Problem 7 - Applied Statistics and Probability for Engineers

Question

Suppose that \(f(x)=1.5 x^{2}\) for \(-1-0.5)\) (f) \(x\) such that \(P(x

Step-by-Step Solution

Answer

(a) 0.5, (b) 0.4375, (c) 0.03125, (d) 0, (e) 0.96875, (f) x ≈ 0.9655.

Step 1: 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.

Step 2: Calculate P(0<X)

To find \(P(0

Step 3: Calculate P(0.5<X)

To find \(P(0.5

Step 4: Calculate P(-0.5 ≤ X ≤ 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:\[ P = \left(0.5(0.5^3) - 0.5(-0.5^3)\right) = (0.015625 - (-0.015625)) = 0.03125 \].

Step 5: Calculate P(X<-2)

Since \(f(x)\) is defined only for \(-1 < x < 1\), and \(X<-2\) is outside this range, \[ P(X<-2) = 0 \].

Step 6: 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:\[ P(-0.5-0.5) = 1-0.015625=0.96875 \].

Step 7: Find x such that P(x<X)=0.05

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: \[ 0.5 - 0.5x^3 = 0.05 \]\[ 0.5x^3 = 0.45 \]\[ x^3 = 0.9 \]\[ x = \sqrt[3]{0.9} \approx 0.9655 \].

Key Concepts

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.

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:

• \(P(0
• \(P(0.5
• \(P(-0.5 \leq X \leq 0.5)\): The integral from -0.5 to 0.5.

These calculations allow us to see the likelihood of the variable within specific sections of its range. Probability calculations are a crucial part of interpreting real-world outcomes in terms of given theoretical models.

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

Previous

Problem 4

Next

Problem 8

Other exercises in this chapter

Problem 4

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function \(f(x)=2 / x^{3}\) for \(x>1 .\) Determine the fol

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Problem 8

The probability density function of the time to failure of an electronic component in a copier (in hours) is \(f(x)=\) \(e^{-x / 1000 / 1000}\) for \(x>0 .\) De

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Problem 10

The probability density function of the length of a cutting blade is \(f(x)=1.25\) for \(74.675.2)\) (c) If the specifications for this process are from 74.7 to

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Problem 11

The probability density function of the length of a metal rod is \(f(x)=2\) for \(2.3

View solution

Key Concepts

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.

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:

• \(P(0
• \(P(0.5
• \(P(-0.5 \leq X \leq 0.5)\): The integral from -0.5 to 0.5.

These calculations allow us to see the likelihood of the variable within specific sections of its range. Probability calculations are a crucial part of interpreting real-world outcomes in terms of given theoretical models.

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

Previous

Problem 4

Next

Problem 8

Other exercises in this chapter

Problem 4

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function \(f(x)=2 / x^{3}\) for \(x>1 .\) Determine the fol

View solution

Problem 8

The probability density function of the time to failure of an electronic component in a copier (in hours) is \(f(x)=\) \(e^{-x / 1000 / 1000}\) for \(x>0 .\) De

View solution

Problem 10

The probability density function of the length of a cutting blade is \(f(x)=1.25\) for \(74.675.2)\) (c) If the specifications for this process are from 74.7 to

View solution

Problem 11

The probability density function of the length of a metal rod is \(f(x)=2\) for \(2.3

View solution