91Ó°ÊÓ

Let \(\alpha\) and \(\beta\) be any two constants such that \(\alpha<\beta .\) Suppose we choose a point \(x\) at random in the interval from \(\alpha\) to \(\beta .\) In this context the phrase at random is taken to mean that the point \(x\) is as likely to be chosen from one particular part of the interval as any other part. Consider the rectangle. The base of the rectangle has length \(\beta-\alpha\) and the height of the rectangle is \(1 /(\beta-\alpha),\) so the area of the rectangle is \(1 .\) As such, this rectangle's top can be thought of as part of a probability density curve. since we specify that \(x\) must lie between \(\alpha\) and \(\beta,\) the probability of a point occurring outside the interval \([\alpha, \beta]\) is, by definition, \(0 .\) From a geometric point of view, \(x\) chosen at random from \(\alpha\) to \(\beta\) means we are equally likely to land anywhere in the interval from \(\alpha\) to \(\beta .\) For this reason, the top of the (rectangle's) density curve is flat or uniform. Now suppose that \(a\) and \(b\) are numbers such that \(\alpha \leq a

Step by step solution

01

Understand the problem

We are given a continuous uniform distribution for pellet diameters between 0.015 mm and 0.065 mm. We are tasked with finding the probability of certain diameter ranges and calculating the mean and standard deviation of this distribution.

02

Define the interval

For this uniform distribution, the interval's lower bound is \(\alpha = 0.015\) mm and the upper bound is \(\beta = 0.065\) mm. The length of the interval \(\beta - \alpha = 0.065 - 0.015 = 0.050\) mm.

03

Calculate probability for size 0.050 mm or larger

The interval for pellets 0.050 mm or larger is \([0.050, 0.065]\). The length of this interval is \(0.065 - 0.050 = 0.015\). The probability is \(P(0.050 \leq x \leq 0.065) = \frac{0.015}{0.050} = 0.3\).

04

Calculate probability for size 0.040 mm or smaller

The interval for pellets 0.040 mm or smaller is \([0.015, 0.040]\). The length of this interval is \(0.040 - 0.015 = 0.025\). The probability is \(P(0.015 \leq x \leq 0.040) = \frac{0.025}{0.050} = 0.5\).

05

Calculate probability for size between 0.035 mm and 0.055 mm

The interval for pellets between 0.035 mm and 0.055 mm is \([0.035, 0.055]\). The length of this interval is \(0.055 - 0.035 = 0.020\). The probability is \(P(0.035 \leq x \leq 0.055) = \frac{0.020}{0.050} = 0.4\).

06

Compute the mean particle size

The mean \(\mu\) of a uniform distribution is calculated as \(\mu = \frac{\alpha + \beta}{2}\). For our interval, \(\mu = \frac{0.015 + 0.065}{2} = 0.040\) mm.

07

Compute the standard deviation of particle size

The standard deviation \(\sigma\) of a uniform distribution is calculated as \(\sigma = \frac{\beta - \alpha}{\sqrt{12}}\). For our interval, \(\sigma = \frac{0.050}{\sqrt{12}} \approx 0.0144\) mm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability

Probability is a fundamental concept in statistics and represents the likelihood of a specific event occurring within a defined set of possibilities. In the context of a continuous uniform distribution—such as the range of pellet diameters mentioned in the exercise—each outcome within the specified interval is equally probable. This means any point or sub-interval selected from the larger interval has an equal chance of being chosen.

To determine the probability for any specific sub-interval in a continuous uniform distribution, we consider the length of the sub-interval. For example:

• The probability of a pellet diameter being between 0.050 mm and 0.065 mm is computed by dividing the length of this sub-interval (\[0.065 - 0.050 = 0.015\]) by the total length of the interval (\[0.065 - 0.015 = 0.050\]), resulting in a probability of\[ \frac{0.015}{0.050} = 0.3 \].
• Similarly, for a pellet diameter between 0.035 mm and 0.055 mm, the probability is calculated as\[ \frac{0.020}{0.050} = 0.4 \].
  Understanding probability in uniform distributions is central to predicting outcomes in a given range.

Mean and Standard Deviation

The mean and standard deviation are key indicators in any probability distribution, providing insight into the central tendency and variability of the distribution, respectively.

For a continuous uniform distribution, the mean represents the midpoint of the interval. It is calculated as:\[\mu = \frac{\alpha + \beta}{2}\]. In the given problem with pellet diameters between 0.015 mm and 0.065 mm, the mean diameter comes out to\[\mu = \frac{0.015 + 0.065}{2} = 0.040\] mm.

The standard deviation in a uniform distribution measures the spread of the data around this mean, capturing the variability of the pellet sizes. The formula for the standard deviation is:\[\sigma = \frac{\beta - \alpha}{\sqrt{12}}\]. For the example provided, this calculation yields\[\sigma \approx 0.0144\] mm.

• The mean tells us that the average pellet size expected in the given range is 0.040 mm.
• The standard deviation shows that the pellet sizes deviate from the mean by approximately 0.0144 mm, indicating how spread out the sizes are around the mean.

Continuous Uniform Distribution

The continuous uniform distribution, often called a rectangular distribution, represents scenarios where all outcomes within a certain interval are equally likely. This distribution is characterized by its flat probability density function (PDF), which results in a constant height across the interval

This distribution is commonly used in situations involving ranges of values over a fixed interval. For example, in the exercise provided, the pellet diameters range uniformly between 0.015 mm and 0.065 mm further exemplifies this. Each pellet size has the same chance of being selected compared to any other size in this range.

• The uniform distribution can be quickly identified by its evenly distributed probabilities across its defined range.
• Its simplicity often makes it a starting point for understanding more complex distributions, as it clearly depicts random variability within indicated bounds.

When applying the concept of continuity to a uniform distribution, keep in mind: the entirety of the interval should sum to a probability of 1, as seen by the integral of the PDF over the interval. This rule ensures that probability mass is uniformly distributed across every potential outcome within the specified range.