Question

A parabolic satellite dish reflects signals to the dish’s focal point. An antenna designer analyzed signals transmitted to a satellite dish and obtained the probability density function f(x) = c(1 - 1/16x^2) for 0 < x < 3, where X is the distance (in meters) from the centroid of the dish surface to a reflection point at which a signal arrives. Determine the value of c that makes f(x) a valid probability density function

Answer 1

Answer:

$c=\frac{16}{39}$

Step-by-step explanation:

The probability density function is :

$f(x)=c(1-\frac{1}{16}x^{2})$

With 0 < x < 3

To be a valid probability density function :

$\int\limits^b_a {f(x)} \,dx=1$

Where a < x < b

And also

f(x) ≥ 0 for a < x < b

Applying this to the probability density function of the exercise :

$\int\limits^3_0 {c(1-\frac{1}{16}x^{2})} \, dx=1$

$c\int\limits^3_0 {(1-\frac{1}{16}x^{2})} \, dx=1$

$c(3-\frac{1}{16}\frac{3^{3}}{3})=1$

$c(\frac{39}{16})=1$

$c=\frac{16}{39}$

We can verify by replacing ''c'' in the original probability density function and integrating :

$\int\limits^3_0 {\frac{16}{39}(1-\frac{1}{16}x^{2})} \, dx=$

$=\frac{16}{39}.(3)-\frac{1}{39}.(\frac{3^{3}}{3})=\frac{16}{13}-\frac{3}{13}=\frac{13}{13}=1$

Also, f(x) ≥ 0 for 0 < x < 3