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 <em>X</em> = number of planes whose engines will start at the first attempt.
The probability of <em>X</em> is:
P (A plane starting immediately) = 1 - P (A plane not starting at the 1st attempt)
= 1 - 0.25
<em>p </em>= 0.75
The number of planes at the air force intercept squadron is, <em>n</em> = 16.
The random variable <em>X</em> follows a Binomial distribution, i.e.
(a)
The expected value of a Binomial distribution is:
Compute the expected number of planes that would be successfully launched as follows:
The variance of a Binomial distribution is:
Compute the variance of planes that would be successfully launched as follows:
Compute the standard deviation of planes that would be successfully launched as follows:
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:
Compute the probability that exactly 12 planes scramble successfully as follows:
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)
Thus, the probability that at least 14 planes scramble successfully is 0.1971.