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Suppose \(X\) has a continuous uniform distribution over the interval [-1,1] . (a) Determine the mean, variance, and standard deviation of \(X\). (b) Determine the value for \(x\) such that \(P(-x

Short Answer

Expert verified
(a) Mean = 0, Variance = 1/3, Standard deviation = 1/√3. (b) x = 0.9. (c) CDF: F(x) = 0 for x < -1, (x+1)/2 for -1 ≤ x ≤ 1, 1 for x > 1.

Step by step solution

01

Mean of Uniform Distribution

The mean (expected value) of a continuous uniform distribution over an interval \([a, b]\) is given by the formula \(\mu = \frac{a+b}{2}\). In this case, \(a = -1\) and \(b = 1\). Thus, the mean \(\mu = \frac{-1+1}{2} = 0\).
02

Variance of Uniform Distribution

The variance of a uniform distribution over an interval \([a, b]\) is given by \(\sigma^2 = \frac{(b-a)^2}{12}\). Here, \(a = -1\) and \(b = 1\). So the variance \(\sigma^2 = \frac{(1-(-1))^2}{12} = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3}\).
03

Standard Deviation of Uniform Distribution

The standard deviation is the square root of the variance. Thus, \(\sigma = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\).
04

Symmetric Interval for Probability

We want to find the value of \(x\) such that \(P(-x < X < x) = 0.90\). The length of this interval is \(2x\), and since the distribution is uniform over \([-1, 1]\), the probability density is constant. The total length of the possible interval is 2 (from -1 to 1). This means \(2x \cdot \frac{1}{2} = 0.90\), leading to \(x = 0.90\). Thus, \(P(-x < X < x) = 0.90\).
05

Cumulative Distribution Function (CDF)

The cumulative distribution function \(F(x)\) for a continuous uniform distribution over \([-1, 1]\) is defined piecewise as follows: \[ F(x) = \begin{cases} 0, & x < -1, \ \frac{x+1}{2}, & -1 \leq x \leq 1, \ 1, & x > 1. \end{cases} \] This expresses the probability that \(X\) is less than or equal to \(x\).

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Key Concepts

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

Mean of Uniform Distribution
In a continuous uniform distribution, every outcome in a given interval is equally likely to occur. Let's focus on calculating the mean of a uniform distribution, which is a vital measure of central tendency. For a uniform distribution defined over the interval \([a, b]\), the mean or expected value \(\mu\) can be found using the straightforward formula:\[\mu = \frac{a+b}{2}\]This formula tells us that the mean is simply the midpoint of the interval.
It's easy to remember because you're just averaging the lower and upper bounds.
  • In the exercise, we have \(a = -1\) and \(b = 1\).
  • Thus, \(\mu = \frac{-1+1}{2} = 0\).
This is because the symmetry of the interval around zero means the average will logically be zero. Understanding the mean helps provide insight into the balance point of the distribution.
Variance of Uniform Distribution
The variance of a uniform distribution measures how much values deviate from the mean.
It's a key indicator of the distribution's spread. For a uniform distribution on an interval \([a, b]\), variance \(\sigma^2\) is calculated with the formula:\[\sigma^2 = \frac{(b-a)^2}{12}\]Here's how to interpret it:
  • The term \(b-a\) is the interval's length, indicating the amount of total spread.
  • Dividing by 12 gives a measure of average squared deviation from the mean.
Applying this to our interval of \([-1, 1]\), the calculation proceeds as:
  • Length of the interval \((b-a) = 2\).
  • Variance \(\sigma^2 = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3}\).
Understanding variance helps assess how spread out the distribution is from the mean, complementing the information given by the mean itself.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a uniform distribution shows how probabilities accumulate over the interval.
It represents the probability that a random variable \(X\) will take a value less than or equal to \(x\). For our uniform distribution over the interval \([-1, 1]\), the CDF \(F(x)\) is defined piecewise as follows:\[F(x) = \begin{cases} 0, & x < -1, \\frac{x+1}{2}, & -1 \leq x \leq 1, \1, & x > 1.\end{cases}\]This piecewise function explains:
  • For \(x < -1\), the probability is 0 since \(x\) falls outside the interval.
  • For \(-1 \leq x \leq 1\), the probabilities accumulate linearly. The transformation \(\frac{x+1}{2}\) accounts for the proportion of the way through the interval \([-1, 1]\).
  • For \(x > 1\), the probability is 1, covering all possible outcomes.
The CDF is invaluable for determining probabilities over ranges of values within a distribution, giving you a full picture of the distribution dynamics.

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