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Suppose a manufacturer makes disposable peppercorn grinders. The number of peppercorns in the grinders is normally distributed with a mean of 322 peppercorns and a standard deviation of 5.3 peppercorns. The manufacturer will only sell peppercorn grinders with a z-score between –0.9 and 0.9. What are the least and most peppercorns a grinder can contain?

Answer :

Final answer:

The least and most peppercorns a disposable grinder can contain based on the manufacturer’s criteria of z-scores between -0.9 and 0.9 are 317 and 327 peppercorns, respectively.

Explanation:

To calculate the least and most peppercorns a grinder can contain based on the z-score constraints, you need to use the mean and standard deviation along with the z-score formula:

Z = (X - μ) / σ

where Z is the z-score, X is the number of peppercorns, μ (mu) is the mean, and σ (sigma) is the standard deviation.

For a z-score of -0.9:

-0.9 = (X - 322) / 5.3

X = -0.9 * 5.3 + 322

X ≈ 317.23 peppercorns (rounded to the nearest whole number)

For a z-score of 0.9:

0.9 = (X - 322) / 5.3

X = 0.9 * 5.3 + 322

X ≈ 326.77 peppercorns (rounded to the nearest whole number)

Therefore, the grinder can contain between 317 and 327 peppercorns to meet the manufacturer's criteria.

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