## (i)

**Answer:** Errors have a **uniform average rate of occurrence**

and occur **randomly and independently**

**Marks:** E1 | E1

**Guidance:**
- E1 must refer to 'errors' not 'events', 'data' or 'conditions'.
- Condone 'constant/fixed average/mean rate/per page' but not 'constant average', 'constant rate' or 'uniform rate', etc.
- Allow large n and small p if both defined
- E1 for randomly **and independently**
- If 'errors' not referred to then SC1 if otherwise correct. Condone 'the number of errors'

---

## (ii)(A)

**Answer:**
$$P(X = 1) = e^{-0.85} \frac{0.85^1}{1!} = 0.3633$$

**Marks:** M1 | A1

**Guidance:**
- M1 for attempt to find P(X = 1) either by Poisson p.d.f. or use of tables.
- A1 CAO 3s.f. for answers which round to 0.363
- www
- NOTE If P(X ≤ 1) used for final answer, award M0A0.
- Interpolation gives 0.79065 − 0.42795 = 0.3627
- www

---

## (ii)(B)

**Answer:**
$$P(X \geq 2) = 1 - P(X \leq 1) = 1 - e^{-0.85}\frac{0.85^0}{0!} - e^{-0.85}\frac{0.85^1}{1!}$$

$$= 1 - 0.4274 - 0.3633 = 0.2093$$

**Marks:** M1 | M1 | A1

**Guidance:**
- M1 for method for P(X = 0)
- M1 for correct structure used
- A1 CAO 3s.f for answers which round to 0.209. Allow 0.2094 if interpolation used. [Interpolation gives 0.42795 for P(X = 0) and 0.20935 for P(X ≤ 1)]

---

## (iii)

**Answer:**
New $\lambda = 10 \times 0.85 = 8.5$

P(Exactly 10 in 10 pages) = $0.7634 - 0.6530 = 0.1104$

Or $= e^{-8.5}\frac{8.5^{10}}{10!} = 0.1104$

**Marks:** B1 | M1 | A1

**Guidance:**
- B1 for 8.5
- M1 for P(X = 10) calculation using $\lambda = 8.5$
- CAO Allow 0.110 and 0.11
- www
- Award M1 only if $\lambda = 8.5$ used

---

# Question 2(iv)

**Answer:** So P(k − 1 or less in 10 pages) > 99%

From tables
$$P(X \leq 15) = 0.9862, \quad P(X \leq 16) = 0.9934$$

$$P(X \geq 16) = 1 - P(X \leq 15) = 0.0138 > 1\%$$
$$P(X \geq 17) = 1 - P(X \leq 16) = 0.0066 < 1\%$$

P(k or more in 10 pages) < 1% means
k − 1 = 16, k = 17

**Marks:** M1 | A1 | A1

**Guidance:**
- M1 for P(X ≤ k − 1) > 0.99 seen, or evidence of a search for values > 0.99 from cumulative Poisson tables seen.
- A1 for finding either one of 0.9862 and 0.9934 (or either one of 0.0138 and 0.0066)
- A1 for both (3s.f.)
- A1 CAO for k = 17
- SC1 for evidence of a search for values > **0.9** from cumulative Poisson tables seen. Or for k = 17with no supporting evidence seen.

---

# Question 2(v)

**Answer:** 
Mean number in 30 pages = $30 \times 0.85 = 25.5$

Using Normal approx. to the Poisson,
$$X \sim N(25.5, 25.5)$$

$$P(X \leq 30) = P\left(Z \leq \frac{30.5-25.5}{\sqrt{25.5}}\right)$$

$$= P(Z < 0.9901) = \Phi(0.9901) = 0.8389$$

**Marks:** B1 | B1 | B1 | M1 | A1

**Guidance:**
- B1 for Normal approx attempted.(SOI)
- B1 for correct parameters (SOI)
- B1 for correct continuity correction.
- M1 for correct structure with their parameters
- CAO (Do not FT wrong or omitted CC) Allow 0.839
- NOTE Using B(100, 0.4873) gives 0.0494. which gets 0/5
- SC If p small enough to justify a Poisson approximation, e.g. 0.05, then B1 for Poisson used, B1 ft for parameter, M1 for structure,M1 attempt at summation, A0

---