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

Answer:

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.

Step-by-step 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.