Problem 92 - Applied Statistics and Probability for Engineers

Question

An article in Microelectronics Reliability ["Advanced Electronic Prognostics through System Telemetry and Pattern Recognition Methods" (2007, Vol.47(12), pp. \(1865-1873\) ) ] presented an example of electronic prognosis. The objective was to detect faults to decrease the system downtime and the number of unplanned repairs in high-reliability systems. Previous measurements of the power supply indicated that the signal is normally distributed with a mean of \(1.5 \mathrm{~V}\) and a standard deviation of \(0.02 \mathrm{~V}\). (a) Suppose that lower and upper limits of the predetermined specifications are \(1.45 \mathrm{~V}\) and \(1.55 \mathrm{~V},\) respectively. What is the probability that a signal is within these specifications? (b) What is the signal value that is exceeded with \(95 \%\) probability? (c) What is the probability that a signal value exceeds the mean by two or more standard deviations?

Step-by-Step Solution

Answer

(a) 0.9876; (b) 1.4671 V; (c) 0.0228.

Step 1: Understanding the Normal Distribution

The signal is normally distributed with a mean \( \mu = 1.5 \mathrm{~V} \) and a standard deviation \( \sigma = 0.02 \mathrm{~V} \). We will use the properties of the normal distribution to solve the problems part (a), (b), and (c).

Step 2: Calculating Probability within Given Specifications (a)

We need to calculate the probability that the signal is between 1.45 V and 1.55 V. This involves finding the probability that a normal random variable falls within these limits. We use the standard normal distribution (Z) which is given by \( Z = \frac{X - \mu}{\sigma} \).\\[ Z_{lower} = \frac{1.45 - 1.5}{0.02} = -2.5 \] \\[ Z_{upper} = \frac{1.55 - 1.5}{0.02} = 2.5 \] \Using standard normal distribution tables or software, we find \( P(Z < 2.5) \approx 0.9938 \) and \( P(Z < -2.5) \approx 0.0062 \).\The probability \( P(1.45 < X < 1.55) = P(Z < 2.5) - P(Z < -2.5) = 0.9938 - 0.0062 = 0.9876 \).

Step 3: Finding Signal Value Exceeded with 95% Probability (b)

We need to find the signal value that is exceeded with 95% probability, i.e., the 5th percentile of the distribution. The corresponding Z-score for the 5th percentile is \(-1.645\) from the Z-table. Using the Z-score formula: \[ X = \mu + Z \cdot \sigma = 1.5 + (-1.645) \cdot 0.02 = 1.4671 \mathrm{~V} \].

Step 4: Calculating Probability of Exceeding the Mean by Two Standard Deviations (c)

We need to calculate the probability that the signal exceeds the mean by at least two standard deviations, i.e., \( X > 1.5 + 2 \times 0.02 = 1.54 \mathrm{~V} \). First, find the Z-score:\[ Z = \frac{1.54 - 1.5}{0.02} = 2 \].Using the standard normal distribution table, \( P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \).

Key Concepts

Probability Calculation

Probability calculation is a fundamental concept in statistics. It involves determining the likelihood of certain outcomes or events occurring. When dealing with normally distributed data, this becomes crucial in applications like electronic prognosis, where you might want to understand the performance and reliability of a system.

To calculate probabilities with normal distributions, you often use the properties of the curve that represents the distribution. For a normally distributed variable with a mean (\( \mu \)) and a standard deviation (\( \sigma \)), probabilities can be calculated for values within a certain range.

• In exercise (a), we calculated the probability that the signal is between 1.45 V and 1.55 V. By converting these values into Z-scores and using standard normal tables, we could determine the likelihood of the signal falling between those limits.
• Probability calculations involving normal distribution often use Z-scores to transform the variable into a standard normal variable, making it easier to use statistical tables or software for finding probabilities.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. This distribution is crucial because it allows us to use Z-scores to convert any normal distribution into the standard form.

The Z-score is calculated by:\[ Z = \frac{X - \mu}{\sigma} \]
This helps in making comparisons or finding probabilities, as shown in step (b) and (c) of the original exercise. By converting the normal distribution to a standard one, using Z-scores, you can easily locate probabilities on the standard normal distribution table.

• In step (b), to find the signal value exceeded with 95% probability, we used the Z-score for the 5th percentile of the standard normal distribution.
• The standard normal distribution simplifies the process of finding probabilities for a wide range of values from other normal distributions by using standardized Z-scores.

Z-score

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It signifies how many standard deviations an element is from the mean. To calculate the Z-score, use the formula:\(Z = \frac{X - \mu}{\sigma}\)

Z-scores are extremely useful because they allow for the standardization of measurements from different normal distributions. You can use Z-scores to determine the probability of a value occurring within a normal distribution by referencing the standard normal distribution table.

• For example, in part (a) of the exercise, we calculated Z-scores to find the probability of the signal falling within a certain range.
• Z-scores also allow us to find percentile ranks, as illustrated in part (b) when determining the signal value exceeded at a particular probability level.
• In part (c), we used the Z-score to calculate the probability that the signal exceeds the mean by specific multiples of the standard deviation.