## Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed

Question

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.8 millimeters (mm) and a standard deviation of 1.4 mm. For a randomly found shard, find the following probabilities.
a) the thickness is less than 3.0 mm.
b) the thickness is more than 7.0 mm.
c) the thickness is between 3.0 mm and 7.0 mm.

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4 days 2021-07-21T21:31:12+00:00 1 Answers 0 views 0

a) 0.0985 = 9.85% probability that the thickness is less than 3.0 mm.

b) 0.0582 = 5.82% probability that the thickness is more than 7.0 mm.

c) 0.8433 = 84.33% probability that the thickness is between 3.0 mm and 7.0 mm.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Normally distributed, with a mean of 4.8 millimeters (mm) and a standard deviation of 1.4 mm.

This means that

a) the thickness is less than 3.0 mm.

This is the pvalue of Z when X = 3. So

has a pvalue of 0.0985

0.0985 = 9.85% probability that the thickness is less than 3.0 mm.

b) the thickness is more than 7.0 mm.

This is 1 subtracted by the pvalue of Z when X = 7. So

has a pvalue of 0.9418

1 – 0.9418 = 0.0582

0.0582 = 5.82% probability that the thickness is more than 7.0 mm.

c) the thickness is between 3.0 mm and 7.0 mm.

This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3. From questions a and b, we have these pvalues. So

0.9418 – 0.0985 = 0.8433

0.8433 = 84.33% probability that the thickness is between 3.0 mm and 7.0 mm.