Question

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development is the Space-Based Infrared System (SBIRS), which uses satellite imagery to detect and track missiles (Chance, Summer 2005). The probability that an intruding object (e.g. a missile) will be detected on a ight track by SBIRS is 0.8. consider a sample of 20 simulated tracks, each with an intruding object. Let x equal the number of these tracks where SBIRS detects the object.

Required:
a. Demonstrate that x is (approximately) a binomial random variable.
b. Give the values of p and n for the binomial distribution.
c. Find P(x = 15), the probability that SBIRS will detect the object on exactly 15 tracks.
d. Find P(x≥15) the probability that SBIRS will detect the object on at least 15 tracks.
e. Find E(x) and interpret the result.

Answer 1

Part A:

`P(X=x)=n_{C_(x)}*p^x*(1-p)^(n-x)`

x is a binomial Random Variable.

Part B:

Value of p=0.8

Value of n=20

Part C:

P(X=15)=0.17456

Part D:

P(X≥15)=0.80417

Part E:

E(x)=16

E(x) tells us that out of 20 tracks the SBIRS will detect from 16 tracks.

Explanation:

Part A:

Binomial Distribution is used because the number of tracks and the probability to find the intruding object is constant for all tracks

It means:

`P(X=x)=n_{C_(x)}*p^x*(1-p)^(n-x)`

x is a binomial Random Variable.

Part B:

Value of p=0.8

Value of n=20

Part C:

x=15

`P(X=15)=20_{C_(15)}*0.8^(15)*(1-0.8)^(20-15)\\`

By solving above Expression:

P(X=15)=0.17456

Part D:

P(X≥15)=P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)

`P(X=15)=20_{C_(15)}*0.8^(15)*(1-0.8)^(20-15)\\`

P(X=15)=0.17456

`P(X=16)=20_{C_(16)}*0.8^(16)*(1-0.8)^(20-16)\\P(X=16)=0.21819`

`P(X=17)=20_{C_(17)}*0.8^(17)*(1-0.8)^(20-17)\\P(X=17)=0.20536`

`P(X=18)=20_{C_(18)}*0.8^(18)*(1-0.8)^(20-18)\\P(X=18)=0.13691\\\\P(X=19)=20_{C_(19)}*0.8^(19)*(1-0.8)^(20-19)\\P(X=19)=0.05765\\\\P(X=20)=20_{C_(20)}*0.8^(20)*(1-0.8)^(20-20)\\P(X=20)=0.01153`

Now, Adding Probabilities:

`P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)\\=0.17456+0.21819+0.20536+0.13691+0.05765+0.01153\\`

P(X≥15)=0.80417

Part E:

Mean Distribution:

E(x)=n*p

E(x)=20*0.8

E(x)=16

E(x) tells us that out of 20 tracks the SBIRS will detect from 16 tracks.