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Magazine Chapter 4: Problem 7
Suppose that \(f(x)=1.5 x^{2}\) for \(-1
Short Answer
Expert verified
(a) 0.5, (b) 0.4375, (c) 0.03125, (d) 0, (e) 0.96875, (f) x 鈮� 0.9655.
Step by step solution
01
Understand the function
The function given is a probability density function (pdf) of the form \(f(x) = 1.5x^2\) valid for \(-1 < x < 1\). This indicates a probability distribution we need to work with within this interval to solve different probability expressions.
02
Calculate P(0
To find \(P(0<X)\), we need to calculate the area under the curve \(f(x)\) from 0 to 1: \[ P(0<X) = \int_{0}^{1} 1.5x^2 \, dx \].Evaluate the integral:\[ \int 1.5x^2 \, dx = 1.5 \cdot \frac{x^3}{3} = 0.5x^3 \]. Thus, \[ P(0<X) = 0.5(1^3) - 0.5(0^3) = 0.5 \].
03
Calculate P(0.5
To find \(P(0.5<X)\), calculate the area under the curve \(f(x)\) from 0.5 to 1: \[ P(0.5<X) = \int_{0.5}^{1} 1.5x^2 \, dx \].Evaluate the integral:\[ 0.5x^3 \] from 0.5 to 1:\[ P(0.5<X) = 0.5(1^3) - 0.5(0.5^3) = 0.5 - 0.5(0.125) = 0.4375 \].
04
Calculate P(-0.5 鈮� X 鈮� 0.5)
Find the probability of \(-0.5 \leq X \leq 0.5\) using:\[ P(-0.5 \leq X \leq 0.5) = \int_{-0.5}^{0.5} 1.5x^2 \, dx \].Evaluate:\[ 0.5x^3 \] from -0.5 to 0.5:\[ P = \left(0.5(0.5^3) - 0.5(-0.5^3)\right) = (0.015625 - (-0.015625)) = 0.03125 \].
05
Calculate P(X
Since \(f(x)\) is defined only for \(-1 < x < 1\), and \(X<-2\) is outside this range, \[ P(X<-2) = 0 \].
06
Calculate P(X-0.5)
This involves two intervals: 1. From -1 to 0 2. From -0.5 to 1.However, since the whole interval from -1 to 1 is covered with these two subintervals but overlap occurs, we find the probability directly using the complete interval:\[ P(X<0 \text{ or } X>-0.5) = \int_{-1}^{1} 1.5x^2 \, dx - \int_{-0.5}^{0} 1.5x^2 \, dx \].Calculate:From -1 to 1:\[ 0.5x^3 \] from -1 to 1:\[ 0.5(1^3)-0.5(-1^3) = 1 \].Let \(I_1 = P(-0.5 < X < 0)\), evaluate\[ 0.5x^3 \] from -0.5 to 0:\[ P(-0.5-0.5) = 1-0.015625=0.96875 \].
07
Find x such that P(x
To find \( x \) such that \( P(x < X) = 0.05 \), set up the equation with cumulative probabilities:\[ \int_{x}^{1} 1.5t^2 \, dt = 0.05 \].Evaluating \(0.5t^3\) from x to 1:\[ 0.5(1)^3 - 0.5x^3 = 0.05 \].Simplify: \[ 0.5 - 0.5x^3 = 0.05 \]\[ 0.5x^3 = 0.45 \]\[ x^3 = 0.9 \]\[ x = \sqrt[3]{0.9} \approx 0.9655 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Probability Distributions
Continuous probability distributions are essential in understanding how probabilities are spread over a continuous range of values. Unlike discrete distributions that handle distinct data points, continuous distributions manage outcomes spread over a real number line, such as within an interval. This is why we need functions like the probability density function (pdf) to describe the likelihood of a random variable assuming a certain range of values. The area under the pdf curve within a particular interval gives us the probability for that range. For instance, with a function like \(f(x) = 1.5 x^2\), defined for \(-1 < x < 1\), we seek these areas to find probabilities associated with different events.
Integration in Probability
Integration is a powerful tool in probability, especially when dealing with continuous probability distributions. When we calculate probabilities for continuous random variables, we are essentially finding the area under the pdf curve over a given interval. This involves integrating the pdf over this range of interest.
For example, if we want to determine \(P(0 < X < 1)\) for the given function \(f(x) = 1.5 x^2\), we integrate from 0 to 1. The integral \(\int_{0}^{1} 1.5x^2 \, dx = 0.5 x^3\) shows the area under this curve, which yields the total probability of the event occurring in this interval. Integrals help us compute such areas precisely, giving complete distributions of probabilities across continuous spaces.
For example, if we want to determine \(P(0 < X < 1)\) for the given function \(f(x) = 1.5 x^2\), we integrate from 0 to 1. The integral \(\int_{0}^{1} 1.5x^2 \, dx = 0.5 x^3\) shows the area under this curve, which yields the total probability of the event occurring in this interval. Integrals help us compute such areas precisely, giving complete distributions of probabilities across continuous spaces.
Probability Calculations
Probability calculations using continuous probability distributions often involve analyzing specific intervals. To find the probability that a continuous random variable falls between two values, we can calculate the integral of the pdf over that interval.
For example, for our function \(f(x) = 1.5 x^2\), we calculated different probabilities:
For example, for our function \(f(x) = 1.5 x^2\), we calculated different probabilities:
- \(P(0
- \(P(0.5
- \(P(-0.5 \leq X \leq 0.5)\): The integral from -0.5 to 0.5.
- \(P(0.5
Cumulative Distribution Function
The cumulative distribution function (CDF) is a concept that outlines the probability that a random variable is less than or equal to a particular value. Essentially, it is the cumulative probability up to a point. It is derived by integrating the pdf from the lower bound of the distribution to the point of interest.
For example, in finding a value \(x\) where \(P(x
For example, in finding a value \(x\) where \(P(x
