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 $P(z<-\frac{5}{3})+P(z>\frac{5}{3})$ using normal distribution table.

$P(z<-\frac{5}{3})+P(z>\frac{5}{3})=P(z<-1.66)+P(z>1.66)$

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

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

$P(z>1.66)=0.04846$

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

$P(z<-\frac{5}{3})+P(z>\frac{5}{3})=0.09692$

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.

Answer 2

About 0% of the parts would be​ rejected, on​ average.

Z score

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (x - μ) / σ

where μ is the mean, x = raw score and σ is the standard deviation.

Given μ = 120, σ = 3. For z > 5:

P(z > 5) = 1 - P(z < -38.3) = 1 - 1 = 1 = 0%

About 0% of the parts would be​ rejected, on​ average.

Find out more on Z score at: https://brainly.com/question/25638875