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Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5\). a. Compute \(P(X<0)\). b. Compute \(P(-2.5

Short Answer

Expert verified
a. \(\frac{1}{2}\), b. \(\frac{1}{2}\), c. \(\frac{1}{2}\), d. \(\frac{2}{5}\).

Step by step solution

01

Understanding uniform distribution

In a uniform distribution, each value between the minimum and maximum is equally likely. This means the probability density function (pdf) is constant across the range. For this exercise, the range is from \( A = -5 \) to \( B = 5 \). The pdf can be defined as \( f(x) = \frac{1}{B - A} \) for \( A \leq x \leq B \).
02

Calculating probability P(X

The probability \( P(X<0) \) can be found by integrating the probability density function from \( A = -5 \) to 0. Since the pdf is \( f(x) = \frac{1}{10} \), the calculation is:\[P(X < 0) = \int_{-5}^{0} \frac{1}{10} \, dx = \frac{1}{10} \times (0 - (-5)) = \frac{1}{2}\]
03

Calculating probability P(-2.5

Given the constant pdf for a uniform distribution, the probability \( P(-2.5 < X < 2.5) \) is calculated as the length of the interval over the total range length. Hence,\[P(-2.5 < X < 2.5) = \frac{1}{10} \times (2.5 - (-2.5)) = \frac{1}{10} \times 5 = \frac{1}{2}\]
04

Calculating probability P(-2 ≤ X ≤ 3)

To find \( P(-2 \leq X \leq 3) \), we calculate the proportion of the interval length to the entire distribution. Thus,\[P(-2 \leq X \leq 3) = \frac{1}{10} \times (3 - (-2)) = \frac{1}{10} \times 5 = \frac{1}{2}\]
05

Calculating probability (general form) P(k

Given \(-5 < k < k+4 < 5\), the probability \( P(k < X < k+4) \) is determined by the length of \( k+4 - k = 4 \), so\[P(k < X < k+4) = \frac{1}{10} \times 4 = \frac{2}{5}\]

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Key Concepts

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

Probability Density Function
In the realm of probability and statistics, a Probability Density Function, often abbreviated as pdf, plays a crucial role in understanding continuous random variables. Unlike discrete probability distributions, where probability is concentrated at individual points, the pdf of a continuous random variable defines a curve where the area under the curve represents probability.

For a uniform distribution, this is particularly straightforward. The probability is spread evenly across the range between a minimum value, denoted as A, and a maximum value, denoted as B. In our case, we have a reaction temperature that ranges uniformly from -5 to 5 degrees Celsius. The function can be written as:
\[ f(x) = \frac{1}{B - A}\] for \(A \leq x \leq B\).

Since each value between A and B has an equal likelihood of occurring, the pdf becomes a horizontal line in a graph representing the uniform distribution. The simplicity of the uniform pdf - being constant across the interval - makes it especially useful for introductory discussions of continuous probabilities.
Integration in Probability
Integration is a fundamental concept in probability, particularly when dealing with continuous distributions. When we want to calculate the probability of a range of values for a continuous random variable, we integrate the probability density function across that range.

This integration process involves summing up infinitesimally small pieces over the interval of interest, yielding the total probability. For example, to calculate the probability that the random variable X is less than zero, given that X uniformly distributes between A and B, we would integrate the pdf from A to 0:
\[ P(X < 0) = \int_{A}^{0} f(x) \, dx\]
Since the pdf for our uniform distribution is \(\frac{1}{10}\), the integration simply becomes \(\frac{1}{10} \times\) the length of the interval, which in this case computes to \(\frac{1}{2}\).

Through integration, we can obtain probabilities for any sub-range in the whole distribution. This method essentially finds the area under the curve of the pdf, offering a visual interpretation of probability as an area on a graph.
Continuous Random Variables
Continuous random variables are a fundamental concept in probability theory, different from their discrete counterparts. These variables can take on any value within a given range, allowing for infinite possibilities even within the smallest interval. This idea is crucial in situations like our chemical process scenario, where the reaction temperature is not limited to set points but can vary fluidly from -5 to 5 degrees Celsius.

A uniform distribution as applied to continuous random variables suggests that every outcome in a certain interval is equally likely. Thus, the probability of the variable taking any specific value is essentially zero since there are infinitely many values possible.
Instead, we talk about the probability of the variable falling within an interval, which is calculated using integration. This leads directly to the concept of a probability density function, which provides a probability spread across a continuum of values.

Understanding continuous random variables enables a deeper insight into real-world scenarios, as most natural and technical phenomena cannot be effectively described by discrete outcomes alone.
Probability Calculation
Probability calculation is an essential skill in both everyday decision-making and scientific analysis. In the context of continuous distributions like the uniform distribution we've been discussing, probability is calculated over an interval where the random variable may lie.

For instance, let's look at the probability of the reaction temperature, X, being between -2 and 3 degrees Celsius. We calculate this by considering the interval size, which is calculated as follows:
\[ P(-2 \leq X \leq 3) = \frac{1}{B - A} \times (3 - (-2))\]
The result, \(\frac{1}{2}\), comes from the interval length relative to the total range from A to B. The general rule with uniform distributions is:
- Identify the interval of interest.- Recognize that the probability density function is \(\frac{1}{B - A}\).- Calculate probability as the product of the interval's length and the pdf.
This approach is simple due to the uniform distribution's characteristics and offers a clear method for determining how likely an event is within the specified constraints. Calculating probabilities in this manner is fundamental in probabilistic modeling and risk assessment.

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Most popular questions from this chapter

Let \(\mathrm{Z}\) have a standard normal distribution and define a new rv \(Y\) by \(Y=\sigma Z+\mu\). Show that \(Y\) has a normal distribution with parameters \(\mu\) and \(\sigma\). [Hint: \(Y \leq y\) iff \(Z \leq\) ? Use this to find the cdf of \(Y\) and then differentiate it with respect to \(y\).]

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