Answer to Question #251194 in Statistics and Probability for cosine

Question

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.

Answer 1

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

Given $\mu=1.41\ cm, \sigma=0.01\ cm.$

(a) What is the probability that a thickness is greater than 1.42 cm?

$P(X>1.42)=1-P(X\leq 1.42)$

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

$\approx0.158655$

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

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

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

$P(Z\leq\dfrac{x-1.41}{0.01})=0.05$

$\dfrac{x-1.41}{0.01}\approx-1.6449$

$x=1.41-0.16449$

$x=1.2455$

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?

$P(1.39<X<1.43)=P(X<1.43)-P(X\leq 1.39)$

$=P(Z<\dfrac{1.43-1.41}{0.01})-P(Z\leq\dfrac{1.39-1.41}{0.01})$

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

$=0.97725-0.02275=0.9545$

95.45 % of the samples meet specifications.