Answer:
the percentage of bearings that will not be acceptable = 7.3%
Step-by-step explanation:
Given that:
Mean = 0.499
standard deviation = 0.002
if the true average diameter of the bearings it produces is 0.500 in and bearing is acceptable if its diameter is within 0.004 in.
Then the ball bearing acceptable range = (0.500 - 0.004, 0.500 + 0.004 )
= ( 0.496 , 0.504)
If x represents the diameter of the bearing , then the probability for the z value for the random variable x with a mean and standard deviation can be computed as follows:
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - \mu}{\sigma} \leq \dfrac{X -\mu}{\sigma} \leq \dfrac{0.504 - \mu}{\sigma})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - 0.499}{0.002} \leq \dfrac{X -0.499}{0.002} \leq \dfrac{0.504 - 0.499}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{-0.003}{0.002} \leq Z \leq \dfrac{0.005}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P(Z \leq 2.5) - P(Z \leq -1.5)[/tex]
From the standard normal tables
[tex]P(0.496\leq X \leq 0.504) = 0.9938-0.0668[/tex]
[tex]P(0.496\leq X \leq 0.504) = 0.927[/tex]
By applying the concept of probability of a complement , the percentage of bearings will now not be acceptable
P(not be acceptable) = 1 - P(acceptable)
P(not be acceptable) = 1 - 0.927
P(not be acceptable) = 0.073
Thus, the percentage of bearings that will not be acceptable = 7.3%
