Problem 91
Question
A signal in a communication channel is detected when the voltage is higher than 1.5 volts in absolute value. Assume that the voltage is normally distributed with a mean of \(0 .\) What is the standard deviation of voltage such that the probability of a false signal is \(0.005 ?\)
Step-by-Step Solution
Verified Answer
The standard deviation is approximately 0.534.
1Step 1: Understanding the Problem
We have a normal distribution of voltage with a mean of \(0\). A false signal is detected when the absolute value of the voltage exceeds 1.5 volts. We need to find the standard deviation \( \sigma \) such that the probability of this happening is \(0.005\).
2Step 2: Translating to Probability
The problem asks about the probability that the absolute value of the voltage is greater than 1.5, which is equivalent to \( P(|X| > 1.5) = 0.005 \). This can be expressed as \( P(X > 1.5) + P(X < -1.5) = 0.005 \.\) Due to symmetry, \( P(X > 1.5) = P(X < -1.5) \, \) giving us \( 2P(X > 1.5) = 0.005 \).
3Step 3: Finding One-sided Probability
Divide the probability by 2: \( P(X > 1.5) = 0.0025 \).
4Step 4: Using the Z-Score Formula
We use the Z-score formula \( Z = \frac{x - \mu}{\sigma} \). Here, \( x = 1.5, \; \mu = 0 \) (mean), and we need to find \( \sigma \). First, find \( Z \) for a probability of \( 0.0025 \).
5Step 5: Finding the Z-Score from the Z-Table
Look up the probability \( 0.0025 \) in the standard normal distribution table (Z-table) and find the corresponding Z-score. This translates to \( Z \approx 2.807 \) or \( Z \approx -2.807 \).
6Step 6: Solving for Standard Deviation
Use \( Z = \frac{x - \mu}{\sigma} \), which gives \( 2.807 = \frac{1.5 - 0}{\sigma} \). Solve for \( \sigma \): \( \sigma = \frac{1.5}{2.807} \). Calculating this gives \( \sigma \approx 0.534\).
Key Concepts
Standard DeviationZ-scoreProbabilityFalse Signal Detection
Standard Deviation
Standard deviation is a measure of how spread out the values in a data set are from the mean. In a normal distribution, most of your data points will fall within a certain distance from the mean, which is set by the standard deviation.
It informs us about the variation or dispersion of a set of values.
In this exercise, we needed to determine the standard deviation of voltage that leads to a small probability of detecting a false signal. This false signal detection corresponds to the voltage exceeding a certain threshold (in this case, 1.5 volts) because of random noise rather than an actual signal.
By calculating the standard deviation, we ensure that we correctly understand the likelihood of going past this threshold under normal circumstances.
It informs us about the variation or dispersion of a set of values.
In this exercise, we needed to determine the standard deviation of voltage that leads to a small probability of detecting a false signal. This false signal detection corresponds to the voltage exceeding a certain threshold (in this case, 1.5 volts) because of random noise rather than an actual signal.
By calculating the standard deviation, we ensure that we correctly understand the likelihood of going past this threshold under normal circumstances.
Z-score
The Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. It's expressed in terms of standard deviations from the mean.
Essentially, it tells you how many standard deviations you are away from the mean.
If the Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score can be positive or negative, indicating whether the data point is above or below the mean, respectively.
In our problem, with a mean of 0, the voltage of 1.5 volts was converted into a Z-score using the formula:
From the Z-table, we found the Z-score that corresponds to our desired probability of detecting a false signal, which was critical in solving for the standard deviation.
Essentially, it tells you how many standard deviations you are away from the mean.
If the Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score can be positive or negative, indicating whether the data point is above or below the mean, respectively.
In our problem, with a mean of 0, the voltage of 1.5 volts was converted into a Z-score using the formula:
- \( Z = \frac{x - \mu}{\sigma} \)
From the Z-table, we found the Z-score that corresponds to our desired probability of detecting a false signal, which was critical in solving for the standard deviation.
Probability
Probability is a measure that quantifies the likelihood of certain events occurring.
For normally distributed data, probability helps in understanding and predicting how often certain events will fall within a particular range.
In this scenario, we looked at the probability of the voltage exceeding the 1.5 volts threshold. The exercise specified that this probability is 0.005, representing a 0.5% chance of false signal detection. This probability translates directly into our exercise as how often we detect voltage as being significant (a false signal) when it's merely random noise.
This concept was a guide for us to find the correct threshold and to make sure that our system remained accurate and reliable over time.
For normally distributed data, probability helps in understanding and predicting how often certain events will fall within a particular range.
In this scenario, we looked at the probability of the voltage exceeding the 1.5 volts threshold. The exercise specified that this probability is 0.005, representing a 0.5% chance of false signal detection. This probability translates directly into our exercise as how often we detect voltage as being significant (a false signal) when it's merely random noise.
This concept was a guide for us to find the correct threshold and to make sure that our system remained accurate and reliable over time.
False Signal Detection
False signal detection refers to the incorrect identification of a signal when there is actually none. In statistical terms, this can often be connected to Type I errors, where something is detected by a system due to noise or error rather than a true signal.
In our normal distribution scenario, a false signal was registered when the voltage exceeded 1.5 volts. Because of the random distribution of voltage, occasionally we exceeded this threshold, even though no true signal was present.
By setting the probability (at 0.005 in this case) and calculating the ideal standard deviation, we were able to understand and minimize false detection and ensure the system is responding appropriately only when actual signals occur. This is crucial for maintaining the integrity and reliability of communication systems.
In our normal distribution scenario, a false signal was registered when the voltage exceeded 1.5 volts. Because of the random distribution of voltage, occasionally we exceeded this threshold, even though no true signal was present.
By setting the probability (at 0.005 in this case) and calculating the ideal standard deviation, we were able to understand and minimize false detection and ensure the system is responding appropriately only when actual signals occur. This is crucial for maintaining the integrity and reliability of communication systems.
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