Answer:
Step-by-step explanation:
given that 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
[tex]f(x)= c(1- x^2/16) ,0<x<2[/tex]
For f(x) to be a pdf total prob = 1
i.e.
[tex]c\int\limits^2_0 {1-x^2/16} \, dx =1\\c(2-\frac{1}{6} )=1\\c=\frac{6}{11} =0.5455[/tex]
b) [tex]P(X\leq 0.4) = \frac{6}{11} \int\limits^0.4_0 {1-\frac{16}{x^2} } \, dx \\=0.5455*0.3989\\=0.2176[/tex]
c) P(0.1<x<0.4) =[tex]0.5455\int_{0. 1} ^{0.4} (1 - x^2/16) dx = 0.298688*0.5455\\=0.1629[/tex]
