A fiber-spinning process currently produces a fiber whose strength is normally distributed with a mean of 75 N/m2. The minimum acceptable strength is 65 N/m2.

a.Ten percent of the fiber produced by the current method fails to meet the minimum specification. What is the standard deviation of fiber strengths in the current process?

b. If the mean remains at 75 N/m2, what must the standard deviation be so that only 1 % of the fiber will fail to meet the specification?

c. If the standard deviation is 5 N/m2, to what value must the mean be set so that only 1% of the fiber will fail to meet the specification?

Answer:

Step 1 of 3:

(a)

In this question, given that the 10% of the fiber produced by the current method fails to meet the minimum specification. we are asked to find the standard deviation of fiber strengths in the fiber-spinning process.

Fiber strength is normally distributed with a mean of 75.

The minimum acceptable strength is 65.

Let be the strength of the fiber.

The proportion of strengths that are less than 65 is 0.10 means 10th percentile of strength.

From the table, the closest area to 0.1 is 0.1014, and corresponding is .

Let be the required standard deviation.

The z-score of 65 is,

z = =

=

Hence the standard deviation of fiber strengths is 0.78125 .

Step 2 of 3:

(b)

In this question, we are asked to find the standard deviation be so that only 1 % of the fiber will fail to meet the specification.

If mean remains same.

From the table, the closest area to 0.01 is 0.0099, and corresponding is .

The z-score of 65 is,

z = =

=

Hence the standard deviation of fiber strengths is 0.4.291 .