Question

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.6 millimeters (mm) and a standard deviation of 1.5 mm. For a randomly found shard, find the probability that the thickness is between 3.0 and 7.0 mm. (Round your answer to four decimal places.)

Answer 1

Answer:

0.8029 = 80.29% probability that the thickness is between 3.0 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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 4.6 millimeters (mm) and a standard deviation of 1.5 mm.

This means that $\mu = 4.6, \sigma = 1.5$

Find the probability that the thickness is between 3.0 and 7.0 mm.

This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3. So

X = 7

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7 - 4.6}{1.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

X = 3

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 4.6}{1.5}$

$Z = -1.07$

$Z = -1.07$ has a pvalue of 0.1423

0.9452 - 0.1423 = 0.8029

0.8029 = 80.29% probability that the thickness is between 3.0 and 7.0 mm.