Question

In a certain process, steel bars are manufactured and subjected to a torsion test. Bars are considered acceptable if they support at least 15 lb-ft. Currently the process produces bars that support on average 15.23 lb-ft with a variance of 0.04 lb-ft^2.

a. What proportion of the bars will not be able to meet the specifications?

b. If we take a sample of 50 bars, how many are expected to fail the test?

c. If we want no more than 5% of the bars to fail the specifications, where should we place the average?

Answer 1

In a certain process, steel bars are manufactured and subjected to a torsion test. Bars are considered acceptable if they support at least 15 lb-ft. Currently the process produces bars that support on average 15.23 lb-ft with a variance of 0.04 lb-ft^2.

Let 'X' represent the wieght length of bar

X: follows N( =15.23, ² = 0.04)   

= 0.2

z-score = = (x - 15.23) / 0.2

So we convert 'x' to z-score and then using normal tables we find the probability

a. What proportion of the bars will not be able to meet the specifications?

The bar will be accpeted if it is at least 15 lb-ft. So it will not be able to meet specification if less than 15 lb ft.

P(X: not acceptable) = P( X < 15)

=P(Z < -1.15)

=1- P( Z < 1.15)

=1 - 0.87493 ...................using normal distribution tables

Ans: 0.12507

b. If we take a sample of 50 bars, how many are expected to fail the test?

Proportion of failing = 0.12507

n = 50

Expected failures = np

= 50*0.12507

Ans:6.254

=6

c. If we want no more than 5% of the bars to fail the specifications, where should we place the average?

Here we want to know at what value of average we will not reject more than 5%. So we find the z-score at 5% and then the average.

Te propbability that bars below 15 are only5%

P( X < 15) = 5%

P( Z < (15-)/ 0.2) = 5%

We use normal percentage tables for this. The tables provide value for greater than 'z' so our sign should be '>' and for 'p < 50%' so that is appropriate. If they weren't we would have to rearrange.

P( Z > -(15-)/ 0.2 )= 5%

z-score = 1.6449

-(15-)/ 0.2 = 1.6449

(15-) / 0.2 = -.1.6449

= 15.329