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Chapter 4: Problem 42

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean \(\mu\), the actual temperature of the medium, and standard deviation \(\sigma\). What would the value of \(\sigma\) have to be to ensure that \(95 \%\) of all readings are within \(.1^{\circ}\) of \(\mu\) ?

Short Answer

Expert verified
\(\sigma \approx 0.051\).

Step by step solution

01

Understanding the Problem

First, we need to find the standard deviation \(\sigma\) given that the readings should be within \(0.1^\circ\) of the mean \(\mu\) with a 95% confidence interval.
02

Concept of Normal Distribution

The readings from the thermocouple follow a normal distribution. We want 95% of the data to fall within a certain range around the mean. This corresponds to the interval \([\mu - 0.1, \mu + 0.1]\) for the normal distribution.
03

Using Properties of the Normal Distribution

For a normal distribution, approximately 95% of data falls within \(\pm 1.96\sigma\) of the mean. Here, 95% of readings should fall within \(0.1^\circ\), so we set up the equation \(1.96\sigma = 0.1\).
04

Solving for \(\sigma\)

Now, solve the equation for \(\sigma\): \(\sigma = \frac{0.1}{1.96}\).
05

Calculate \(\sigma\)

Perform the division: \(\sigma = \frac{0.1}{1.96} \approx 0.051\). This is the value of \(\sigma\) required to meet the condition.

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Key Concepts

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

Understanding Standard Deviation
The concept of standard deviation is fundamental to the analysis of statistical data. In simple terms, it measures how spread out the numbers in a dataset are. To imagine it more vividly, you can think of standard deviation as the "average distance" that a data point is from the mean or average number.

Here's why it's important:
  • A small standard deviation means that most of the numbers are close to the mean. Imagine scores in a class, if most students scored around the same mark, the standard deviation is small.
  • A large standard deviation, on the other hand, means the numbers are more spread out. This would be like a class where some students scored very high and others very low.
In our thermocouple problem, we needed to find a specific standard deviation, \( \sigma \), that ensures 95% of temperature readings fall within 0.1 degrees of the mean reading. This involves a deeper understanding of how normal distribution connects to standard deviation.
Confidence Interval Explained
A confidence interval provides a range of values that you can be certain, or confident, contains the true mean of the population. It's like saying "we鈥檙e 95% sure that the true value lies within this range."

Confidence intervals are constructed using a key feature of normal distribution. Specifically, if you want a 95% confidence interval, you will observe data to fall within roughly 1.96 standard deviations from the mean on both sides.
  • This translates to saying in a normally distributed dataset, 95% of it will be within \( \pm 1.96\sigma \) from the mean.
  • In our exercise, the mean temperature readings are intended to fall within the narrow band of 0.1 degrees, which determines the calculation \( 1.96\sigma = 0.1 \).
Finding the right \( \sigma \) here means shaping an interval small enough to capture the precise needed range around our mean but wide enough to allow for standard reading variability.
Understanding the Mean in Probability Distributions
The mean is the average of all data points in a dataset. It鈥檚 calculated by summing all values and dividing by the number of values. In normally distributed data, the mean is also the peak of the bell-shaped curve.

Here's why the mean is so fundamental:
  • The mean provides a central value that serves as a reference for assessing other values in the distribution.
  • A normal distribution is symmetrical about its mean. This means that anything you calculate using this distribution, like standard deviation or confidence intervals, fundamentally revolves around this number.
In the context of our exercise, the mean \( \mu \) represents the true temperature of the medium. We use this central value to gauge how dispersed readings are with the help of standard deviation. The goal is to maintain readings that scatter close to this mean, which enables reliability in measurements by ensuring minor fluctuations stay within an acceptable range of the mean.

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Most popular questions from this chapter

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