Question 1186386: Usually the size of sediment particles follows a uniform distribution (Reference: Y. Zimmels, “Theory of Hindered Sedimentation of Polydisperse Mixtures,” AIchE journal, Vol.29, No. 4, pp.669-676).Suppose a veterinary science experiment injects very small, spherical pellets of low-level radiation directly into an animal’s bloodstream. The purpose is to attempt to cure a form of recurring
cancer. The pellets eventually dissolve and pass through the animal’s system.
Diameters of the pellets are uniformly distributed from 0.010 mm to 0.065 mm.
What is the probability that it will be 0.055 mm or larger? Hint: all particles are between
0.010 mm and 0.065 mm so larger than 0.055 mm means 0.055 ≤ x ≤ 0.065.
b. What is the probability that it will be 0.035 mm or smaller?
c. What is the probability that it will be between 0.040 mm and 0.050 mm?
Compute the following for the distribution of the diameter of the pellets:
d. The mean. Round the final answer to 4 decimals.
e. The variance. Round the final answer to 6 decimals.
f. The standard deviation. Round the final answer to 4 decimals.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem about the uniform distribution of pellet diameters:
**Understanding Uniform Distribution**
In a uniform distribution over an interval [a, b], the probability density function is constant. The probability of a value falling within any subinterval is proportional to the length of that subinterval.
* a = 0.010 mm (minimum diameter)
* b = 0.065 mm (maximum diameter)
**a. Probability of 0.055 mm or larger:**
We want P(0.055 ≤ x ≤ 0.065). The length of this interval is 0.065 - 0.055 = 0.010 mm. The length of the entire interval is 0.065 - 0.010 = 0.055 mm.
P(0.055 ≤ x ≤ 0.065) = (0.010 mm) / (0.055 mm) = 10/55 = 2/11 ≈ 0.1818
**b. Probability of 0.035 mm or smaller:**
We want P(0.010 ≤ x ≤ 0.035). The length of this interval is 0.035 - 0.010 = 0.025 mm.
P(0.010 ≤ x ≤ 0.035) = (0.025 mm) / (0.055 mm) = 25/55 = 5/11 ≈ 0.4545
**c. Probability between 0.040 mm and 0.050 mm:**
We want P(0.040 ≤ x ≤ 0.050). The length of this interval is 0.050 - 0.040 = 0.010 mm.
P(0.040 ≤ x ≤ 0.050) = (0.010 mm) / (0.055 mm) = 10/55 = 2/11 ≈ 0.1818
**Calculations for the Distribution:**
**d. Mean:**
The mean of a uniform distribution is given by:
μ = (a + b) / 2
μ = (0.010 + 0.065) / 2
μ = 0.075 / 2
μ = 0.0375 mm
**e. Variance:**
The variance of a uniform distribution is given by:
σ² = (b - a)² / 12
σ² = (0.065 - 0.010)² / 12
σ² = (0.055)² / 12
σ² = 0.003025 / 12
σ² ≈ 0.000252083
**f. Standard Deviation:**
The standard deviation is the square root of the variance:
σ = sqrt(σ²)
σ = sqrt(0.000252083)
σ ≈ 0.015877 mm
**Summary of Answers:**
* a. 0.1818
* b. 0.4545
* c. 0.1818
* d. 0.0375 mm
* e. 0.000252
* f. 0.0159 mm
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