91Ó°ÊÓ

Let \(X_{1}, X_{2}\), and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=15\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 200\right)\). What is \(P\left(150 \leq X_{1}+X_{2}+\right.\) \(\left.X_{3} \leq 200\right) ?\) b. Using the \(\mu_{i}\) s and \(\sigma_{i}\) s given in part (a), calculate \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i} \mathrm{~s}\) and \(\sigma_{i} \mathrm{~s}\) given in part (a), calculate \(P\left(-10 \leq X_{1}-.5 X_{2}-.5 X_{3} \leq 5\right)\) d. If \(\mu_{1}=40, \mu_{2}=50, \mu_{3}=60, \sigma_{1}^{2}=10, \sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 160\right)\) and \(P\left(X_{1}+\right.\) \(\left.X_{2} \geq 2 X_{3}\right)\)

Step by step solution

01

Understand Random Variable Sum

Given three independent normal random variables \(X_1, X_2,\) and \(X_3\), the sum \(X_1 + X_2 + X_3\) is also a normal random variable. The expected value of the sum is \(\mu_1 + \mu_2 + \mu_3\) and the variance is \(\sigma_1^2 + \sigma_2^2 + \sigma_3^2\). For part (a), we have: \(\mu_1 = \mu_2 = \mu_3 = 60\) and \(\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 15\). Hence, \(\mu = 60 + 60 + 60 = 180\) and \(\sigma^2 = 15 + 15 + 15 = 45\).

02

Calculate Probability for Part a

For part (a): including \(P(X_1 + X_2 + X_3 \leq 200)\), determine \[ Z = \frac{(X_1 + X_2 + X_3) - 180}{\sqrt{45}} \]. Then calculate \[ Z = \frac{200 - 180}{\sqrt{45}} = \frac{20}{\sqrt{45}} \approx 2.98 \]. Use a standard normal table to find \(P(Z \leq 2.98)\). Also calculate \(P(150 \leq X_1 + X_2 + X_3 \leq 200)\) similarly with \(Z = \frac{150 - 180}{\sqrt{45}} = -4.47\), and find \(P(-4.47 \leq Z \leq 2.98)\).

03

Calculate Sample Mean Probability for Part b

The sample mean \(\bar{X} = \frac{X_1 + X_2 + X_3}{3}\) has mean \(\mu = 60\) and variance \(\sigma^2 = \frac{15}{3} = 5\). For \(P(55 \leq \bar{X})\), use \[ Z = \frac{55 - 60}{\sqrt{5}} \approx -2.24 \]. For \(P(58 \leq \bar{X} \leq 62)\), use \[ Z_{58} = \frac{58 - 60}{\sqrt{5}}, Z_{62} = \frac{62 - 60}{\sqrt{5}}\]. Then calculate these probabilities using the standard normal distribution.

04

Calculate Linear Combination for Part c

Linear combination: \(Y = X_1 - 0.5X_2 - 0.5X_3\) has expectation \(\mu_Y = 60 - 0.5 \times 60 - 0.5 \times 60 = 0\) and variance \(\sigma_Y^2 = 15 + 0.25 \times 15 + 0.25 \times 15 = 30\). For \(P(-10 \leq Y \leq 5)\), calculate \[ Z_{-10} = \frac{-10 - 0}{\sqrt{30}}, Z_5 = \frac{5 - 0}{\sqrt{30}} \].

05

Non-Symmetric Case in Part d

For part (d): Expected value and variance of \(X_1 + X_2 + X_3\) when \(\mu_1 = 40, \mu_2 = 50, \mu_3 = 60\) are \(150\) and variance \(36\). Hence, \[ Z = \frac{160 - 150}{\sqrt{36}} = 1.67 \] to find \(P(X_1 + X_2 + X_3 \leq 160)\). Also calculate \(P(X_1 + X_2 \geq 2X_3)\) with means and variance.

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Key Concepts

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

Normal Distribution

The normal distribution is a critical concept in probability and statistics. It describes how the values of a variable are distributed. Imagine a symmetrical bell-shaped curve where most of the data points cluster around a central mean value, spreading symmetrically towards the ends, or 'tails.' This is the normal distribution, often referred to as a Gaussian distribution.
A key property is that the mean, median, and mode are all located at the center of the distribution. Even though real-world data cannot strictly adhere to this exact form, the normal distribution is a powerful tool in statistics. It's frequently used in hypothesis testing and is the foundation of many statistical methods. When dealing with independent random variables, like times for repair tasks, if each is normally distributed, their combined sum is also normally distributed.

Random Variables

Random variables are variables that take on random values. In statistical terms, a random variable represents outcomes of a probabilistic event. These outcomes can be numbers, categories, or any other quantifiable data. There are two types: discrete and continuous.
In the context of our problem, random variables like \(X_{1}, X_{2}, \text{ and } X_{3}\) represent times taken for tasks. They are continuous because they can take any real value within a range. Each random variable has an expected value or mean (\(\mu\)) and variance (\(\sigma^2\)). These values help in understanding the average behavior and spread of the random variable, respectively.
When dealing with random variables in a normal distribution, you can predict how data behaves, which is crucial for calculating probabilities.

Probability Calculation

Probability calculation involves finding the likelihood that a particular event will occur. In mathematics, probability values lie between 0 (impossible event) and 1 (certain event). For random variables that follow a normal distribution, specific techniques are used to calculate probabilities.
To find the probability of a sum of independent normal variables, like \(X_1 + X_2 + X_3\), we use the properties of the normal distribution. First, determine the expected mean \((\mu_1 + \mu_2 + \mu_3)\) and variance \((\sigma_1^2 + \sigma_2^2 + \sigma_3^2)\). Then convert the problem into a standard form using \(Z\)-scores. \(Z\)-scores transform data to a common scale, facilitating the use of standard normal distribution tables or software for probability determination.
This method is powerful across statistical analyses for estimating the likelihood of events.

Linear Combinations

A linear combination involves constructing a new variable by multiplying each variable by a constant and summing them. It is a fundamental concept in statistics, widely used in regression analysis and other statistical methods.
For example, in the exercise, a linear combination is formed by combining the repair times \((X_1, X_2, X_3)\) using different weights \(Y = X_1 - 0.5X_2 - 0.5X_3\). The mean of this linear combination can be found by taking the weighted sum of the means of the individual variables. Similarly, its variance is calculated using the variances and covariances of those variables, considering their weights.
Understanding linear combinations is essential in analyzing complex data structures and simplifying calculations in probability and statistics. They allow for a nuanced analysis of data by adjusting contributions of different components, resulting in meaningful interpretations.