Answer :
Given :
Population proportion $(p) = 0.52$
Sample size $(n) = 775$
Solution :
→ The probability that the sample proportion of girls will differ from the population proportion by more than 5% :
$$\text{P}(|\hat{p} – p| > 0.05) = \text{P}(\hat{p} – p < -0.05 \text{ or } \hat{p} – p > 0.05)$$ $$= \text{P}(\hat{p} < p – 0.05 \text{ or } \hat{p} > p + 0.05)$$ $$\text{P}(\hat{p} < 0.52 – 0.05 \text{ or } \hat{p} > 0.52 + 0.05)$$ $$= \text{P}(\hat{p} < 0.47 \text{ or } \hat{p} > 0.57)$$ $$= 1 – \text{P}(0.47 < \hat{p} < 0.57)$$ $$=1 – \text{P}\left(\frac{0.47 – p}{\sqrt{\frac{p(1-p)}{n}}} < z < \frac{0.57 – p}{\sqrt{\frac{p(1-p)}{n}}}\right)$$ $$=1 – \text{P}\left(\frac{0.47 – 0.52}{\sqrt{\frac{0.52(1 – 0.52)}{775}}} < z < \frac{0.57 – 0.52}{\sqrt{\frac{0.52(1 – 0.52)}{775}}}\right)$$ $$= 1 – \text{P}(-2.786 < z < 2.786)$$ $$=1 – (\text{P}(z < 2.786) – \text{P}(z < -2.786))$$ $$=1 – (0.9973 – 0.0027)$$ $$=1-0.9946$$ $$= 0.0054$$
