Consider the following number line
2 answers:
You might be interested in
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
now, by traditional method, as "x" progresses towards the positive infinitity, it becomes 100, 10000, 10000000, 1000000000 and so on, and notice, the limit of the numerator becomes large.
BUT, notice the denominator, for the same values of "x", the denominator becomes larg"er" than the numerator on every iteration, ever becoming larger and larger, and yielding a fraction whose denominator is larger than the numerator.
as the denominator increases faster, since as the lingo goes, "reaches the limit faster than the numerator", the fraction becomes ever smaller an smaller ever going towards 0.
now, we could just use L'Hopital rule to check on that.
notice those derivatives atop and bottom, the top is static, whilst the bottom is racing away to infinity, ever going towards 0.