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.