Answer:
7.3% of the bearings produced will not be acceptable.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.499 inch
Standard Deviation, σ = 0.002 inch
We are given that the distribution of the diameters is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
A bearing is acceptable if its diameter is within 0.004 inches that is the range is from 0.496 inch to 0.504 inches.
P(diameter between 0.496 inch and 0.504 inch)
[tex]P(0.496 \leq x \leq 0.504) = P(\displaystyle\frac{0.496 - 0.499}{0.002} \leq z \leq \displaystyle\frac{0.504-0.499}{0.002}) = P(-1.5 \leq z \leq 2.5)\\\\= P(z \leq 2.5) - P(z < -1.5)\\= 0.994 - 0.067 = 0.927= 92.7\%[/tex]
Percentage of the bearings produced will not be acceptable =
[tex]100-92.7\\=7.3\%[/tex]
Thus, 7.3% of the bearings produced will not be acceptable.