Checkpoint: Application of z-Scores Solution

Checkpoint: Application of z-Scores

In the 20th century, the United States Mint began manufacturing pennies with an average diameter of 19.05 millimeters. To ensure consistent dimensions for all pennies, the standard deviation was set to 0.015 millimeters. What is the likelihood that a randomly selected penny has a diameter between 19.01 and 19.044 mm?

$z = \frac{19.01 - 19.05}{0.015} = - 2.67 P (penny < 19.01) = 0.0038$

$z = \frac{19.044 - 19.05}{0.015} = - 0.40 P (penny < 19.044) = 0.3446 0.3446 - 0.0038 = 0.3408$

There is a 34.08% chance that a randomly selected penny will be between 19.01 and 19.044 mm .

Note

Q: How do you find the percent chance between two values?

A: Calculate the z-score, find the area on the z-table, and then subtract the areas from one another.