$ awrt $0.13$ or $P(A \geq 7) = 0.8699\ldots$ awrt $0.87$ | M1 | Allow $P(A < n) < 0.2$ or $P(A < 7) =$ awrt $0.13$ or $P(A > 6) =$ awrt $0.87$ |
| $n = 7$ | A1cao | $n = 7$ which must come from use of $Po(10)$ or $N(10, 9.5)$ |
| **Note:** Use of normal approx. with $\mu = 10$ and $\sigma^2 = 9.5$ leading to $n < 7.4\ldots$ can score M1; Exact binomial gives $P(A \leq 6) = 0.14$ / $P(A \geq 7) = 0.86$ scores B0M0A0 | | |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.05$, $H_1: p > 0.05$ | B1 | Both hypotheses correct (allow use of $p$ or $\pi$). Allow $H_0: p = 0.95$, $H_1: p < 0.95$ |
| Using $C \sim B(25, 0.05)$ and $P(C \geq 4)$ \| Using $D \sim B(25, 0.95)$ and $P(D \leq 21)$ | M1 | Using $B(25, 0.05)$ and writing/using $P(C \geq 4)$ or if CR given $P(C \geq 3)$ using $B(25, 0.95)$ and writing/using $P(D \leq 21)$ or if CR given $P(D \leq 20)$ |
| $P(C \geq 4) = 0.0341$ / CR $C \geq 4$ \| $P(D \leq 21) = 0.0341$ / CR $D \leq 21$ | A1 | Correct probability to 3sf (must not go on and give incorrect CR) or correct CR (ignore upper tail) |
| Evidence to reject $H_0$, in the CR, significant | dM1 | (dep on 1st M1) A correct non-contextual statement (do not allow contradicting non-contextual comments) which is consistent with their prob and $0.05$ (If not stated, may be implied by A1) |
| There is evidence that the proportion of **oysters** not surviving has **increased**/Jim's **belief** is supported | A1cso | All previous marks must be awarded. Correct contextual conclusion with bold words (oe) |
| **SC: 2-tail** — Use of two-tailed test can score max: B1M1A1M1A0, but must **not reject** $H_0$ for 2nd M1 | | |

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