Question

The tolerance limits for a circuit breaker are Amperes. This means that any circuit breaker that breaks at an amperage less than 39.5 breaks at too low a level and any that breaks at an amperage greater than 40.5 breaks at too high a level. If a shipment of circuit breakers possesses break points that are normally distributed with mean 39.3 and standard deviation 0.2, then what percentage of the shipment is defective

Answer 1

84.13% of the shipment is defective

Explanation:

When the distribution is normal, we use the z-score formula.

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

`Z = (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 pvalue, we get the probability that the value of the measure is greater than X.

Normally distributed with mean 39.3 and standard deviation 0.2

This means that
`\mu = 39.3, \sigma = 0.2`

What percentage of the shipment is defective

Less than 39.5 or greater than 40.5.

Less than 39.5:

This is the pvalue of Z when X = 39.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (39.5 - 39.3)/(0.2)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413.

Greater than 40.5:

1 subtracted by the pvalue of Z when X = 40.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (40.5 - 39.3)/(0.2)`

`Z = 6`

`Z = 6` has a pvalue of 1

1 - 1 = 0

Total:

0.8413 + 0 = 0.8413

84.13% of the shipment is defective