/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Five automobiles of the same typ... [FREE SOLUTION] | 91Ó°ÊÓ

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Five automobiles of the same type are to be driven on a 300-mile trip. The first two will use an economy brand of gasoline, and the other three will use a name brand. Let \(X_{1}, X_{2}\), \(X_{3}, X_{4}\), and \(X_{5}\) be the observed fuel efficiencies (mpg) for the five cars. Suppose these variables are independent and normally distributed with \(\mu_{1}=\mu_{2}=20, \mu_{3}=\mu_{4}=\mu_{5}=21\), and \(\sigma^{2}=4\) for the economy brand and \(3.5\) for the name brand. Define an rv \(Y\) by $$ Y=\frac{X_{1}+X_{2}}{2}-\frac{X_{3}+X_{4}+X_{5}}{3} $$ so that \(Y\) is a measure of the difference in efficiency between economy gas and name-brand gas. Compute \(P(0 \leq Y)\) and \(P(-1 \leq Y \leq 1) .\left[\right.\)

Short Answer

Expert verified
\(P(0 \leq Y) \approx 0.2877\); \(P(-1 \leq Y \leq 1) \approx 0.3686\).

Step by step solution

01

Identify Random Variable Distributions

First, recognize that each observed fuel efficiency, \(X_i\), follows a normal distribution. Specifically, \(X_1\) and \(X_2\) have a distribution of \(N(20, 4)\), while \(X_3, X_4, X_5\) have a distribution of \(N(21, 3.5)\).
02

Mean and Variance of Y

Compute the mean, \(\mu_Y\), of \(Y\):\[ \mu_Y = \frac{\mu_1 + \mu_2}{2} - \frac{\mu_3 + \mu_4 + \mu_5}{3} = \frac{20 + 20}{2} - \frac{21 + 21 + 21}{3} = 20 - 21 = -1 \]Compute the variance, \(\sigma_Y^2\), of \(Y\):\[ \sigma_Y^2 = \frac{\sigma_1^2 + \sigma_2^2}{4} + \frac{\sigma_3^2 + \sigma_4^2 + \sigma_5^2}{9} = \frac{4 + 4}{4} + \frac{3.5 + 3.5 + 3.5}{9} = 2 + \frac{10.5}{9} \approx 3.17 \]
03

Normal Distribution of Y

From the previous calculations, \(Y\) is normally distributed with \(Y \sim N(-1, 3.17)\). We can now transform \(Y\) into a standard normal variable \(Z\) by using \(Z = \frac{Y + 1}{\sqrt{3.17}}\).
04

Calculate P(0 ≤ Y)

Standardize the inequality 0 ≤ Y using \(Z\):\[ P(0 \leq Y) = P \left( \frac{0 + 1}{\sqrt{3.17}} \leq Z \right) = P \left( 0.561 \leq Z \right)\]. Using the standard normal distribution table or calculator, \(P(0 \leq Y) = 1 - P(Z < 0.561) \approx 1 - 0.7123 = 0.2877\).
05

Calculate P(-1 ≤ Y ≤ 1)

Standardize the inequality \(-1 \leq Y \leq 1\) using \(Z\):\[ P(-1 \leq Y \leq 1) = P\left(\frac{-1 + 1}{\sqrt{3.17}} \leq Z \leq \frac{1 + 1}{\sqrt{3.17}}\right) \]. Simplify to \(P(0 \leq Z \leq 1.121)\). Calculate the probabilities:\[ P(0 \leq Z \leq 1.121) = P(Z < 1.121) - P(Z < 0) \approx 0.8686 - 0.5 = 0.3686 \].

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Key Concepts

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

Random variable
A random variable is a fundamental concept in probability and statistics. It is essentially a variable that takes on different values based on the outcomes of a random phenomenon.

In the context of our exercise, the random variables are the fuel efficiencies of the five cars, denoted as \(X_1, X_2, X_3, X_4,\) and \(X_5\). Each encompasses a range of possible fuel efficiency values that these cars can exhibit during a 300-mile journey. Given that these variables are associated with the natural variation in fuel consumption, they are designed to account for randomness in measurements.

Key points about random variables:
  • Random variables can be either discrete or continuous. In this exercise, they are continuous, as fuel consumption can vary without fixed steps.
  • They often follow certain probability distributions, which dictate how likely specific outcomes are.
  • Random variables are used to make probabilistic statements about the outcomes of experiments or processes.
Understanding random variables helps in modeling and investigating various processes in scientific research and practical applications.
Normal distribution
The normal distribution is a cornerstone in statistics, known for its bell-shaped curve. It helps describe how many natural phenomena distribute themselves across a population without any specific pattern. It is symmetric around the mean, exhibiting the property that data within one standard deviation of the mean accounts for about 68% of the dataset.

In this exercise, the fuel efficiencies \(X_1\) and \(X_2\) follow a normal distribution with a mean (\(\mu\)) of 20 and variance (\(\sigma^2\)) of 4, making them \(N(20, 4)\). Similarly, \(X_3, X_4,\) and \(X_5\) follow \(N(21, 3.5)\).

Characteristics of the normal distribution:
  • Symmetric and peaking at the mean value.
  • Fully characterized by its mean and variance.
  • Allows for the computation of probabilities based on the area under the curve.
Utilizing normal distribution, statisticians can estimate probabilities of events and study the behavior of data sets.
Standard normal variable
A standard normal variable, often denoted as \(Z\), is essentially a normal distribution that has been modified to have a mean of 0 and a standard deviation of 1. This transformation is particularly helpful in simplifying the process of finding probabilities and comparing different normal distributions.

In our solution, to compute probabilities relating to \(Y\) (the difference in fuel efficiencies), we transformed \(Y\) into a standard normal variable \(Z\) using the formula:\[Z = \frac{Y + 1}{\sqrt{3.17}}\]This transformation allows us to use standard normal distribution tables or calculators to find probabilities.

Benefits of using a standard normal variable:
  • Simplifies the calculation of probabilities since you can use a universal Z-table.
  • Makes it easier to compare and interpret different normal distributions by standardizing them.
  • Reduces complexity of dealing with multiple normal distributions by turning them into a common framework.
  • Essential for performing z-tests in hypothesis testing.
Understanding and using the standard normal variable is crucial for anyone working with statistical data analysis.
Variance
Variance is a vital statistical measure that represents the extent to which numbers in a data set differ from the mean value. It essentially gives users insight into how spread out the values of a random variable are.

In the exercise, the variance helps us understand the variability in fuel efficiency among the cars. For the economy brand, a variance of 4 is given, whereas for the name brand, it is 3.5. The calculations further extend to finding the variance of \(Y\), highlighting how the difference in fuel efficiencies can vary.

Key points on variance:
  • Variance is always non-negative, as it is derived from squaring the deviations from the mean.
  • It is expressed in the units squared, e.g., mpg² in our context.
  • A higher variance indicates greater variability in data.
  • Variance plays a critical role in risk assessment, data analysis, and decision-making processes.
Calculating variance is a fundamental step in understanding the behavior of random variables across various applications.

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