Information for Problems 11 15 A fiber spinning process cur

Information for Problems 11 - 15: 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 65N/m². (2 points each)

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

2.) If the mean remains at 75N/m² and the minimum acceptable strength remains at 65N/m², what must the standard deviation be so that only 5% of the fiber will fail to meet the specification?

3.) If the mean remains at 75N/m² and the minimum acceptable strength remains at 65N/m², what must the standard deviation be so that only 1% of the fiber will fail to meet the specification?

4.) If the standard deviation is 5N/m², to what value must the mean be set so that only 1% of the fiber will fail to meet the minimum specification of 65N/m²?

5.) If the standard deviation is 3N/m², to what value must the mean be set so that only 1% of the fiber will fail to meet the specification of 65N/m²?       

Solution

a)
Normal Distribution
Mean ( u ) =250
Standard Deviation ( sd )=80
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

the 10th percentile is z = -1.28
the value of sd must be chosen so that z-score of 65 is -1.28
-1.28 = (65-75)/sd => sd = 7.8125
b)

the value chosen that the 5% of percentile of the distribution is 65.
The z-score of the 5% percentile = -1.64
(65-75)/sd = -1.64 => 6.0975

c)
the value chosen that the 5% of percentile of the distribution is 65.
The z-score of the 5% percentile = -2.33
(65-75)/sd = -2.33 = 4.2918

d)
the value chosen that the 1% of percentile of the distribution is 65.
The z-score of the 5% percentile = -2.33
(65-u)/5 = -2.33 => Mean = 76.65