Calculating σ for Woolen Scarves: Normal Distribution Help

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SUMMARY

The discussion focuses on calculating the standard deviation (σ) for the lengths of woolen scarves produced by a mill, specifically for a sample of 20 scarves. The total length of the scarves is 1428 cm, and the sum of the squares of the lengths is 102286 cm². To meet the requirement that 90% of scarves fall within 10 cm of the mean length, the calculated value of σ is 6.079. This is derived from the equation 10 = 1.6449σ, where 1.6449 corresponds to the z-score for a 90% confidence interval.

PREREQUISITES
  • Understanding of normal distribution and z-scores
  • Familiarity with basic statistics concepts such as mean and standard deviation
  • Knowledge of confidence intervals and their calculations
  • Ability to perform calculations involving sums and sums of squares
NEXT STEPS
  • Study the properties of the normal distribution and how to calculate z-scores
  • Learn about confidence intervals and their application in statistical analysis
  • Explore the concept of standard deviation and its significance in data analysis
  • Review statistical software tools for performing calculations, such as R or Python's SciPy library
USEFUL FOR

This discussion is beneficial for statisticians, data analysts, and anyone involved in quality control in manufacturing, particularly in textile production.

FeDeX_LaTeX
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A woollen mill produces scarves. The mill has several machines each operated by a different person. Jane has recently started working at the mill and the supervisor wishes to check the lengths of the scarves Jane is producing. A random sample of 20 scarves is taken and the length, x cm, of each scarf is recorded. The results are summarised as:

∑x = 1428, ∑x² = 102286

The mill's owners require that 90% of scarves should be within 10 cm of the mean length.

Find the value of σ that would satisfy this condition.

I considered

[tex]P(|\overline{X}| < 10) = 0.9[/tex]

but that didn't get me anywhere. And I tried confidence intervals with the t-distribution but that didn't work. Their answer is 6.079. How did they get it?

Their only line of working is "10 = 1.6449σ". I recognise that 1.6449 is the value of z for which P(Z > z) = 0.05, but I don't understand how they formed that equation.
 
Last edited:
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Never mind, got it...
 

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