Answer:
[tex]c=\frac{16}{39}[/tex]
Step-by-step explanation:
The probability density function is :
[tex]f(x)=c(1-\frac{1}{16}x^{2})[/tex]
With 0 < x < 3
To be a valid probability density function :
[tex]\int\limits^b_a {f(x)} \,dx=1[/tex]
Where a < x < b
And also
f(x) ≥ 0 for a < x < b
Applying this to the probability density function of the exercise :
[tex]\int\limits^3_0 {c(1-\frac{1}{16}x^{2})} \, dx=1[/tex]
[tex]c\int\limits^3_0 {(1-\frac{1}{16}x^{2})} \, dx=1[/tex]
[tex]c(3-\frac{1}{16}\frac{3^{3}}{3})=1[/tex]
[tex]c(\frac{39}{16})=1[/tex]
[tex]c=\frac{16}{39}[/tex]
We can verify by replacing ''c'' in the original probability density function and integrating :
[tex]\int\limits^3_0 {\frac{16}{39}(1-\frac{1}{16}x^{2})} \, dx=[/tex]
[tex]=\frac{16}{39}.(3)-\frac{1}{39}.(\frac{3^{3}}{3})=\frac{16}{13}-\frac{3}{13}=\frac{13}{13}=1[/tex]
Also, f(x) ≥ 0 for 0 < x < 3
The value of c that makes f(x) a valid probability density function is 16/39
The probability density function is given as:
f(x) = c(1 - 1/16x^2) for 0 < x < 3
For f(x) to be valid, the following must be true
[tex]\int\limits^b_a f(x) \ dx= 1[/tex]
So, we have:
[tex]\int\limits^3_0 c(1 - \frac{1}{16}x^2) \ dx= 1[/tex]
Factor out c
[tex]c\int\limits^3_0 (1 - \frac{1}{16}x^2) \ dx= 1[/tex]
Integrate the function
[tex]c[ x - \frac{1}{48}x^3]|\limits^3_0= 1[/tex]
Expand
[tex]c[3 - \frac{1}{48}*3^3]= 1[/tex]
[tex]c[3 - \frac{27}{48}]= 1[/tex]
[tex]\frac{117}{48}c= 1[/tex]
Solve for c
[tex]c= \frac{48}{117}[/tex]
Simplify
[tex]c= \frac{16}{39}[/tex]
Hence, the value of c is 16/39
Read more about probability at:
brainly.com/question/25638875
