Answer:
we can conclude that 91.04% population of shafts are usable
Explanation:
Given that;
X[tex]_{min}[/tex] = 37.50 mm
X[tex]_{max}[/tex] = 38.20 mm
mean μ = 37.80
standard deviation σ = 0.20 mm
This problem is based on normal probability distribution so, to get the probability, we must calculate; z = x-μ / σ
so;
Z[tex]_{min}[/tex] = (X[tex]_{min}[/tex] - μ) / μ = (37.50 - 37.80) / 0.20 = -1.5
Z[tex]_{max}[/tex] = (X[tex]_{max}[/tex] - μ) / μ = (38.20 - 37.80) / 0.20 = 2
Hence;
P( 37.50 < X < 38.20 ) = P( -1.5 < Z < 2 )
P( 37.50 < X < 38.20 ) = P( 0 < Z < 1.5 ) + P( 0 < Z < 2 )
from the standard normal table table;
P( 37.50 < X < 38.20 ) = 0.4332 + 0.4772
P( 37.50 < X < 38.20 ) = 0.9104 = 91.40%
Hence, we can conclude that 91.04% population of shafts are usable
