## (i)

**Answer:**
$$P(1100 < X < 1200) = P\left(\frac{1100-1100}{\sqrt{2000}} < Z < \frac{1200-1100}{\sqrt{2000}}\right)$$

$$= P(0 < Z < 2.236) = \Phi(2.236) - 0.5 = 0.9873 - 0.5 = 0.4873$$

**Marks:** M1 | M1 | A1

**Guidance:**
- M1 for for standardising
- M0 if 'continuity correction' applied
- M1 for for correct structure
- A1 CAO do not allow 0.4871, 0.48713, 0.48745 or 0.4875

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## (ii)

**Answer:** Use Normal approx with

$$\mu = np = 100 \times 0.4873 = 48.73$$

$$\sigma^2 = npq = 100 \times 0.4873 \times 0.5127 = 24.98$$

$$P(X \leq 40) = P\left(Z \leq \frac{40.5-48.73}{\sqrt{24.98}}\right)$$

$$= P(Z \leq -1.647) = 1 - \Phi(1.647) = 1 - 0.9502 = 0.0498$$

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

**Guidance:**
- B1 B1 for $\mu$ & $\sigma^2$ It their answer to part (i) provided that a normal approximation is appropriate from their part (i).
- B1 for continuity correction 40.5
- M1 correct structure using appropriate Normal approximation
- CAO 3s.f.
- 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

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## (iii)

**Answer:** From tables $\Phi^{-1}(0.01) = -2.326$

$$\frac{k-1100}{\sqrt{2000}} = -2.326$$

$$k = 1100 - 2.326 \times \sqrt{2000}$$
$$k = 996$$

**Marks:** B1 | M1 | A1

**Guidance:**
- B1 ±2.326 seen
- M1 correct equation as seen or equivalent
- CAO Allow 996.0

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# Question 3(iv)

**Answer:**
$$H_0: \mu = 7000;$$
$$H_1: \mu \neq 7000$$

Where $\mu$ denotes the population mean lifetime of low energy bulbs

$$\text{Test statistic} = \frac{6972 - 7000}{100/\sqrt{25}} = \frac{-28}{20} = -1.4$$

Lower 10% level 2 tailed critical value of z = − 1.645

$-1.4 > -1.645$ so not significant.
There is not sufficient evidence to reject $H_0$
There is insufficient evidence to conclude that the manufacturer is wrong.

**Marks:** B1 | B1 | B1 | M1 | A1 | M1 | A1 | [8]

**Guidance:**
- B1 For use of 7000 in hypotheses
- B1 For correct hypotheses given in terms of $\mu$ (not p or x, etc. unless letter used is clearly defined as population mean.) If hypotheses are reversed lose second B1 and final A1
- B1 for definition of $\mu$.
- M1 calculation of test statistic with a divisor of $100/\sqrt{25}$. Condone numerator reversed.
- A1 CAO for − 1.4. Allow +1.4 for A1 only if this is later compared with +1.645
- B1 For −1.645. Must be negative unless it is clear that absolute values are being used. NB: FT a 1-tail test (c.v. = −1.282)
- M1 for a sensible comparison leading to a conclusion.
- A1 For correct conclusion in context. **FT only their test statistic if c.v. correct and both M marks earned.**

**Critical Value Method**
7000 − 1.645 × 100 ÷ √25 gets M1B1 = 6967.1 gets A1
6972 > 6967.1 gets M1for sensible comparison A1 still available for correct conclusion in words & context

**Confidence Interval Method**
CI centred on 6972 ± or − 1.645 × 100 ÷ √25 gets M1 B1 = (6939.1, 7004.9) gets A1
contains 7000 gets M1 A1 still available for correct conclusion in words & context

**Probability Method**
Finding P(sample mean < 6972) = 0.0808 gets M1 A1 B1
0.0808 > 0.05 gets M1 for a sensible comparison if a conclusion is made.
0.0808 < 0.10 gets M1 A0 unless using one-tailed test A1 still available for correct (one-tailed) conclusion in words & context.
Condone P(sample mean > 6972) = 0.9192 for M1 but only allow A1 B1 if later compared with 0.95, at which point the final M1and A1 are still available

**One-tailed test**
Max B1 B0 B1 M1 A1 B1 (for cv = -1.282) M1 A0

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