Answer :
Given :
Population proportion $(p) = 0.08$
Sample size $(n) = 596$
Solution :
→ The probability that the proportion of no-show would differ from the population proportion by more than 3% :
$$\text{P}(|\hat{p} – p| > 0.03) = \text{P}(\hat{p} – p < -0.03 \ \text{or} \ \hat{p} – p > 0.03)$$ $$= \text{P}(\hat{p} < p – 0.03 \ \text{or} \ \hat{p} > p + 0.03)$$ $$= \text{P}(\hat{p} < 0.08 – 0.03 \ \text{or} \ \hat{p} > 0.08 + 0.03)$$ $$= \text{P}(\hat{p} < 0.05 \ \text{or} \ \hat{p} > 0.11)$$ $$= 1 – \text{P}(0.05 < \hat{p} < 0.11)$$ $$= 1 – \text{P}\left(\frac{0.05 – p}{\sqrt{\frac{p(1-p)}{n}}} < z < \frac{0.11 – p}{\sqrt{\frac{p(1-p)}{n}}}\right)$$ $$= 1 – \text{P}\left(\frac{0.05 – 0.08}{\sqrt{\frac{0.08(1-0.08)}{596}}} < z < \frac{0.11 – 0.08}{\sqrt{\frac{0.08(1-0.08)}{596}}}\right)$$ $$= 1 – \text{P}(-2.7 < z < 2.7)$$ $$= 1 – \left(\text{P}(z < 2.7) – \text{P}(z < -2.7)\right)$$ $$= 1 – (0.9965 – 0.0035)$$ $$= 1 – 0.9930$$ $$= 0.0070$$
