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Chapter 1: Problem 3

For each of the following distributions, compute \(P(\mu-2 \sigma

Short Answer

Expert verified
The resulting probability \(P(\mu-2\sigma<X<\mu+2\sigma)\) for the first function (a) is 1, while calculating the actual value for the second function (b) requires summing the probabilities in the range, which depends on the specific values of \(\mu\) and \(\sigma\).

Step by step solution

01

Identify the Distribution of (a)

The distribution function \(f(x)=6x(1-x)\) is a continuous random variable, i.e. it represents the probability density function (PDF) of a Beta Distribution. Beta Distribution is defined in the range 0 < x < 1.
02

Apply the Property of Beta Distribution for (a)

The Beta Distribution has the property that its standard deviation \(\sigma\) is always less than half the mean \(\mu\). Therefore, \(\mu-2\sigma > 0\) and \(\mu+2\sigma < 1\). So \(P(\mu-2\sigma<X<\mu+2\sigma) = 1 \) for the Beta Distribution.
03

Identify the Distribution of (b)

The function \(p(x)=(1/2)^x\), where x = 1,2,3,... represents a discrete random variable. This is the probability mass function (PMF) of a Geometric Distribution.
04

Apply the Property of Geometric Distribution for (b)

In the case of a Geometric Distribution, the mean \(\mu = 1/p\), and the standard deviation \(\sigma = \sqrt{(1-p)/p^2}\) where p = 1/2. Calculation of \(P(\mu-2\sigma<X<\mu+2\sigma)\) will need the sum of probabilities for values of x within the range \([\mu-2\sigma, \mu+2\sigma]\). Here, \(P(\mu-2\sigma<X<\mu+2\sigma)\) can be calculated by summing the probabilities for which x falls in this range.
05

Calculation of (b)

Substitute values for \(\mu\) and \(\sigma\), calculate the range \([\mu-2\sigma, \mu+2\sigma]\). Then, add up probabilities of the calculated range. This can be done using the sum formula for geometric sequences.

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