Answer:
A
Step-by-step explanation:
Answer:
i dont know im sry
Step-by-step explanation:
The original number would be 7.5
Answer:The square root of 436149 is 660.41577812769
Use the definitions of expectation and variance.
![E(X) = \displaystyle \int_{-\infty}^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^{-x/4} \, dx](https://tex.z-dn.net/?f=E%28X%29%20%3D%20%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%20f_X%28x%29%20%5C%2C%20dx%20%3D%20%5Cfrac14%20%5Cint_0%5E%5Cinfty%20x%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx)
Integrate by parts,
![\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_a%5Eb%20u%20%5C%2C%20dv%20%3D%20uv%20%5Cbigg%7C_a%5Eb%20-%20%5Cint_a%5Eb%20v%20%5C%2C%20du)
with
![u = x \implies du = dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}](https://tex.z-dn.net/?f=u%20%3D%20x%20%5Cimplies%20du%20%3D%20dx%20%5C%5C%5C%5C%20dv%20%3D%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx%20%5Cimplies%20v%20%3D%20-4%20e%5E%7B-x%2F4%7D)
Then
![E(X) = \displaystyle \frac14 \left(\left(-4x e^{-x/4}\right)\bigg|_0^\infty + 4 \int_0^\infty e^{-x/4} \, dx\right)](https://tex.z-dn.net/?f=E%28X%29%20%3D%20%5Cdisplaystyle%20%5Cfrac14%20%5Cleft%28%5Cleft%28-4x%20e%5E%7B-x%2F4%7D%5Cright%29%5Cbigg%7C_0%5E%5Cinfty%20%2B%204%20%5Cint_0%5E%5Cinfty%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx%5Cright%29)
![E(X) = \displaystyle \int_0^\infty e^{-x/4} \, dx = \boxed{4}](https://tex.z-dn.net/?f=E%28X%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E%5Cinfty%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx%20%3D%20%5Cboxed%7B4%7D)
(since the integral of the PDF is 1, and this integral is 4 times that)
![V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%5Cbigg%28%28X%20-%20E%28X%29%29%5E2%5Cbigg%29%20%3D%20E%28X%5E2%29%20-%20E%28X%29%5E2)
Compute the so-called second moment.
![E(X^2) = \displaystyle \int_{-\infty}^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^{-x/4} \, dx](https://tex.z-dn.net/?f=E%28X%5E2%29%20%3D%20%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2%20f_X%28x%29%5C%2C%20dx%20%3D%20%5Cfrac14%20%5Cint_0%5E%5Cinfty%20x%5E2%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx)
Integrate by parts, with
![u = x^2 \implies du = 2x \, dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}](https://tex.z-dn.net/?f=u%20%3D%20x%5E2%20%5Cimplies%20du%20%3D%202x%20%5C%2C%20dx%20%5C%5C%5C%5C%20dv%20%3D%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx%20%5Cimplies%20v%20%3D%20-4%20e%5E%7B-x%2F4%7D)
Then
![E(X^2) = \displaystyle \frac14 \left(\left(-4x^2 e^{-x/4}\right)\bigg|_0^\infty + 8 \int_0^\infty x e^{-x/4} \, dx\right)](https://tex.z-dn.net/?f=E%28X%5E2%29%20%3D%20%5Cdisplaystyle%20%5Cfrac14%20%5Cleft%28%5Cleft%28-4x%5E2%20e%5E%7B-x%2F4%7D%5Cright%29%5Cbigg%7C_0%5E%5Cinfty%20%2B%208%20%5Cint_0%5E%5Cinfty%20x%20e%5E%7B-x%2F4%7D%20%5C%2C%20dx%5Cright%29)
![E(X^2) = 8 E(X) = 32](https://tex.z-dn.net/?f=E%28X%5E2%29%20%3D%208%20E%28X%29%20%3D%2032)
and the variance is
![V(X) = 32 - 4^2 = \boxed{16}](https://tex.z-dn.net/?f=V%28X%29%20%3D%2032%20-%204%5E2%20%3D%20%5Cboxed%7B16%7D)