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Suppose that \(f(x)=x / 8\) for \(33.5)\) c. \(P(44.5)\)

Short Answer

Expert verified
a) \(P(X<4)=\frac{7}{16}\); b) \(P(X>3.5)=\frac{51}{64}\); c) \(P(44.5)=\frac{1}{2}\)."

Step by step solution

01

Understand the function

The given function is a probability density function, where \( f(x) = \frac{x}{8} \) for \( x \) in the range \( 3 < x < 5 \). This means that the probability is proportional to the value of \( x \) in this range.
02

Identify the ranges for integration

For each probability query, identify the limits of integration. - For \( P(X < 4) \), we adjust the range to \((3, 4)\).- For \( P(X > 3.5) \), the range changes to \((3.5, 5)\).- For \( P(4 < X < 5) \), the range is \((4, 5)\).- For \( P(X < 4.5) \), the range is \((3, 4.5)\).- For \( P(X < 3.5 \text{ or } X > 4.5) \), calculate separately for \((3, 3.5)\) and \((4.5, 5)\).
03

Set up the integrals

Set up integrals for each probability:- \( P(X < 4) = \int_{3}^{4} \frac{x}{8} \, dx \)- \( P(X > 3.5) = \int_{3.5}^{5} \frac{x}{8} \, dx \)- \( P(4 < X < 5) = \int_{4}^{5} \frac{x}{8} \, dx \)- \( P(X < 4.5) = \int_{3}^{4.5} \frac{x}{8} \, dx \)- \( P(X < 3.5 \text{ or } X > 4.5) = \left(\int_{3}^{3.5} \frac{x}{8} \, dx + \int_{4.5}^{5} \frac{x}{8} \, dx \right)\)
04

Integrate each expression

Calculate each integral:- \( \int_{3}^{4} \frac{x}{8} \, dx = \left[ \frac{x^2}{16} \right]_{3}^{4} = \frac{4^2}{16} - \frac{3^2}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}\)- \( \int_{3.5}^{5} \frac{x}{8} \, dx = \left[ \frac{x^2}{16} \right]_{3.5}^{5} = \frac{25}{16} - \frac{12.25}{16} = \frac{12.75}{16}\)- \( \int_{4}^{5} \frac{x}{8} \, dx = \left[ \frac{x^2}{16} \right]_{4}^{5} = \frac{25}{16} - \frac{16}{16} = \frac{9}{16}\)- \( \int_{3}^{4.5} \frac{x}{8} \, dx = \left[ \frac{x^2}{16} \right]_{3}^{4.5} = \frac{20.25}{16} - \frac{9}{16} = \frac{11.25}{16}\)- \( \int_{3}^{3.5} \frac{x}{8} \, dx + \int_{4.5}^{5} \frac{x}{8} \, dx = \left(\frac{12.25}{16} - \frac{9}{16}\right) + \left(\frac{25}{16} - \frac{20.25}{16}\right) = \frac{3.25}{16} + \frac{4.75}{16} = \frac{8}{16} = \frac{1}{2}\)
05

Finalize the calculations

Compile all your results:- \(P(X<4) = \frac{7}{16}\)- \(P(X>3.5) = \frac{12.75}{16}\)- \(P(44.5) = \frac{1}{2}\)

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Key Concepts

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

Probability Density Function
The concept of a probability density function (PDF) is central to understanding continuous random variables. In simple terms, a PDF helps us understand how the probabilities of a continuous random variable are distributed along its possible values. For the function to qualify as a probability density function, it must satisfy two main criteria:

  • The function must be non-negative; that is, it should be equal to or greater than zero for all possible values within its domain.
  • The area under the entire function over its domain must equal 1. This property ensures the probabilities for all different values of the random variable sum up to a whole or complete probability.
For any given range within this domain, the definite integral of the PDF over that range gives us the probability that the random variable falls within it. Similarly, in our original problem, the function given is \( f(x) = \frac{x}{8} \) for the range \( 3 < x < 5 \), which satisfies the non-negative condition and can be integrated over its interval to find probabilities for various conditions.
Definite Integrals
Definite integrals are an integral part of calculus used to calculate the net area under a curve on a real number line. In the context of probability density functions, they are employed to calculate the probability that a continuous random variable falls within a specific range.

Each query for probability is set up as a definite integral with specific limits of integration, representing the range of interest. For example, the probability that a random variable is less than 4 is found by integrating the PDF from the lower limit of 3 up to 4. Mathematically, this is expressed as:
\[ P(X < 4) = \int_{3}^{4} \frac{x}{8} \, dx \]
The integration process involves finding the antiderivative and evaluating it at the boundaries (upper and lower limits). This becomes the result of the integral, representing the probability for the described condition. These results can be directly utilized to interpret and discern the likelihood of different occurrences within the designated interval.
Continuous Random Variables
Continuous random variables are variables that can take on an infinite number of possible values within a given range. They are the opposite of discrete random variables, which only assume distinct values. Because they can take any value within an interval, the probability of the variable assuming a specific value is practically zero.

Instead of considering the likelihood of a single value, probabilities for continuous random variables are determined over an interval. For example, in our exercise, the variable \( X \) can assume any value between 3 and 5, making it continuous. The function \( f(x) \) represents the probability distribution across these values.

By employing PDFs and definite integrals, we can acquire meaningful probabilities that help in predicting the occurrence of events. This is done by determining the area under the curve over our interval of interest, hence referring to the application of continuous distributions.

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