Respuesta :
Answer:
(a) The mean, variance and standard deviation of planes that would be successfully launched are 12, 3 and 1.732 respectively.
(b) The probability that exactly 12 planes scramble successfully is 0.225.
(c) The probability that at least 14 planes scramble successfully is 0.1971.
Step-by-step explanation:
Let X = number of planes whose engines will start at the first attempt.
The probability of X is:
P (A plane starting immediately) = 1 - P (A plane not starting at the 1st attempt)
= 1 - 0.25
p = 0.75
The number of planes at the air force intercept squadron is, n = 16.
The random variable X follows a Binomial distribution, i.e.[tex]X\sim Bin(16,0.75)[/tex]
(a)
The expected value of a Binomial distribution is:
[tex]E(X)=np[/tex]
Compute the expected number of planes that would be successfully launched as follows:
[tex]E(X)=np=16\times0.75=12[/tex]
The variance of a Binomial distribution is:
[tex]V(X)=np(1-p)[/tex]
Compute the variance of planes that would be successfully launched as follows:
[tex]V(X)=np(1-p)=16\times0.75\times (1-0.75)=3[/tex]
Compute the standard deviation of planes that would be successfully launched as follows:
[tex]SD(X)=\sqrt{V(X)}=\sqrt{3}=1.732[/tex]
Thus, the mean, variance and standard deviation of planes that would be successfully launched are 12, 3 and 1.732 respectively.
(b)
The probability function of a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]
Compute the probability that exactly 12 planes scramble successfully as follows:
[tex]P(X=12)={16\choose 12}(0.75)^{12}(1-0.75)^{16-12}=1820\times0.0317\times0.0039=0.225[/tex]
Thus, the probability that exactly 12 planes scramble successfully is 0.225.
(c)
Compute the probability that at least 14 planes scramble successfully as follows:
P (X ≥ 14) = P (X = 14) + P (X = 15) + P (X = 16)
[tex]={16\choose 14}(0.75)^{14}(1-0.75)^{16-14}+{16\choose 15}(0.75)^{15}(1-0.75)^{16-15}\\+{16\choose 16}(0.75)^{16}(1-0.75)^{16-16}\\=0.1336+0.0535+0.01\\=0.1971[/tex]
Thus, the probability that at least 14 planes scramble successfully is 0.1971.
