A times a to the 7 power
1 answer:
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Answer:
7.3% of the bearings produced will not be acceptable
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.
So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.
Larger than 0.504
1 subtracted by the pvalue of Z when X = 0.504.
has a pvalue of 0.9938
1 - 0.9938= 0.0062
Smaller than 0.496
pvalue of Z when X = -1.5
has a pvalue of 0.0668
0.0668 + 0.0062 = 0.073
7.3% of the bearings produced will not be acceptable