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Suppose that \(f(x)=0.25\) for \(0

Short Answer

Expert verified
The mean is 2, and the variance is \(\frac{4}{3}\).

Step by step solution

01

Understand the Distribution

The function given, \(f(x) = 0.25\) for \(0 < x < 4\), represents a uniform distribution over the interval \((0, 4)\). In a uniform distribution \(a\) and \(b\) are the boundaries of the interval, hence here \(a = 0\) and \(b = 4\).
02

Calculate the Mean of X

For a uniform distribution over the interval \((a, b)\), the mean \(\mu\) is given by the formula: \[ \mu = \frac{a + b}{2} \]Substitute \(a = 0\) and \(b = 4\) to find:\[ \mu = \frac{0 + 4}{2} = 2 \]
03

Calculate the Variance of X

The variance \(\sigma^2\) for a uniform distribution is given by:\[ \sigma^2 = \frac{(b - a)^2}{12} \]Substitute \(a = 0\) and \(b = 4\) to find:\[ \sigma^2 = \frac{(4 - 0)^2}{12} = \frac{16}{12} = \frac{4}{3} \]

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Key Concepts

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

Understanding the Mean of Uniform Distribution
The mean of a uniform distribution is a simple concept. When you have a uniform distribution, all outcomes in the interval are equally likely. In a mathematical sense, this is evenly spread across the interval.
For a distribution defined over an interval aFor example, if your interval is from 0 to 4, you calculate it like this: Given by \(\mu = \frac{a + b}{2}\) a = 0 and b = 4,the mean, or average, is calculated as:
\[ \mu = \frac{0 + 4}{2} = 2 \].
This tells you that the center of the distribution, or the expected average outcome, is at the midpoint between your boundary values of 0 and 4.
Examining the Variance of Uniform Distribution
The variance in a uniform distribution tells you how spread out the values are. Variance provides a measure of how much the values deviate from the mean.
In the case of the uniform distribution, the formula to calculate variance is:\[ \sigma^2 = \frac{(b - a)^2}{12} \], where 'a' and 'b' are the boundaries.
The formula for the variance of a uniform distribution might seem complex, but it's primarily about checking how far apart the endpoints of the interval are, compared to 12.
For our interval from 0 to 4, the variance is calculated as:
\[ \sigma^2 = \frac{(4 - 0)^2}{12} = \frac{16}{12} = \frac{4}{3} \].
This tells you that the values of the uniform distribution vary around the mean by an amount given by the calculated variance.
The Basics of Probability Distribution
A probability distribution is essentially a map of all the possible outcomes of a random variable
and the likelihood of each outcome happening. In a uniform distribution, this map is particularly simple.
The outcomes are evenly spread across the interval, meaning each outcome is equally likely.
In mathematical terms, for a uniform distribution between given boundaries aFor the function \(f(x) = 0.25\) in the interval \(0 < x < 4\), each value is equally probable, creating a flat, rectangle-like distribution when graphed. This is why it's termed 'uniform'.
A uniform probability distribution helps you understand how the total probability or certainty of 1 is evenly allocated across the interval. This simplicity can be powerful, allowing for straightforward computations of statistics like mean and variance.

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Most popular questions from this chapter

(a) Calculate the mode, mean, and variance of the distribution for \(\alpha=3\) and \(\beta=1.4\) (b) Calculate the mode, mean, and variance of the distribution for \(\alpha=10\) and \(\beta=6.25\). (c) Comment on the difference in dispersion in the distribution from parts (a) and (b).A European standard value for a low-emission window glazing uses 0.59 as the proportion of solar energy that enters a room. Suppose that the distribution of the proportion of solar energy that enters a room is a beta random variable.

Provide approximate sketches for beta probability density functions with the following parameters. Comment on any symmetries and show any peaks in the probability density functions in the sketches. (a) \(\alpha=\beta<1\). (b) \(\alpha=\beta=1\). (c) \(\alpha=\beta>1\).

The lifetime of a mechanical assembly in a vibration test is exponentially distributed with a mean of 400 hours. (a) What is the probability that an assembly on test fails in less than 100 hours? (b) What is the probability that an assembly operates for more than 500 hours before failure? (c) If an assembly has been on test for 400 hours without a failure, what is the probability of a failure in the next 100 hours? (d) If 10 assemblies are tested, what is the probability that at least one fails in less than 100 hours? Assume that the assemblies fail independently. (e) If 10 assemblies are tested, what is the probability that all have failed by 800 hours? Assume that the assemblies fail independently.

An airline makes 200 reservations for a flight that holds 185 passengers. The probability that a passenger arrives for the flight is \(0.9,\) and the passengers are assumed to be independent. (a) Approximate the probability that all the passengers who arrive can be seated. (b) Approximate the probability that the flight has empty seats. (c) Approximate the number of reservations that the airline should allow so that the probability that everyone who arrives can be seated is \(0.95 .\) [Hint: Successively try values for the number of reservations.]

Suppose that \(X\) has a lognormal distribution with parameters \(\theta=10\) and \(\omega^{2}=16\). Determine the following: (a) \(P(X<2000)\) (b) \(P(X>1500)\) (c) Value exceeded with probability 0.7

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