The tensile strength of a group of shock absorbers is normally distributed with

Question

The tensile strength of a group of shock absorbers is normally distributed with a mean of 1,000 lb and a std. dev. of 40 lb. The shock absorbers are proof tested at 950 lb.

a) What fraction will survive the proof test?

b) If it is decided to increase the strength of the shock absorbers (i.e. to increase the mean strength while leaving the std. dev. unchanged) so that 99% pass the test, what must the new value of the mean strength be?

c) If it is decided to improve the quality control, i.e. to decrease the variance while leaving the mean strength unchanged, so that 99% pass the test, what must the new std. dev. be?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    950      
u = mean =    1000      
          
s = standard deviation =    40      
          
Thus,          
          
z = (x - u) / s =    -1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.25   ) =    0.894350226 [ANSWER]

*****************

b)

The z score for the left tailed area of 1 - 0.99 = 0.01

z = -2.326347874

Hence, the new mean is

u = x - z*sigma = 950 - (-2.326347874)*40 = 1043.053915 [ANSWER]

****************

c)

The z score for the left tailed area of 1 - 0.99 = 0.01

z = -2.326347874

Hence, the new standard deviation is

sigma = (x-u)/z = (950-1000)/(-2.326347874) = 21.49291624 [ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.