Let X= the thickness of a cylinder: X∼N(μ,σ2).
Given μ=1.41 cm,σ=0.01 cm.
(a) What is the probability that a thickness is greater than 1.42 cm?
P(X>1.42)=1−P(X≤1.42)
=1−P(Z≤0.011.42−1.41)=1−P(Z≤1)
≈0.158655
(b) What thickness is exceeded by 95% of the samples?
P(X>x)=1−P(X≤x)
=1−P(Z≤0.01x−1.41)=0.95
P(Z≤0.01x−1.41)=0.05
0.01x−1.41≈−1.6449
x=1.41−0.16449
x=1.2455 Thickness 1.2455 cm is exceeded by 95% of the samples.
(c) If the specifications require that the thickness is between 1.39 cm and 1.43 cm, what proportion of the samples meets specifications?
P(1.39<X<1.43)=P(X<1.43)−P(X≤1.39)
=P(Z<0.011.43−1.41)−P(Z≤0.011.39−1.41)
=P(X<2)−P(Z≤−2)
=0.97725−0.02275=0.9545 95.45 % of the samples meet specifications.
Comments
Leave a comment