if part manufactring data was 10 50 and 90 percentile were 1

if part manufactring data was 10% 50% and 90% percentile were 120 mm , 150 mm, and 180 mm. and the lower spec limit were 130 mm. what % would we expect to see fall below the lower spec limit?

Solution

As the 50th percentile is 150 mm, then it is the mean,

u = 150.

For the 90th percentile, the corresponding z score is

z = 1.281551566

Thus, as

sigma = (x-u)/z

Then for x = 180 mm,

sigma = (180-150)/1.281551566 = 23.40912437

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    130      
u = mean =    150      
          
s = standard deviation =    23.40912437      
          
Thus,          
          
z = (x - u) / s =    -0.854367711      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.854367711   ) =    0.19645064 or 19.645064% [ANSWER]