Answer to Question #320756 in Statistics and Probability for Gaby

Question

According to a survey, 32% of drivers tailgate other vehicles before passing them (thinking it will incite them to speed up). Consider 45 drivers selected at random. Use a normal approximation to find the probability that:

(a) at least 17 drivers tailgate?  (Use only 3 decimal places)

(b) more than 9 but less than 25 drivers tailgate?  (Use only 3 decimal places) ans_a is 0.251

Answer 1

$\hat{p}-the\,\,sample\,\,proportion\\a:\\P\left( \hat{p}\geqslant \frac{17}{45} \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}\geqslant \sqrt{n}\frac{\frac{17}{45}-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( Z\geqslant \sqrt{45}\frac{\frac{17}{45}-0.32}{\sqrt{0.32\left( 1-0.32 \right)}} \right) =P\left( Z\geqslant 0.830879 \right) =\\=1-\varPhi \left( 0.830879 \right) =1-0.7970=0.203\\a:\\P\left( \frac{9}{45}<\hat{p}<\frac{25}{45} \right) =P\left( \sqrt{n}\frac{\frac{9}{45}-p}{\sqrt{p\left( 1-p \right)}}<\sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}<\sqrt{n}\frac{\frac{25}{45}-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( \sqrt{45}\frac{\frac{9}{45}-0.32}{\sqrt{0.32\left( 1-0.32 \right)}}<Z<\sqrt{45}\frac{\frac{25}{45}-0.32}{\sqrt{0.32\left( 1-0.32 \right)}} \right) =\\=P\left( -1.72567<Z<3.38743 \right) =\varPhi \left( 3.38743 \right) -\varPhi \left( -1.72567 \right) =\\=0.9996-0.0422=0.9574$