Statistics: sample mean of normal distribution

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SUMMARY

The discussion centers on calculating the fraction of shafts that conform to design specifications for an optical storage drive, where the diameter is normally distributed as N(μ,σ²). The specified diameter is 0.2500 ± 0.0015 in, with a population mean (μ) of 0.2508 in and a standard deviation (σ) of 0.0005 in. The correct approach involves finding the probability that the shaft diameter falls between 0.2485 in and 0.2515 in, confirming that the specified diameter is the target mean, not the sample mean.

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Knowledge of statistical terminology, including population mean and standard deviation
  • Familiarity with probability calculations for continuous random variables
  • Ability to use statistical tools or software for calculations (e.g., Python, R)
NEXT STEPS
  • Learn how to calculate probabilities using the normal distribution in Python with libraries like SciPy
  • Study the Central Limit Theorem and its implications for sample means
  • Explore confidence intervals and their applications in quality control
  • Investigate hypothesis testing related to population means and variances
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musicmar
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Homework Statement


The diameter of a shaft in an optical storage drive is normally distributed N(μ,σ2). The drive specifies that the shaft be 0.2500 ± 0.0015 in. Suppose μ= 0.2508 in and σ = 0.0005 in. What fraction of shafts conform to the design specifications?


The Attempt at a Solution


I am confused with the terminology of this problem. Is 0.2500 the population mean and 0.2508 is the sample mean?
Or is this completely unrelated, and I have to find the probability that a length be between 0.2485 and 0.2515 in?
Some help in getting started would be appreciated.
Thank you!
 
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musicmar said:

Homework Statement


The diameter of a shaft in an optical storage drive is normally distributed N(μ,σ2). The drive specifies that the shaft be 0.2500 ± 0.0015 in. Suppose μ= 0.2508 in and σ = 0.0005 in. What fraction of shafts conform to the design specifications?

The Attempt at a Solution


I am confused with the terminology of this problem. Is 0.2500 the population mean and 0.2508 is the sample mean?
Or is this completely unrelated, and I have to find the probability that a length be between 0.2485 and 0.2515 in?
Some help in getting started would be appreciated.
Thank you!

It is crystal clear from the wording that your second interpretation is the correct one. (When I say crystal clear I mean: don't read more into the problem than what it says.)

RGV
 
Last edited:

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