Tracking missiles with satellite imagery.The Space-BasedInfrared System (SBIRS) uses satellite imagery to detect andtrack missiles (Chance, Summer 2005). The probability thatan intruding object (e.g., a missile) will be detected on aflight track by SBIRS is .8. Consider a sample of 20 simulated tracks, each with an intruding object. Let x equal the numberof these tracks where SBIRS detects the object.

a. Demonstrate that x is (approximately) a binomial randomvariable.

b. Give the values of p and n for the binomial distribution.

c. Find $\mathbf{P}\left(x=15\right)$, the probability that SBIRS will detect the object on exactly 15 tracks.

d. Find $\mathit{P}\left(x\ge 15\right)$, the probability that SBIRS will detect the object on at least 15 tracks.

e. Find$\mathit{E}\left(x\right)$ and interpret the result.

The probability that an intruding object will be detected on a flight track by SBIRS is 8. Let X be the number of these tracks where SBIRS detects the object.

a.

Given that p=0.8 and n=20

Hence, x follows a binomial distribution with parameters n=20 and p=0.8

And pdf can be given by:

$P\left(X=x\right)=\left(\begin{array}{c}20\\ x\end{array}\right){\left(0.8\right)}^x{\left(1-0.8\right)}^{20-x}$

b.

Here the values of n and p are

p=0.8

n=20

c.

We have to calculate the probability that SBIRS will detect the object on exactly 15 tracks that is P(X=15)

$$\begin{gathered} P\left(X=5\right)=\left(\begin{array}{c}20\\ 15\end{array}\right){\left(0.8\right)}^{15}{\left(0.2\right)}^5 \\ =15504\times 0.0351\times 0.00032 \\ =0.1741 \end{gathered}$$

Therefore$P\left(X=15\right)=0.1741$

d.

To find the probability that SBIRS will detect the object on at least 15 tracks is $\mathrm{P}\left(\mathrm{X}\ge 15\right)$:

$$\begin{gathered} p\left(x\ge 15\right)=p\left(x=15\right)+p\left(x=16\right)+p\left(x=17\right)+p\left(x=18\right)+p\left(x=19\right)+ \\ p\left(x=20\right) \\ p\left(x\ge 15\right)=\left(\begin{array}{c}20\\ 15\end{array}\right){\left(0.8\right)}^{15}{\left(0.20\right)}^5+\left(\begin{array}{c}20\\ 16\end{array}\right){\left(0.8\right)}^{16}{\left(0.20\right)}^4+\left(\begin{array}{c}20\\ 17\end{array}\right){\left(0.8\right)}^{17}{\left(0.20\right)}^3 \\ \left(\begin{array}{c}20\\ 18\end{array}\right){\left(0.8\right)}^{18}{\left(0.20\right)}^2+\left(\begin{array}{c}20\\ 19\end{array}\right){\left(0.8\right)}^{19}{\left(0.20\right)}^1+\left(\begin{array}{c}20\\ 15\end{array}\right){\left(0.8\right)}^{20}{\left(0.20\right)}^0 \\ p\left(x\ge 15\right)=0.1746+0.2182+0.2054+0.1369+0.0576+0.0115 \\ =0.8042 \end{gathered}$$

Therefore

$p\left(x\ge 15\right)=0.8042$

e.

Mean of x that is $E\left(X\right)$can be given as:

$$\begin{gathered} E\left(X\right)=n\times p \\ =20\times 0.8 \\ =16 \end{gathered}$$

Therefore,

$E\left(X\right)=16$.