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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

We need to find the standard deviation \(\sigma\) such that \(95\%\) of the temperature readings fall within \(\pm 0.1^\circ\) of the mean temperature \(\mu\). This involves understanding the properties of a normal distribution.
02

Properties of the Normal Distribution

A normal distribution is symmetric around its mean. For a normal distribution, approximately \(68.27\%\) of the measurements fall within one standard deviation of the mean, \(95.45\%\) within two, and \(99.73\%\) within three. To ensure that about \(95\%\) of readings fall within a certain range, we can use the concept of percentiles and z-scores.
03

Using Z-scores for 95% Confidence

Since we want \(95\%\) of the readings within \(\pm 0.1^\circ\), we need to find the z-score corresponding to \(95\%\), which is \(1.96\) (as it is \(2.5\%\) on each tail of the distribution). This translates to \(P(-1.96 < Z < 1.96) = 0.95\).
04

Setting Up the Inequality

The condition can be expressed as: \(-0.1 = 1.96 \times \sigma\) and \(0.1 = 1.96 \times \sigma\). Thus, we solve for \(\sigma\) using one of these equations to find the absolute value of \(\sigma\) because it affects both sides equally.
05

Solving for \(\sigma\)

Rearrange the equation \(0.1 = 1.96 \times \sigma\) to solve for \(\sigma\). By dividing both sides by \(1.96\), we have \(\sigma = \frac{0.1}{1.96}\).
06

Calculating the Standard Deviation

Now, perform the calculation: \(\sigma = \frac{0.1}{1.96} \approx 0.051\).
07

Conclusion

Therefore, the standard deviation \(\sigma\) must be approximately \(0.051\) degrees to ensure \(95\%\) of readings are within \(\pm 0.1^\circ\) of the actual temperature \(\mu\).

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In a normal distribution, most values cluster around the mean, creating a bell-shaped curve. The standard deviation tells us how tightly the data points are bunched around the mean. This is crucial for understanding how variable a set of data is.

When working with normal distributions, the standard deviation (\(\sigma\)) plays an essential role in predicting the likelihood of a data point falling within a certain range around the mean value. For instance:
  • About 68% of the data falls within one standard deviation (\(\pm \sigma\)) of the mean.
  • About 95% falls within two standard deviations (\(\pm 2\sigma\)).
  • Almost all (99.7%) is within three standard deviations (\(\pm 3\sigma\)).
In our example, to maintain 95% of the readings within \(\pm 0.1^\circ\) of the mean temperature \(\mu\), determining the correct standard deviation is key. We found it to be around 0.051, meaning the temperature readings should variate by approximately 0.051 degrees around the mean to satisfy the 95% condition.
Confidence Interval
A confidence interval gives a range of values that is likely to contain the true value of an unknown parameter. For normal distributions, these intervals are often centered around the mean and extend a distance on either side. The width of this interval depends on the standard deviation and how confident you wish to be in your interval.

When we say a confidence interval covers 95% of possible readings, we mean that if we took many samples, 95% of them would fall within this range around the mean. This percentage is known as the confidence level. To construct such an interval:
  • Identify the mean (\(\mu\)) of the distribution, which is the center of the interval.
  • Determine the corresponding z-score (in our problem, 1.96 is used for 95% confidence).
  • Calculate the margin of error by multiplying the z-score by the standard deviation (\(\sigma\)).
With these steps, one can compute that the symmetric interval ranging \(\pm 0.1^\circ\) around the mean should hold 95% of the data when the standard deviation is set accurately.
Z-score
A z-score measures how far a specific data point is from the mean, in terms of standard deviations. It provides a way to say how unusual or typical a data point is within the distribution:

For example, a z-score of 0 means the data point is exactly at the mean, whereas a z-score of 1 or -1 means it is one standard deviation above or below the mean, respectively. A high absolute z-score indicates that a point is far from the mean.

Z-scores are integral in finding confidence intervals because they help determine how wide the interval should be. In a normal distribution, precise z-scores can be used to set boundaries for a given confidence level. For our situation:
  • A z-score of 1.96 allows us to find the interval for 95% confidence.
  • We calculated the standard deviation required to keep readings within \(\pm 0.1^\circ\)
Understanding z-scores helps to precisely locate where most data will fall in a normal distribution, aiding in quality control and predictions.

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