A manufacturer of bolts has a​ quality-control policy that requires it to destroy any bolts that are more than 44 standard deviations from the mean. The​ quality-control engineer knows that the bolts coming off the assembly line have mean length of 1515 cm with a standard deviation of 0.100.10 cm. For what lengths will a bolt be​ destroyed?

The problem is asking us to determine the lengths of the bolts that will be destroyed based on the manufacturer's quality control policy. The policy states that any bolt more than 4 standard deviations from the mean will be destroyed.

We are given:

- Mean length \( \mu = 1515 \) cm
- Standard deviation \( \sigma = 0.10 \) cm
- The threshold for destruction is \( 4 \) standard deviations from the mean.

To calculate the lengths at which the bolts will be destroyed, we need to consider both directions:

1. The **upper limit** will be 4 standard deviations **above** the mean.
   \[
   \text{Upper limit} = \mu + 4\sigma = 1515 + 4(0.10) = 1515 + 0.40 = 1515.40 \, \text{cm}
   \]

2. The **lower limit** will be 4 standard deviations **below

A manufacturer of bolts has a​ quality-control policy that requires it to destroy any bolts that are more than 4 standard deviations from the mean. The​ quality-control engineer knows that the bolts coming off the assembly line have a mean length of 1515 cm with a standard deviation of 0.10 cm. For what lengths will a bolt be​ destroyed?

Learn how to apply statistics and standard deviation to determine bolt lengths that will be destroyed under a quality-control policy. This example uses a mean length of 1515 cm and a standard deviation of 0.10 cm to calculate the destruction limits at 4 standard deviations from the mean. Suitable for Grades 10-12.

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