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Find the mean and variance, if they exist, of each of the following distributions. (a) \(p(x)=\frac{3 !}{x(3-x) !}\left(\frac{1}{2}\right)^{3}, x=0,1,2,3\), zero elsewhere. (b) \(f(x)=6 x(1-x), 0

Short Answer

Expert verified
(a) Mean: 1.5, Variance: 0.75; (b) Mean: 0, Variance: 1/10; (c) Mean: 2, Variance: Does not exist

Step by step solution

01

Identify the type of distribution for (a)

The distribution in (a) is a binomial distribution because it follows the form \(p(x) = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\). In this problem, \(n = 3\) and \(p = \frac{1}{2}\), therefore the mean is \(np = 3*\frac{1}{2} = 1.5\) and the variance is \(np(1 - p) = 3*\frac{1}{2}*\frac{1}{2} = 0.75\).
02

Calculate the mean and variance for distribution (b)

For the continuous distribution \(f(x) = 6x(1 - x)\), the mean (or expectation) is calculated as \( E[X] = \int xf(x) dx\) over the range of \(x\), and the variance is calculated as \(E[X^2] - (E[X])^2\). So start by calculating the mean \( E[X] = \int_{0}^{1} x * 6x(1 - x) dx = \int_{0}^{1} 6x^{2} - 6x^{3} dx = 1 - 1 = 0\). Next calculate \(E[X^2] = \int_{0}^{1} x^{2} * 6x(1 - x) dx = \int_{0}^{1} 6x^{3} - 6x^{4} dx = \frac{6}{4} - \frac{6}{5} = \frac{1}{10}\). Hence, the variance is \(\frac{1}{10}\) - \(0^2 = \frac{1}{10}\)
03

Calculate the mean and variance for distribution (c)

For the continuous distribution \(f(x) = 2/x^{3}\), the mean (or expectation) is calculated as \( E[X] = \int xf(x) dx\) over the range of \(x\), and the variance is calculated as \(E[X^2] - (E[X])^2\). So start by calculating the mean \(E[X] = \int_{1}^{\infty} x * \frac{2}{x^{3}} dx = \int_{1}^{\infty} \frac{2}{x^{2}} dx = 2 * [-\frac{1}{x} ]_{1}^{\infty} = 2\). Next calculate \(E[X^2] = \int_{1}^{\infty} x^{2} * \frac{2}{x^{3}} dx = \int_{1}^{\infty} \frac{2}{x} dx = 2 * [ln|x|]_{1}^{\infty}\). Since \(ln(x)\) is not defined for \(x = \infty\) (it diverges), the variance does not exist for this distribution. If it did exist, it would be calculated as \(E[X^2] - (E[X])^2\).

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Key Concepts

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

Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials of a binary experiment. An experiment is called binary if it results in one of two outcomes, 'success' or 'failure'. For instance, flipping a coin results in either heads or tails - a binary outcome.

The probability mass function for a binomial distribution is given by:
\[ p(x) = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x} \]
where:\begin{itemize}\item \( n \) is the number of trials,\item \( x \) is the number of successful outcomes (successes),\item \( p \) is the probability of success on an individual trial, and\item \( 1-p \) is the probability of failure.\end{itemize}
The mean (also known as expectation) of a binomial distribution is \( np \), revealing how many successes we would expect on average. The variance, which measures the spread of the distribution, is \( np(1-p) \). These measures help to understand the behavior of the distribution and to predict probabilities of different outcomes.
Continuous Distribution
In contrast to discrete distributions that deal with distinct or separate outcomes, continuous distributions are used when dealing with data that can take on any value within a given range. This could include measurements like height, weight, or time.

A continuous distribution is characterized by a probability density function (pdf), which is used to find the probabilities of a random variable falling within a particular range of values. The pdf for a continuous random variable is a function that satisfies:
  • The function must be non-negative.
  • The total area under the curve of the function and above the x-axis is equal to one.

The mean or expected value of a continuous distribution is found by integrating the product of the variable and its pdf over the range of possible values. As with the discrete case, the variance of a continuous distribution is the expected value of the squared deviation of the variable from its mean, giving a sense of the 'spread' of the distribution.

For example, a continuous random variable with a range from 0 to 1 might have a pdf represented as \( f(x) = 6x(1-x) \), which implies that values near 0 or 1 are less likely than those near the midpoint.
Probability Density Function (PDF)
The probability density function (PDF) is at the heart of continuous distributions. It's important to understand that while a pdf can tell us about the relative likelihood of a random variable's outcomes, it does not give the probability of a single outcome. Instead, the area under the curve of the pdf over an interval gives the probability that the random variable falls within that interval.

For a function to qualify as a pdf, it must meet two criteria:
  • The function must integrate to one over the range of possible values, ensuring that the total probability of all possible outcomes is 100%.
  • The function must always be greater than or equal to zero, because probabilities cannot be negative.

To calculate the mean of a distribution with pdf \( f(x) \), we integrate \( x \) times the pdf over the possible values:
\[ E[X] = \int xf(x) dx \]
Variance is slightly more complex, requiring the calculation of the expected value of the squared variable, and then subtracting the square of the expected value of the variable:
\[ \text{Variance} = E[X^2] - (E[X])^2 \]
This process can be applied to any continuous probability distribution to determine its mean and variance, as long as the necessary integrals converge.

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