Question

A manufacturing process produces a critical part of average length 120 ​millimeters, with a standard deviation of 3 millimeters. All parts deviating by more than 5 millimeters from the mean must be rejected. What percentage of the parts must be​ rejected, on​ average? Assume a normal distribution.

Answer 1

Answer:

9.692%

Step-by-step explanation:

We have been given that a manufacturing process produces a critical part of average length 120 ​millimeters, with a standard deviation of 3 millimeters. All parts deviating by more than 5 millimeters from the mean must be rejected.

5 millimeters below mean would be $115$ and 5 millimeters above mean would be $125$.

Corresponding z values for 115 and 125 would be:

$z=\frac{x-\mu}{\sigma}$

$z=\frac{115-120}{3}$

$z=\frac{-5}{3}$

$z=-\frac{5}{3}$

$z=\frac{125-120}{3}$

$z=\frac{5}{3}$

Now, we need to find $P(z\frac{5}{3})$ using normal distribution table.

$P(z\frac{5}{3})=P(z1.66)$

We know that $P(z>1.66)=1-P(z$.

$P(z>1.66)=1-0.95154$

$P(z>1.66)=0.04846$

$P(z\frac{5}{3})=0.04846+0.04846$

$P(z\frac{5}{3})=0.09692$

Now, we need to convert 0.09692 into percentage as:

$0.09692\times 100\%=9.692\%$

Therefore, 9.692% of parts must be​ rejected on​ average.