Respuesta :
Answer:
7.3% percentage of the bearings produced will not be acceptable.
Step-by-step explanation:
Consider the provided information.
Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.
Let X is the normal random variable which represents the diameter of bearing.
Thus, 0.500-0.004<X<0.500+0.004
0.496<X<0.504
The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.
Use the Z score formula: [tex]\frac{X-\mu}{\sigma}[/tex]
Therefore
[tex]\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}[/tex]
[tex]\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}[/tex]
[tex]-1.5\leq z\leq 2.5[/tex]
Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.
[tex]P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270[/tex]
We need to find the percentage of the bearings produced will not be acceptable.
So subtract it from 1 as shown.
1-0.9270=0.073
Hence, 7.3% percentage of the bearings produced will not be acceptable.
