Answer:
a. p=1.000
b. p=0.2924
c. p=0.7358
d. No
Step-by-step explanation:
a. This problem satisfies all the criteria for a binomial experiment expressed as:
[tex]P(X=x){n\choose x}p^x(1-p)^{n-x}[/tex]
-Given that p=0.85, n=14, the probability that exactly all 14 were on time is calculated as:
[tex]P(X\geq 1)=1-P(X=0)\\\\=1-{12\choose 0}0.85^0(1-0.85)^{12}\\\\=1-1.297\times 10^{-10}\\\\=1.0000[/tex]
Hence, the probability that all 12 flights are on time is 1.0000
b. Given that n=12, and p=0.85
-The probability that exactly 10 flights are on time is calculated as;
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924[/tex]
Hence, the probability that exactly 10 flights are on time is 0.2924
c. Given that n=12, and p=0.85
-The probability that more of 10 or more flights are on time:
[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358[/tex]
Hence, the probability of 10+ flights being on time is 0.7358
d. We first find the mean of the distribution:
[tex]\mu=E(X)=0.85\times 14\\\\=11.9[/tex]
#We then find the probability of 11+=0.3012+0.1422=0.4434
-We compare the expectation to the probability of 11+ flights being on time.
No. Since the probability [tex]P(X\geq 10)[/tex]=0.4434 < that the expectation, 11.9, it is not unusual for 11+ flights to be on time.
*I have used a sample size of n=12 since there are two separate n values:
