In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel. It was found that 102 adults in the sample travel by bus. Find an approximate $$\(90 \%\)$$ confidence interval for the proportion of all the adults in Fromleigh who travel by bus. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_63 Year：2020 Question No：2(a)
Answer:
\( \begin{array}{l}
\frac{\frac{102}{250} \times \frac{250-102}{250}}{250}(=0.000966144) \\ \frac{102}{250} \pm z \sqrt{1^{\prime} 0.00096614^{\prime}} \\z=1.645
\end{array} \)
Confident Interval is \(0.357\) to \(0.459(3 \mathrm{sf})\)
\frac{\frac{102}{250} \times \frac{250-102}{250}}{250}(=0.000966144) \\ \frac{102}{250} \pm z \sqrt{1^{\prime} 0.00096614^{\prime}} \\z=1.645
\end{array} \)
Confident Interval is \(0.357\) to \(0.459(3 \mathrm{sf})\)
Knowledge points:
6.4.8 determine, from a large sample, an approximate confidence interval for a population proportion.
Solution:
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