Question 1185623: A package delivery service divides their packages into weight classes. Suppose that package in the 14 to 20 pound class are uniformly distributed, meaning that all weights within that class are equally likely to occur.
In the light of the above case, you are required to answer the following questions:
(a) Provide a sketch diagram to summarise the above information. Show precisely f(X), the mean value and the boundaries.
(b) Find the probability that a package weighs between 15 and 16.5 pounds.
(c) Find the probability that a package weighs less that 15 pounds.
(d) Find the probability that a package weighs at least 18 pounds
(e) Calculate the variance and the standard deviation of the distribution
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **(a) Sketch Diagram:**
A sketch diagram for a uniform distribution is a rectangle.
* **Boundaries:** The rectangle starts at 14 on the x-axis (representing weight) and ends at 20.
* **f(X):** The height of the rectangle is constant and represents the probability density function f(x). Since the total area under the rectangle must equal 1 (representing the total probability), and the width is 6 (20 - 14), the height f(x) is 1/6.
* **Mean Value:** The mean of a uniform distribution is the midpoint of the interval. So, the mean is (14 + 20) / 2 = 17. Mark this point on the x-axis within the rectangle.
**(b) Probability between 15 and 16.5 pounds:**
In a uniform distribution, probability is represented by the area. The area between 15 and 16.5 is a smaller rectangle within the larger one.
* **Width:** 16.5 - 15 = 1.5
* **Height:** 1/6
* **Area (Probability):** 1.5 * (1/6) = 1.5 / 6 = 0.25
**(c) Probability less than 15 pounds:**
This is the area of the rectangle from 14 to 15.
* **Width:** 15 - 14 = 1
* **Height:** 1/6
* **Area (Probability):** 1 * (1/6) = 1/6 ≈ 0.1667
**(d) Probability at least 18 pounds:**
This is the area of the rectangle from 18 to 20.
* **Width:** 20 - 18 = 2
* **Height:** 1/6
* **Area (Probability):** 2 * (1/6) = 2/6 = 1/3 ≈ 0.3333
**(e) Variance and Standard Deviation:**
For a uniform distribution over the interval [a, b]:
* **Variance (σ²):** (b - a)² / 12
* **Standard Deviation (σ):** √(Variance)
In this case, a = 14 and b = 20.
* **Variance:** (20 - 14)² / 12 = 6² / 12 = 36 / 12 = 3
* **Standard Deviation:** √3 ≈ 1.732
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