/*! 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 55 A balanced scale You have two sc... [FREE SOLUTION] | 91Ó°ÊÓ

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A balanced scale You have two scales for measuring weights in a chemistry lab. Both scales give answers that vary a bit in repeated weighings of the same item. If the true weight of a compound is 2.00 grams (g), the first scale produces readings \(X\) that have mean 2.000 g and standard deviation 0.002 g. The second scale's readings Y have mean 2.001 \(\mathrm{g}\) and standard deviation 0.001 \(\mathrm{g} .\) The readings \(X\) and \(\mathrm{Y}\) are independent. Find the mean and standard deviation of the difference \(Y-\mathrm{X}\) between the readings. Interpret each value in context.

Short Answer

Expert verified
The mean of \( Y - X \) is 0.001 g, and the standard deviation is approximately 0.00224 g, indicating that the second scale reads slightly higher on average with moderate variability.

Step by step solution

01

Understanding the Problem

We need to find the mean and standard deviation of the difference between the readings of two independent scales for a compound that has a true weight of 2.00 grams.
02

Calculate the Mean of the Difference

The mean of the difference of two independent random variables is the difference between their means. For the scales, this is:\[\text{Mean of } Y - X = E(Y) - E(X) = 2.001 - 2.000 = 0.001 \, \text{g}\]
03

Calculate the Variance of the Difference

For independent random variables, the variance of their difference is the sum of their variances. Given variances \( V(X) = 0.002^2 = 0.000004 \, \text{g}^2 \) and \( V(Y) = 0.001^2 = 0.000001 \, \text{g}^2 \), the variance is:\[V(Y - X) = 0.000001 + 0.000004 = 0.000005 \, \text{g}^2\]
04

Calculate the Standard Deviation of the Difference

The standard deviation is the square root of the variance:\[\text{SD}(Y - X) = \sqrt{0.000005} \approx 0.00224 \, \text{g}\]
05

Interpret the Results

The mean of 0.001 g indicates that, on average, the second scale reads 0.001 g higher than the first scale when measuring the same object. The standard deviation of approximately 0.00224 g means that the variability in the difference of the two scales' readings tends to be about 0.00224 g.

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Key Concepts

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

Random Variables
In statistics, a random variable is a numerical description of the outcome of a random process or experiment. Random variables can be either discrete or continuous. Discrete random variables have specific, distinct values, like the outcome of rolling a die. Continuous random variables, on the other hand, can take any value within a certain range; for instance, the weight of an object measured with a scale.

In the context of the exercise, the readings from the scales, represented by the variables \( X \) and \( Y \), are examples of continuous random variables. Each measurement by the scales is a potential outcome of the weighing process, which might vary each time due to small measurement inconsistencies. The idea is to understand how the variables \( X \) and \( Y \) behave, considering they fluctuate slightly around their given mean values.
Mean and Standard Deviation
The mean and standard deviation are essential measures in statistics that summarize characteristics of data distributions. The mean, often called the expected value, provides a central value for a random variable. It's calculated by adding all possible values, each multiplied by its probability, and then summing these values. In simpler terms, it's like the average value that a random variable takes.

The standard deviation, on the other hand, shows us how much values in a dataset deviate from the mean, giving insights into the data's spread. It is the square root of the variance and is denoted as \( \sigma \).

In our problem, the mean readings for scales \( X \) and \( Y \) are 2.000 g and 2.001 g, respectively. The standard deviations are 0.002 g and 0.001 g. By calculating the mean and standard deviation of the difference \( Y - X \), we can understand these variations better, helping to ensure more precise measurements.
Independent Events
Independent events in probability are those whose outcomes do not affect each other. When dealing with random variables, independence means that the occurrence of one event does not provide any information about the other.

For the two scales in the lab, the readings \( X \) and \( Y \) are considered independent. This means that the measurement of an item on scale one does not influence the outcome on the second scale. This property is crucial as it simplifies the calculation of variance for the difference \( Y - X \).

When random variables are independent, the variance of their difference (or sum) becomes the sum of their individual variances. This simplification plays an essential role in the calculations within the exercise, showing how independent properties make statistical treatments manageable.
Variance
Variance is a statistical measure that tells us how much a set of values (e.g., readings from a scale) are spread out. It's calculated as the average of the squared differences from the mean. Variance is usually denoted by \( V \) or \( \sigma^2 \).

The purpose of calculating variance is to quantify the degree of variation or dispersion in a set of data points. The higher the variance, the greater the spread between numbers in the dataset. This concept becomes particularly useful when assessing the reliability and accuracy of measurements.

In our scenario, the variance for the first scale \( X \) is 0.000004 \( \, \text{g}^2 \) and for the second scale \( Y \) is 0.000001 \( \, \text{g}^2 \). Given that \( X \) and \( Y \) are independent, the variance of their difference \( Y - X \) is simply the sum of these variances, leading to a total variance of 0.000005 \( \, \text{g}^2 \). Understanding variance helps us to grasp just how much the measurements deviate from their expected means.

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