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Chapter 4: Problem 51

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

Short Answer

Expert verified
(a) Mean = 0, Variance = 1/3, SD 鈮� 0.577 (b) x = 0.9 (c) CDF: 0 for x < -1; (x+1)/2 for -1 鈮� x 鈮� 1; 1 for x > 1

Step by step solution

01

Identify Mean of a Uniform Distribution

The mean of a continuous uniform distribution over an interval \([a, b]\) is given by \(\frac{a+b}{2}\). Here, the interval is \([-1, 1]\), so the mean \(\mu\) is calculated as:\[\mu = \frac{-1 + 1}{2} = 0.\]
02

Calculate Variance of a Uniform Distribution

The variance of a continuous uniform distribution over an interval \([a, b]\) is given by \(\frac{(b-a)^2}{12}\). Using the same interval, \([-1, 1]\), we find:\[\sigma^2 = \frac{(1 - (-1))^2}{12} = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3}.\]
03

Find Standard Deviation

The standard deviation is the square root of the variance. From the variance calculated in step 2:\[\sigma = \sqrt{\frac{1}{3}} \approx 0.577.\]
04

Solve for P(-x < X < x) = 0.90

The probability that \(-x < X < x\) for a uniform distribution is equal to the length of the interval divided by the total length of the distribution. Therefore:\[\frac{2x}{2} = 0.90.\]This simplifies to \(x = 0.9\).
05

Write the Cumulative Distribution Function (CDF)

For a uniform distribution over \([-1, 1]\), the cumulative distribution function \(F(x)\) is defined piecewise:- For \(x < -1\), \(F(x) = 0\).- For \(-1 \le x \le 1\), \(F(x) = \frac{x + 1}{2}\).- For \(x > 1\), \(F(x) = 1\).

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Key Concepts

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

Mean of Uniform Distribution
The mean of a uniform distribution, also known as the expected value, is a measure of the central tendency of the distribution. For a continuous uniform distribution defined over the interval \([a, b]\), the mean is calculated using the formula:
  • \( \mu = \frac{a+b}{2} \)
In this exercise, the interval is \([-1, 1]\).
By applying the formula, you'll find:
  • \( \mu = \frac{-1 + 1}{2} = 0 \)
This means that the average or expected value of the random variable \(X\) is zero.
Variance of Uniform Distribution
The variance of a uniform distribution measures the spread or dispersion of the values around the mean. It provides insight into how much the values differ from the mean on average. The formula for finding the variance in a uniform distribution over an interval \([a, b]\) is:
  • \( \sigma^2 = \frac{(b-a)^2}{12} \)
For the interval \([-1, 1]\), this calculation becomes:
  • \( \sigma^2 = \frac{(1 - (-1))^2}{12} = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3} \)
The variance tells us that the data points are distributed symmetrically around the mean.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) gives the probability that the random variable takes on values less than or equal to \(x\). For uniform distributions, the CDF can be defined using a piecewise function.
Specifically, for the interval \([-1, 1]\):
  • For \(x < -1\), \(F(x) = 0\)
  • For \(-1 \le x \le 1\), \(F(x) = \frac{x + 1}{2}\)
  • For \(x > 1\), \(F(x) = 1\)
This function helps to understand the accumulation of probability across different points in the interval.
Probability Calculation
Calculating probabilities in a uniform distribution often involves understanding the percentage of the interval that a particular condition covers. In this exercise, we determine the value of \(x\) such that \(P(-x < X < x) = 0.90\). This involves finding the length of the interval, \(2x\), and dividing it by the total length of the distribution, \(2\), which gives:
  • \( \frac{2x}{2} = 0.90 \)
Solving for \(x\), we have:
  • \( x = 0.9 \)
Thus, the probability that the random variable falls between \(-x\) and \(x\) with the condition specified is 90%.
Standard Deviation
The standard deviation of a uniform distribution gives a measure of the amount of variation or dispersion in the data. It is simply the square root of the variance. In mathematical terms, the standard deviation \(\sigma\) is given by:
  • \( \sigma = \sqrt{\sigma^2} \)
From our variance calculation, we know that \(\sigma^2 = \frac{1}{3}\). Therefore, the standard deviation is:
  • \( \sigma = \sqrt{\frac{1}{3}} \approx 0.577 \)
The standard deviation provides insight into the average distance of the values from the mean in this uniform distribution.

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