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The life of a semiconductor laser at a constant power is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. a. What is the probability that a laser fails before 5000 hours? b. What is the life in hours that \(95 \%\) of the lasers exceed? c. If three lasers are used in a product and they are assumed to fail independently, what is the probability that all three are still operating after 7000 hours?

Short Answer

Expert verified
a. Probability is 0.0004. b. 6013 hours. c. Probability is 0.125.

Step by step solution

01

Define the Parameters

The life of the semiconductor laser is normally distributed with a mean \( \mu = 7000 \) hours and a standard deviation \( \sigma = 600 \) hours.
02

Calculate Probability for Part (a)

For part (a), we need to find the probability that a laser fails before 5000 hours. First, find the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{5000 - 7000}{600} = -3.33 \] Using a standard normal distribution table or calculator, find the probability corresponding to \( Z = -3.33 \), which is approximately 0.0004.
03

Find Value for Part (b)

For part (b), we need to find the life in hours that 95% of the lasers exceed. This means finding the value for the 5th percentile of the distribution. The Z-score corresponding to 5% in the lower tail is approximately \( Z = -1.645 \). Use the Z-score formula: \[ X = Z \cdot \sigma + \mu = -1.645 \cdot 600 + 7000 = 6013 \text{ hours} \]
04

Calculate Probability for Part (c)

For part (c), find the probability that a single laser operates beyond 7000 hours. The Z-score is: \[ Z = \frac{7000 - 7000}{600} = 0 \] The probability that one laser exceeds 7000 hours is 0.5. Now, calculate the probability that all three lasers exceed 7000 hours, assuming independence: \[ P(\text{all three}) = P(\text{one})^3 = 0.5^3 = 0.125 \]

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in probability and statistics. It's a type of continuous probability distribution that is symmetrical and bell-shaped, often described as the "normal curve." In this distribution:
  • The mean is the center of the distribution.
  • The spread or width of the curve is determined by the standard deviation.
  • Approximately 68% of data falls within one standard deviation of the mean, about 95% within two standard deviations, and nearly 99.7% within three standard deviations.
A key property of the normal distribution is how it handles traits like life expectancy, IQ scores, and measurement errors, which often naturally follow this pattern. In our exercise, the life of a semiconductor laser is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. Understanding this helps to determine probabilities of events measured against this typical distribution.
Z-score
A Z-score is a statistical measure that describes a data point's position within a normal distribution. It shows how many standard deviations a point is from the mean. The formula to calculate the Z-score is:\[Z = \frac{X - \mu}{\sigma}\]where:
  • \(X\) is the value in question.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
In practical terms, a negative Z-score indicates a value below the mean, while a positive Z-score indicates a value above the mean. For instance, in our exercise, the Z-score for a laser failing before 5000 hours is -3.33, which is quite low, indicating this is a rare event in the context of the distribution.
Independent Events
Independent events are those whose outcomes do not affect each other. In probability, if two events A and B are independent, the occurrence of A does not influence the probability of B occurring, and vice versa. Mathematically, this is expressed as:\[P(A \text{ and } B) = P(A) \times P(B)\]In the given problem, we're told that the failure of each laser is independent of the others. This means the likelihood that all three lasers fail is the product of their individual probabilities of failure. Specifically, to calculate the probability that all three lasers function beyond 7000 hours, we assume independence and compute:\[P(\text{all three operational}) = P(\text{one operational})^3\]Understanding this concept is crucial when dealing with multiple trials or components in probability calculations.
Probability Calculation
Probability calculations allow us to determine how likely an event is to occur. For the semiconductor lasers, we used several types of probability calculations.
For example, the probability of a laser failing before 5000 hours involves finding the cumulative probability up to that point in a normal distribution.
When seeking the time that 95% of lasers will exceed, we used percentile calculations on the normal curve, focusing on what's known as the lower tail to find the 5th percentile. Probability calculations often involve steps like using Z-scores to determine positions on a standard normal distribution table, multiplier effects for independent events, or even direct computation when probabilities are straightforward.
  • Mutually exclusive events require simple addition of probabilities if not overlapping.
  • Continuous distributions use density functions derived from the distribution shape, like the bell curve of a normal distribution.
Regular application of these principles lays the foundation for more complex analyses in statistical studies.

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