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The random variable \(x\) is known to be uniformly distributed between 1.0 and 1.5 a. Show the graph of the probability density function. b. Compute \(P(x=1.25)\) c. Compute \(P(1.0 \leq x \leq 1.25)\) d. Compute \(P(1.20

Short Answer

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a. PDF is a horizontal line from 1.0 to 1.5 at height 2. b. 0. c. 0.5. d. 0.6.

Step by step solution

01

Understanding Uniform Distribution

A uniformly distributed random variable means every interval of the same length has the same probability. In this case, the random variable \(x\) ranges from 1.0 to 1.5.
02

Graphing the Probability Density Function (PDF)

The PDF of a uniform distribution is constant between the lower limit \(a = 1.0\) and the upper limit \(b = 1.5\). The height of the PDF graph is \(\frac{1}{b-a}\). Here, it is \(\frac{1}{1.5-1.0} = 2\). The graph is a horizontal line from \(x=1.0\) to \(x=1.5\) at the height 2.
03

Calculating P(x=1.25)

For a continuous random variable, the probability of any single point is zero. Thus, \(P(x=1.25) = 0\).
04

Calculating P(1.0 ≤ x ≤ 1.25)

To find this probability, calculate the area under the PDF between 1.0 and 1.25. The length of this interval is 0.25. Therefore, the probability is \(\text{width} \times \text{height} = 0.25 \times 2 = 0.5\).
05

Calculating P(1.20 < x < 1.5)

Calculate the area under the PDF between 1.20 and 1.5. The length of this interval is 0.3. Thus, the probability is \(0.3 \times 2 = 0.6\).

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Key Concepts

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

Probability Density Function
A Probability Density Function (PDF) is a crucial concept in statistics, especially when dealing with continuous random variables. It represents how the probability is distributed over different values. In the case of a uniform distribution, the PDF is quite straightforward.
If a continuous random variable is uniformly distributed between two endpoints, say from point \(a\) to point \(b\), the PDF will be constant between these points. Mathematically, the PDF for a uniform distribution is given by \( f(x) = \frac{1}{b-a} \). This means the PDF is simply a horizontal line between \(a\) and \(b\), and it holds the same value for any \(x\) within this interval.
  • This constant value of the PDF ensures that all sub-intervals of equal length are equally probable within the range.
  • In a graph, this becomes evident as the area under the PDF over a specific range represents the probability. However, outside the range \(a\) to \(b\), the PDF is zero, indicating no probability.
This uniform behavior makes it easy to graph and calculate probabilities for continuous random variables within the defined bounds.
Continuous Random Variable
A continuous random variable can take on an infinite number of possible values within a specified range. Unlike a discrete random variable, which might only hold countable outcomes, a continuous random variable embraces a continuous range.
Thus, any single specific value has a probability of zero. This might seem counterintuitive at first since we're accustomed to seeing probabilities associated with outcomes. However, in the realm of continuous variables, probability is more aptly represented over intervals rather than at points.
  • The integral of the PDF over an interval gives the probability that the variable falls within that interval.
  • This interval could be infinitesimally small, but never absolutely a single point.
Understanding continuous random variables helps clarify why calculations for probabilities are usually focused on intervals, such as \(P(1.0 \leq x \leq 1.25)\), rather than specific points.
Probability Calculation
Probability calculation in the context of uniformly distributed continuous random variables is straightforward, owing to the simplicity of the uniform distribution's PDF.
For a given interval, the probability is calculated as the product of the interval's length and the height of the PDF over that interval. This is rooted in the principle that probability is the area under the curve of the PDF.
  • For example, if calculating \(P(1.0 \leq x \leq 1.25)\), the length of the interval is 0.25, and since our PDF height within 1.0 to 1.5 is 2, the probability is \(0.25 \times 2 = 0.5\).
  • This straightforward calculation method makes it easy to determine probabilities across different sub-ranges of the defined interval.
With uniform distributions, these calculations provide quick insights into the likelihood of a random variable falling between any two points.

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Most popular questions from this chapter

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