Answer in Statistics and Probability for cosine #251194
In an accelerator center, an experiment needs a 1.41 cm thickaluminum cylinder. Suppose that the thickness of a cylinderhas a normal distribution with a mean 1.41 cm and a standarddeviation of 0.01 cm.
Let “X=” the thickness of a cylinder: “Xsim N(mu, sigma^2).”
Given “mu=1.41 cm, sigma=0.01 cm.”
(a) What is the probability that a thickness is greater than 1.42 cm?
“P(X>1.42)=1-P(Xleq 1.42)”
“=1-P(Zleqdfrac{1.42-1.41}{0.01})=1-P(Zleq1)”
“approx0.158655”
(b) What thickness is exceeded by 95% of the samples?
“P(X>x)=1-P(Xleq x)”
“=1-P(Zleqdfrac{x-1.41}{0.01})=0.95”
“P(Zleqdfrac{x-1.41}{0.01})=0.05”
“dfrac{x-1.41}{0.01}approx-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(Xleq 1.39)”
“=P(Z<dfrac{1.43-1.41}{0.01})-P(Zleqdfrac{1.39-1.41}{0.01})”
“=P(X<2)-P(Zleq-2)”
“=0.97725-0.02275=0.9545”
95.45 % of the samples meet specifications.
