Answer to Question #251194 in Statistics and Probability for cosine

Question #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.


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Expert's answer
2021-10-14T18:15:36-0400

Let X=X= the thickness of a cylinder: XN(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=1.41 cm,σ=0.01 cm.\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)=1P(X1.42)P(X>1.42)=1-P(X\leq 1.42)

=1P(Z1.421.410.01)=1P(Z1)=1-P(Z\leq\dfrac{1.42-1.41}{0.01})=1-P(Z\leq1)


0.158655\approx0.158655

(b) What thickness is exceeded by 95% of the samples?


P(X>x)=1P(Xx)P(X>x)=1-P(X\leq x)

=1P(Zx1.410.01)=0.95=1-P(Z\leq\dfrac{x-1.41}{0.01})=0.95

P(Zx1.410.01)=0.05P(Z\leq\dfrac{x-1.41}{0.01})=0.05

x1.410.011.6449\dfrac{x-1.41}{0.01}\approx-1.6449

x=1.410.16449x=1.41-0.16449

x=1.2455x=1.2455

Thickness 1.24551.2455 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(X1.39)P(1.39<X<1.43)=P(X<1.43)-P(X\leq 1.39)

=P(Z<1.431.410.01)P(Z1.391.410.01)=P(Z<\dfrac{1.43-1.41}{0.01})-P(Z\leq\dfrac{1.39-1.41}{0.01})

=P(X<2)P(Z2)=P(X<2)-P(Z\leq-2)

=0.977250.02275=0.9545=0.97725-0.02275=0.9545

95.45 % of the samples meet specifications.



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