Uses of mean, standard deviation and Z tables

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SUMMARY

The discussion centers on calculating the guaranteed lifetime for light-emitting diodes (LEDs) based on a sample mean lifetime of 2048 hours and a standard deviation of 40 hours. To ensure that only 2% of the bulbs fail to meet this guarantee, the Z-score corresponding to the 98th percentile of the standard normal distribution is utilized. The calculation involves transforming the normal distribution parameters to the standard normal distribution using the formula X = μ + Zσ, where μ is the mean, σ is the standard deviation, and Z is the Z-score for 0.02 failure probability, which is approximately 2.05. The final guaranteed lifetime should be calculated as 2048 + (2.05 * 40).

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Familiarity with Z-scores and Z-tables
  • Basic statistical concepts such as mean and standard deviation
  • Ability to perform calculations involving transformations of distributions
NEXT STEPS
  • Study the properties of normal distribution in statistics
  • Learn how to use Z-tables for various confidence levels
  • Practice calculating guaranteed lifetimes using different mean and standard deviation values
  • Explore applications of statistical methods in quality control and product reliability
USEFUL FOR

Statisticians, quality control engineers, product managers, and anyone involved in reliability testing and warranty analysis for consumer products.

sazzlefrazzle
Hey i was wondering if anyone could help me with this question at all? I don't have a clue about the calculations and would be grateful if it could be explained to me! I'm really struggling! Thanks

Some light emitting diodes are tested and a sample is found to have a mean lifetime of 2048 hours and a standard deviation of 40 hours. The manufacturer wants to give a guaranteed lifetime for the LED’s on the box so that only 2% of the bulbs will be returned for not lasting for the guaranteed lifetime. Using Z table standard normal distribution show your calculation giving the lifetime he should put on the box.
 
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The question is asking you to find the lifetime X such that the probability of failure P(F) is less then or equal to .02.
 
Do you know how to change from a normal distribution with mean \mu and standard deviation \sigma to the standard normal distribution? If not, why in the world are you doing a problem like this?
 

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