homework: section 4.3\nscore: 1/15 answered: 1/15\nquestion 2\namerican airlines flights from chicago to…

Explanation:

Step1: Identify the binomial distribution parameters

Let (n = 13) (number of trials, i.e., number of flights), (p=0.6) (probability of success, i.e., probability of a flight being on - time), and (q = 1 - p=0.4). The binomial probability formula is (P(X = k)=C(n,k)\times p^{k}\times q^{n - k}), where (C(n,k)=\frac{n!}{k!(n - k)!}).

Step2: Calculate the probability that at least 3 flights are on time

(P(X\geq3)=1 - P(X\lt3)=1-(P(X = 0)+P(X = 1)+P(X = 2))) [ \begin{align*} P(X = 0)&=C(13,0)\times(0.6)^{0}\times(0.4)^{13}=\frac{13!}{0!(13 - 0)!}\times1\times(0.4)^{13}= (0.4)^{13}\approx0.000000671\ P(X = 1)&=C(13,1)\times(0.6)^{1}\times(0.4)^{12}=\frac{13!}{1!(13 - 1)!}\times0.6\times(0.4)^{12}=13\times0.6\times(0.4)^{12}\approx0.0000154\ P(X = 2)&=C(13,2)\times(0.6)^{2}\times(0.4)^{11}=\frac{13!}{2!(13 - 2)!}\times0.36\times(0.4)^{11}=78\times0.36\times(0.4)^{11}\approx0.000137\ P(X\lt3)&\approx0.000000671 + 0.0000154+0.000137\approx0.000153\ P(X\geq3)&\approx1 - 0.000153 = 0.999847 \end{align*} ]

Step3: Calculate the probability that at most 12 flights are on time

(P(X\leq12)=1 - P(X = 13)) [ \begin{align*} P(X = 13)&=C(13,13)\times(0.6)^{13}\times(0.4)^{0}=\frac{13!}{13!(13 - 13)!}\times(0.6)^{13}\times1=(0.6)^{13}\approx0.001317 \end{align*} ] (P(X\leq12)=1 - 0.001317 = 0.998683)

Step4: Calculate the probability that exactly 9 flights are on time

[ \begin{align*} P(X = 9)&=C(13,9)\times(0.6)^{9}\times(0.4)^{4}\ &=\frac{13!}{9!(13 - 9)!}\times(0.6)^{9}\times(0.4)^{4}\ &=\frac{13\times12\times11\times10}{4\times3\times2\times1}\times(0.6)^{9}\times(0.4)^{4}\ & = 715\times0.010077696\times0.0256\ &\approx0.184 \end{align*} ]

Answer:

1. (0.999847)
2. (0.998683)
3. (0.184)