Determine the diameter of a hole that is drilled vertically through the center of the solid bounded by the graphs of the equations
z = 0, and
if one-tenth of the volume of the solid is removed. (Round your answer to four decimal places)
The solid takes the form of a scaled (by ` ``80pi`) bivariate bell (Normal/Gaussian) curve where the variance of the two variables `x` and `y` is `sigma^2 =2` and the correlation `rho=0` .
Ignoring the scale factor of `80pi` which only affects the height of the solid and not the radius (which we are interested in), a radius of 4 (the radius of the function `x^2+y^2 = 16`) on this graph is equivalent to a radius of `4/sigma= 4/sqrt(2)` on a standard bivariate bell curve.
The quantile `4/sqrt(2)` gives a one-sided proportion of a standard Normal/Gaussian curve of
`0.5- Phi(-4/sqrt(2)) = 0.498`
where `Phi` is the distribution function of a standard Normal/Gaussian variable.
A twentieth (2 x tenth) of this is 0.0249. The required radius is then at the 100(0.5 - 0.0249) = 47.51th percentile of the standard Normal curve.
Now, using lookup tables for the inverse of the standard Normal distribution function
`Phi^(-1)(0.4751) = -0.06241`
The radius we require is` ` `sigma=sqrt(2)` times this value, working on our standard deviation being `sigma` and not 1 as for the standard Normal distribution.
The required radius of the drilled section is sqrt(2)(-0.06241) = 0.0883.
The required diameter of the drilled section is then 2*0.0883 = 0.1765