Question

According to Zimmels (1983), the sizes of particles used in sedimentation experiments often have a uniform distribution. In sedimentation involving mixtures of particles of various sizes, the larger particles hinder the movements of the smaller ones. Thus, it is important to study both the mean and the variance of particle sizes. Suppose that spherical particles have diameters that are uniformly distributed between 0.02 and 0.08 centimeters. Find the mean and variance of the volumes of these particles. (Recall that the volume of a sphere is (4/3) πr3) Round your answers to four decimal places.
E(Y)= ___ x10−5 cm3
V(Y) = ___ x10−9

Answer 1

The expected value of a normal distribution is the mean of the distribution, while the variance measures the squared deviation of a value from the expected value. The expected value and the variance are: $13.6190 \times 10^{-5}$ and $580.8000 \times 10^{-9}$

Given that, the diameters are:

$d_1 = 0.02$

$d_2 = 0.08$

The radius is:

$r = \frac{d}{2}$

So, we have:

$r_1 = \frac{0.02}{2} = 0.01$

$r_2 = \frac{0.08}{2} = 0.04$

The volume of the sphere is:

$V = \frac{4}{3} \times \pi \times r^3$

For $r_1 = 0.01$, the volume is:

$V_1 = \frac{4}{3} \times \frac{22}{7} \times 0.01^3 = 0.419047 \times 10^{-5}$

For $r_2 = 0.04$, the volume is

$V_2 = \frac{4}{3} \times \frac{22}{7} \times 0.04^3 = 26.819047 \times 10^{-5}$

The mean of a uniform distribution is:

$E(y) = \frac{a + b}{2}$

In this case, the mean is:

$E(y) = \frac{V_1 + V_2}{2}$

So, we have:

$E(y) = \frac{0.419047 \times 10^{-5} + 26.819047 \times 10^{-5}}{2}$

$E(y) = \frac{27.238094\times 10^{-5} }{2}$

$E(y) = 13.619047 \times 10^{-5}$

Approximate

$E(y) = 13.6190 \times 10^{-5}$

The variance of a uniform distribution is:

$V(y) = \frac{(b-a)^2}{12}$

In this case, the volume is:

$V(y) = \frac{(V_2-V_1)^2}{12}$

So, we have:

$V(y) = \frac{(26.819047 \times 10^{-5}- 0.419047 \times 10^{-5})^2}{12}$

$V(y) = \frac{(26.4 \times 10^{-5})^2}{12}$

$V(y) = \frac{(696.96 \times 10^{-10})}{12}$

$V(y) = 58.08000 \times 10^{-10}$

Rewrite as:

$V(y) = 580.8000 \times 10^{-9}$

Hence, the expected value and the variance of the sphere are:$13.6190 \times 10^{-5}$ and $580.8000 \times 10^{-9}$

Learn more about expected values and variance at:

https://brainly.com/question/4470015