Answer:
97.95% of these nuts will meet the specification limits.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The offsetting machine used to make these nuts has been producing nuts with an average locking torque of 8.62 in-lb and variance of 4.49 squared in-lb.
This means that
Torque is normally distributed. If the upper specification is 13.0 in-lb and the lower specification is 2.25 in-lb, what percent of these nuts will meet the specification limits?
The proportion is the p-value of Z when X = 13 subtracted by the p-value of Z when X = 2.25. So
X = 13
has a p-value of 0.9808.
X = 2.25
has a p-value of 0.0013.
0.9808 - 0.0013 = 0.9795
0.9795*100% = 97.95%
97.95% of these nuts will meet the specification limits.