## (i)
**Answer/Working:** $X \sim B(20, 0.15)$

(A) Either $P(X = 1) = \binom{20}{1} \times 0.15^1 \times 0.85^{19} = 0.1368$ or $P(X = 1) = P(X \leq 1) - P(X \leq 0) = 0.1756 - 0.0388 = 0.1368$

**Marks:** M1 $0.15^1 \times 0.85^{19}$, M1 $\binom{20}{1} \times p^1 q^{19}$, A1 CAO | 3

**Guidance:** With $p + q = 1$. Allow answer 0.137 with or without working or 0.14 if correct working shown. See tables at the website http://www.mei.org.uk/files/pdf/formula_book_mf2.pdf. For misread of tables 0.3917 – 0.1216 = 0.2701 allow M1M1A0 also for 0.1304 – 0.0261 = 0.1043

**Answer/Working:** (B) $P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.1756 = 0.8244$

**Marks:** M1 for 1 – their 0.1756, A1 CAO | 2

**Guidance:** Provided 0.1756 comes from $P(X=0) +P(X=1)$. Allow answer 0.824 with or without working or 0.82 if correct working shown. Point probability method: $P(1) = 0.1368$, $P(0) = 0.0388$. So $1 - P(X \leq 1) = 1 - 0.1756$ gets M1 then mark as per scheme. M0A0 for $1 - P(X \leq 1) = 1 - 0.4049 = 0.5951$. For misread of tables $1 - 0.3917 = 0.6083$ allow M1A1 also for $1 - 0.1304 = 0.8696$ provided consistent with part (A) OR M1A0 if formula used in part (A)

---

# Question 7(ii)

**Answer/Working:** Let $X \sim B(n, p)$. Let $p =$ probability of a 'no-show' (for population). $H_0: p = 0.15$, $H_1: p < 0.15$

$H_1$ has this form because the hospital management hopes to reduce the proportion of no-shows.

**Marks:** B1 for definition of $p$, B1 for $H_0$, B1 for $H_1$, E1 | 4

**Guidance:** Allow $p = P(\text{no-show})$ for B1. Definition of $p$ must include word probability (or chance or proportion or percentage or likelihood but NOT possibility). Preferably as a separate comment. However can be at end of $H_0$ as long as it is a clear definition ' $p =$ the probability of no-show, NOT just a sentence 'probability is 0.15'. $H_0 : p(\text{no-show}) = 0.15$, $H_1 : p(\text{no-show}) < 0.15$ gets B0B1B1. Allow $p=15\%$, allow $0$ or $\pi$ and $p$ but not $x$. However allow any single symbol if defined. Allow $H_0 = p=0.15$, Do not allow $H_0: P(X=x) = 0.15$, $H_1 : P(X=x) < 0.15$ (unless $x$ correctly defined as a probability). Do not allow $H_0$ and $H_1$ reversed for B marks but can still get E1. Allow NH and AH in place of $H_0$ and $H_1$. For hypotheses given in words allow Maximum B0B1B1E1 Hypotheses in words must include probability (or chance or proportion or percentage) and the figure 0.15 oe.

---

# Question 7(iii)

**Answer/Working:** $P(X \leq 1) = 0.1756 > 5\%$

So not enough evidence to reject $H_0$. Not significant. Conclude that there is not enough evidence to indicate that the proportion of no-shows has decreased.

**Marks:** M1 for probability seen, but not in calculation for point probability; M1 dep for comparison; A1 | 4

**Guidance:** Zero for use of point prob - $P(X = 1) = 0.1368$. Do NOT FT wrong $H_1$. Allow accept $H_0$, or reject $H_1$. Full marks only available if 'not enough evidence to...' oe mentioned somewhere. Do not allow 'evidence to reject $H_1$' for final mark but can still get 3/4. Upper end comparison: $1 - 0.1756 = 0.8244 < 95\%$ gets

---

# Question 7 continued

**Answer/Working:** Note: use of critical region method scores M1 for region [0], M1 for 1 does not lie in critical region, then A1 E1 as per scheme

**Marks:** | 

**Guidance:** E1 dep for conclusion in context. M2 then A1E1 as per scheme. Line diagram method: M1 for squiggly line between 0 and 1 with arrow pointing to left, M1 0.0388 seen on diagram from squiggly line or from 0, A1E1 for correct conclusion. Bar chart method: M1 for line clearly on boundary between 0 and 1 and arrow pointing to left, M1 0.0388 seen on diagram from boundary line or from 0, A1E1 for correct conclusion

## (iv)
**Answer/Working:** $6 < 8$. So there is sufficient evidence to reject $H_0$. Conclude that there is enough evidence to indicate that the proportion of no-shows appears to have decreased.

**Marks:** M1 for comparison seen, A1, E1 for conclusion in context | 3

**Guidance:** Allow '6 lies in the CR'. Do NOT insist on 'not enough evidence' here. Do not FT wrong $H_1$; p>0.15 but may get M1. In part (iv) ignore any interchanged $H_0$ and $H_1$ seen in part (ii)

## (v)
**Answer/Working:** For $n \leq 18$, $P(X \leq 0) > 0.05$ so the critical region is empty.

**Marks:** E1 for $P(X \leq 0) > 0.05$, E1 indep for critical region is empty | 2

**Guidance:** E1 also for sight of 0.0536. Condone $P(X = 0) > 0.05$ or all probabilities or values, (but not outcomes) in table (for $n \leq 18$) > 0.05. Or 'There is no critical region'. For second E1 accept '$H_0$ would always be accepted'. Do NOT FT wrong $H_1$. Use professional judgement - allow other convincing answers

---